Class 12 : Maths (English) – Chapter 11: Three Dimensional Geometry

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On June 25, 2025

EXPLANATION & SUMMARY

🔷 Explanation Section (~1700 Words)
🔹 Introduction
Three Dimensional Geometry (3D Geometry) deals with the study of points, lines, and planes in three-dimensional space. In two dimensions (2D), we use x and y axes; in 3D, we introduce a third axis — z-axis, perpendicular to both x and y.
A point in 3D space is represented as P(x, y, z).

🔹 1. Coordinate Axes and Coordinate Planes
There are three mutually perpendicular axes:
X-axis
Y-axis
Z-axis
They intersect at the origin (O).
The planes formed:
XY-plane: z = 0
YZ-plane: x = 0
ZX-plane: y = 0
💡 Concept: The 3D coordinate system divides space into 8 octants.

🔹 2. Distance Formula
For two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂),
➡️ Distance (PQ) = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²]
✏️ Note: It generalizes the 2D distance formula by adding the z-component.

🔹 3. Section Formula
If point P(x, y, z) divides the line joining A(x₁, y₁, z₁) and B(x₂, y₂, z₂) in ratio m : n, then:
➡️ x = (m·x₂ + n·x₁) / (m + n)
➡️ y = (m·y₂ + n·y₁) / (m + n)
➡️ z = (m·z₂ + n·z₁) / (m + n)
✔️ If m = n, it’s the midpoint.

🔹 4. Direction Cosines (l, m, n)
The direction cosines of a line are the cosines of the angles (α, β, γ) that the line makes with the x, y, and z axes respectively.
➡️ l = cos α, m = cos β, n = cos γ
✔️ Relation: l² + m² + n² = 1
✏️ Note: Direction cosines help represent a line’s orientation.

🔹 5. Direction Ratios (DRs)
If a, b, c are proportional to direction cosines, they are called direction ratios.
They satisfy: l / a = m / b = n / c
💡 Concept: Any set of numbers proportional to direction cosines can be used as DRs.

🔹 6. Equation of a Line
(i) Vector Form
If a line passes through point A(a₁, a₂, a₃) and is parallel to vector b̅ = b₁î + b₂ĵ + b₃k̂,
➡️ Equation: r̅ = a̅ + λ b̅, where λ ∈ ℝ
(ii) Cartesian Form
From vector form,
➡️ (x − a₁)/b₁ = (y − a₂)/b₂ = (z − a₃)/b₃
✏️ Note: Each ratio equals parameter λ.

🔹 7. Condition for Parallel and Perpendicular Lines
Let DRs of lines L₁ and L₂ be (a₁, b₁, c₁) and (a₂, b₂, c₂):
✔️ Parallel: a₁/a₂ = b₁/b₂ = c₁/c₂
✔️ Perpendicular: a₁a₂ + b₁b₂ + c₁c₂ = 0

🔹 8. Shortest Distance Between Two Lines
If two lines are skew, the shortest distance (d) is given by:
➡️ d = |(a₂ − a₁) ⋅ (b₁ × b₂)| / |b₁ × b₂|
💡 Concept: Skew lines are non-intersecting and non-parallel.

🔹 9. Equation of a Plane
(i) Vector Form
If a plane passes through point A(a₁, a₂, a₃) with normal vector n̅ = n₁î + n₂ĵ + n₃k̂,
➡️ Equation: (r̅ − a̅) ⋅ n̅ = 0
(ii) Cartesian Form
➡️ n₁(x − a₁) + n₂(y − a₂) + n₃(z − a₃) = 0
Or
➡️ n₁x + n₂y + n₃z + D = 0

🔹 10. Normal Form
If the plane is at distance p from origin and normal makes direction cosines (l, m, n),
➡️ lx + my + nz = p

🔹 11. Intercept Form
➡️ x/a + y/b + z/c = 1
where a, b, c are intercepts on the axes.

🔹 12. Angle Between Two Planes
If planes are:
P₁: A₁x + B₁y + C₁z + D₁ = 0
P₂: A₂x + B₂y + C₂z + D₂ = 0
➡️ cos θ = (A₁A₂ + B₁B₂ + C₁C₂) / [√(A₁²+B₁²+C₁²) · √(A₂²+B₂²+C₂²)]

🔹 13. Angle Between a Line and a Plane
If line has DRs (a, b, c) and plane has normal (A, B, C):
➡️ sin θ = (A·a + B·b + C·c) / [√(A²+B²+C²) · √(a²+b²+c²)]

🔹 14. Condition for Line Parallel/Perpendicular to Plane
✔️ Parallel: A·a + B·b + C·c = 0
✔️ Perpendicular: a/A = b/B = c/C

🔹 15. Distance of a Point from a Plane
For point P(x₁, y₁, z₁) and plane Ax + By + Cz + D = 0:
➡️ Distance (d) = |A·x₁ + B·y₁ + C·z₁ + D| / √(A² + B² + C²)

🔹 16. Equation of Plane Passing Through Three Points
Given A(x₁, y₁, z₁), B(x₂, y₂, z₂), C(x₃, y₃, z₃):
Form vectors AB̅ and AC̅,
Normal n̅ = AB̅ × AC̅,
Then plane equation: (r̅ − a̅) ⋅ n̅ = 0

🔹 17. Coplanarity Condition for Lines
Two lines with DRs (a₁, b₁, c₁), (a₂, b₂, c₂) and position vectors a̅₁, a̅₂ are coplanar if:
➡️ (a₂ − a₁) ⋅ (b₁ × b₂) = 0

🔹 18. Distance Between Two Parallel Planes
For Ax + By + Cz + D₁ = 0 and Ax + By + Cz + D₂ = 0:
➡️ Distance = |D₁ − D₂| / √(A² + B² + C²)

🔹 19. Angle Between Line and Line
If direction ratios of two lines are (a₁, b₁, c₁) and (a₂, b₂, c₂):
➡️ cos θ = (a₁a₂ + b₁b₂ + c₁c₂) / [√(a₁² + b₁² + c₁²) · √(a₂² + b₂² + c₂²)]

🔹 20. Miscellaneous Results
If a line lies in a plane, its DRs satisfy plane equation.
Skew lines never intersect and are not parallel.
Intersection line of two planes can be found by cross product of normals.

🔷 Summary (~300 Words)
🔹 Key Formulas:
Distance: √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²]
Section Formula: ((mx₂ + nx₁)/(m+n), …)
Line (Vector): r̅ = a̅ + λb̅
Line (Cartesian): (x−x₁)/a = (y−y₁)/b = (z−z₁)/c
Plane (Vector): (r̅ − a̅) ⋅ n̅ = 0
Plane (Cartesian): Ax + By + Cz + D = 0
Angle (line-line): cos θ = (a₁a₂ + b₁b₂ + c₁c₂)/(…)
Angle (plane-plane): cos θ = (A₁A₂ + B₁B₂ + C₁C₂)/(…)
Point-plane distance: |Ax₁ + By₁ + Cz₁ + D| / √(A² + B² + C²)
Parallel plane distance: |D₁ − D₂| / √(A² + B² + C²)
Shortest distance (skew lines): |(a₂−a₁) ⋅ (b₁×b₂)| / |b₁×b₂|

🔹 Important Concepts:
Direction cosines (l, m, n) satisfy l² + m² + n² = 1
Direction ratios proportional to cosines
Line: single parameter λ; Plane: normal vector defines orientation
Relations between line & plane help analyze geometry in 3D

🔹 Applications:
Used in physics (motion, forces)
Engineering design, computer graphics, navigation, robotics, architecture

📝 Quick Recap:
✔️ 3D point: P(x, y, z)
✔️ Line: r̅ = a̅ + λb̅
✔️ Plane: (r̅ − a̅) ⋅ n̅ = 0
✔️ Distance & angles from formulas
✔️ Direction cosines: l² + m² + n² = 1
✔️ Point-plane distance formula essential
✔️ Skew line shortest distance via vector triple product

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QUESTIONS FROM TEXTBOOK

Exercise 11.1

🔵 Question 1:
If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines.
🟢 Answer:
➡️ Let direction cosines be l = cos α, m = cos β, n = cos γ.
➡️ Given: α = 90°, β = 135°, γ = 45°.
✳️ Step 1:
l = cos 90° = 0
✳️ Step 2:
m = cos 135° = −1 / √2
✳️ Step 3:
n = cos 45° = 1 / √2
✏️ Check:
l² + m² + n² = 0² + (−1/√2)² + (1/√2)² = 0 + 1/2 + 1/2 = 1 ✔️
✔️ Final Direction Cosines: (0, −1 / √2, 1 / √2)
🔵 Question 2:
Find the direction cosines of a line which makes equal angles with the coordinate axes.
🟢 Answer:
➡️ Let the line make equal angles ⇒ α = β = γ.
➡️ Therefore, l = m = n.
✳️ Step 1:
Use identity l² + m² + n² = 1.
✳️ Step 2:
3 l² = 1 ⇒ l² = 1 / 3 ⇒ l = 1 / √3.
✏️ Hence, l = m = n = 1 / √3.
✔️ Final Direction Cosines: (1 / √3, 1 / √3, 1 / √3)
💡 Note: Opposite direction is (−1 / √3, −1 / √3, −1 / √3).
🔵 Question 3:
If a line has the direction ratios −18, 12, −4, then what are its direction cosines?
🟢 Answer:
➡️ Given direction ratios: a = −18, b = 12, c = −4.
✳️ Step 1:
Find magnitude k = √(a² + b² + c²)
= √( (−18)² + 12² + (−4)² )
= √(324 + 144 + 16) = √484 = 22.
✳️ Step 2:
Compute direction cosines:
l = a / k = −18 / 22 = −9 / 11
m = b / k = 12 / 22 = 6 / 11
n = c / k = −4 / 22 = −2 / 11
✏️ Check:
(−9/11)² + (6/11)² + (−2/11)² = 81/121 + 36/121 + 4/121 = 121/121 = 1 ✔️
✔️ Final Direction Cosines: (−9 / 11, 6 / 11, −2 / 11)
🔵 Question 4:
Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.
🟢 Answer:
➡️ Let A(2, 3, 4), B(−1, −2, 1), C(5, 8, 7).
✳️ Step 1:
Find AB = B − A = (−1−2, −2−3, 1−4) = (−3, −5, −3).
✳️ Step 2:
Find AC = C − A = (5−2, 8−3, 7−4) = (3, 5, 3).
✳️ Step 3:
Observe that AB = −1 × AC.
➡️ Therefore, AB and AC are parallel and pass through point A.
✔️ Conclusion: Points A, B, C are collinear.
🔵 Question 5:
Find the direction cosines of the sides of the triangle whose vertices are (3, 5, −4), (−1, 1, 2) and (−5, −5, −2).
🟢 Answer:
➡️ Let A(3, 5, −4), B(−1, 1, 2), C(−5, −5, −2).

✳️ Side AB:
AB = B − A = (−1−3, 1−5, 2−(−4)) = (−4, −4, 6).
|AB| = √( (−4)² + (−4)² + 6² ) = √(16 + 16 + 36) = √68 = 2√17.
➡️ Direction cosines of AB:
l₁ = −4 / 2√17 = −2 / √17
m₁ = −4 / 2√17 = −2 / √17
n₁ = 6 / 2√17 = 3 / √17
✔️ AB → (−2 / √17, −2 / √17, 3 / √17)

✳️ Side BC:
BC = C − B = (−5−(−1), −5−1, −2−2) = (−4, −6, −4).
|BC| = √( (−4)² + (−6)² + (−4)² ) = √(16 + 36 + 16) = √68 = 2√17.
➡️ Direction cosines of BC:
l₂ = −4 / 2√17 = −2 / √17
m₂ = −6 / 2√17 = −3 / √17
n₂ = −4 / 2√17 = −2 / √17
✔️ BC → (−2 / √17, −3 / √17, −2 / √17)

✳️ Side CA:
CA = A − C = (3−(−5), 5−(−5), −4−(−2)) = (8, 10, −2).
|CA| = √(8² + 10² + (−2)²) = √(64 + 100 + 4) = √168 = 2√42.
➡️ Direction cosines of CA:
l₃ = 8 / 2√42 = 4 / √42
m₃ = 10 / 2√42 = 5 / √42
n₃ = −2 / 2√42 = −1 / √42
✔️ CA → (4 / √42, 5 / √42, −1 / √42)

✔️ Final Answers:
🔹 AB → (−2 / √17, −2 / √17, 3 / √17)
🔹 BC → (−2 / √17, −3 / √17, −2 / √17)
🔹 CA → (4 / √42, 5 / √42, −1 / √42)

Exercise 11.2

🔵 Question 1:
Show that the three lines with direction cosines 12/13, −3/13, −4/13; 4/13, 12/13, 3/13; 3/13, −4/13, 12/13 are mutually perpendicular.

🟢 Answer:
➡️ Let the three direction-cosine triplets be
L₁ = (12/13, −3/13, −4/13), L₂ = (4/13, 12/13, 3/13), L₃ = (3/13, −4/13, 12/13).
➡️ Two lines are perpendicular ⇔ dot product of their direction cosines is 0.

✳️ Step 1 (L₁ ⋅ L₂):
(12/13)(4/13) + (−3/13)(12/13) + (−4/13)(3/13)
= 48/169 − 36/169 − 12/169 = 0. ✔️

✳️ Step 2 (L₁ ⋅ L₃):
(12/13)(3/13) + (−3/13)(−4/13) + (−4/13)(12/13)
= 36/169 + 12/169 − 48/169 = 0. ✔️

✳️ Step 3 (L₂ ⋅ L₃):
(4/13)(3/13) + (12/13)(−4/13) + (3/13)(12/13)
= 12/169 − 48/169 + 36/169 = 0. ✔️

✔️ Final: All pairwise dot products are 0 ⇒ the three lines are mutually perpendicular.

🔵 Question 2:
Show that the line through the points (1, −1, 2), (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

🟢 Answer:
➡️ Direction ratios of line ℓ₁ through P₁(1, −1, 2), P₂(3, 4, −2):
✳️ Step 1: P₁P₂ = (3−1, 4−(−1), −2−2) = (2, 5, −4).

➡️ Direction ratios of line ℓ₂ through Q₁(0, 3, 2), Q₂(3, 5, 6):
✳️ Step 2: Q₁Q₂ = (3−0, 5−3, 6−2) = (3, 2, 4).

✳️ Step 3 (Dot product test):
(2, 5, −4) ⋅ (3, 2, 4) = 2×3 + 5×2 + (−4)×4 = 6 + 10 − 16 = 0.

✔️ Final: Dot product = 0 ⇒ the two lines are perpendicular.

🔵 Question 3:
Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5).

🟢 Answer:
➡️ Direction ratios of line ℓ₁ through A(4, 7, 8), B(2, 3, 4):
✳️ Step 1: AB = (2−4, 3−7, 4−8) = (−2, −4, −4).

➡️ Direction ratios of line ℓ₂ through C(−1, −2, 1), D(1, 2, 5):
✳️ Step 2: CD = (1−(−1), 2−(−2), 5−1) = (2, 4, 4).

✳️ Step 3 (Scalar multiple test):
(−2, −4, −4) = −1 × (2, 4, 4).

✔️ Final: Direction ratios are proportional ⇒ the two lines are parallel.

🔵 Question 4:
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3i + 2j − 2k.

🟢 Answer:
➡️ Point vector: r₀ = 1i + 2j + 3k.
➡️ Direction vector: a = 3i + 2j − 2k.

✳️ Step 1 (Vector form):
r⃗ = r₀ + λ a = (i + 2j + 3k) + λ(3i + 2j − 2k), λ ∈ ℝ.

✳️ Step 2 (Cartesian form):
(x − 1)/3 = (y − 2)/2 = (z − 3)/(−2).

✔️ Final: Required line is r⃗ = (i + 2j + 3k) + λ(3i + 2j − 2k) and (x − 1)/3 = (y − 2)/2 = (z − 3)/(−2).

🔵 Question 5:
Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2i − j + 4k and is in the direction i + 2j − k.

🟢 Answer:
➡️ Given point vector r₀ = 2i − j + 4k.
➡️ Given direction vector a = i + 2j − k.

✳️ Step 1 (Vector form):
r⃗ = r₀ + λ a = (2i − j + 4k) + λ(i + 2j − k), λ ∈ ℝ.

✳️ Step 2 (Cartesian form using direction ratios 1, 2, −1):
(x − 2)/1 = (y + 1)/2 = (z − 4)/(−1).

✔️ Final: r⃗ = (2i − j + 4k) + λ(i + 2j − k); (x − 2)/1 = (y + 1)/2 = (z − 4)/(−1).

🔵 Question 6:
Find the cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by (x + 3)/3 = (y − 4)/5 = (z + 8)/6.

🟢 Answer:
➡️ A line parallel to (x + 3)/3 = (y − 4)/5 = (z + 8)/6 has direction ratios 3, 5, 6.
➡️ Required line must pass through (−2, 4, −5) with same direction ratios.

✳️ Step 1 (Cartesian form):
(x − (−2))/3 = (y − 4)/5 = (z − (−5))/6.

✳️ Step 2 (Simplify brackets):
(x + 2)/3 = (y − 4)/5 = (z + 5)/6.

✔️ Final: (x + 2)/3 = (y − 4)/5 = (z + 5)/6.

🔵 Question 7:
The cartesian equation of a line is (x − 5)/3 = (y + 4)/7 = (z − 6)/2. Write its vector form.

🟢 Answer:
➡️ From the given form, a point on the line is (5, −4, 6).
➡️ Direction ratios are 3, 7, 2 ⇒ direction vector a = 3i + 7j + 2k.

✳️ Step 1 (Vector form using r₀ = 5i − 4j + 6k):
r⃗ = r₀ + λ a = (5i − 4j + 6k) + λ(3i + 7j + 2k), λ ∈ ℝ.

✔️ Final: r⃗ = (5i − 4j + 6k) + λ(3i + 7j + 2k).

🔵 Question 8:
Find the angle between the following pairs of lines:

(i) r⃗ = 2i − 5j + k + λ(3i + 2j + 6k) and r⃗ = 7i − 6k + μ(i + 2j + 2k)

(ii) r⃗ = 3i + j − 2k + λ(i − j − 2k) and r⃗ = 2i − j − 56k + μ(3i − 5j − 4k)

🟢 Answer (common idea):
➡️ If a₁ and a₂ are direction vectors, then cos θ = (a₁ ⋅ a₂) / (|a₁| |a₂|), with 0° ≤ θ ≤ 180°.

✳️ Part (i):
➡️ a₁ = 3i + 2j + 6k ⇒ components (3, 2, 6).
➡️ a₂ = i + 2j + 2k ⇒ components (1, 2, 2).
➡️ a₁ ⋅ a₂ = 3×1 + 2×2 + 6×2 = 3 + 4 + 12 = 19.
➡️ |a₁| = √(3² + 2² + 6²) = √(9 + 4 + 36) = √49 = 7.
➡️ |a₂| = √(1² + 2² + 2²) = √(1 + 4 + 4) = √9 = 3.
➡️ cos θ = 19 / (7 × 3) = 19/21.
✔️ Angle: θ = cos⁻¹(19/21).

✳️ Part (ii):
➡️ a₁ = i − j − 2k ⇒ components (1, −1, −2).
➡️ a₂ = 3i − 5j − 4k ⇒ components (3, −5, −4).
➡️ a₁ ⋅ a₂ = 1×3 + (−1)×(−5) + (−2)×(−4) = 3 + 5 + 8 = 16.
➡️ |a₁| = √(1² + (−1)² + (−2)²) = √(1 + 1 + 4) = √6.
➡️ |a₂| = √(3² + (−5)² + (−4)²) = √(9 + 25 + 16) = √50 = 5√2.
➡️ cos θ = 16 / (√6 × 5√2) = 16 / (5 √12) = 16 / (10 √3) = 8 / (5 √3).
✔️ Angle: θ = cos⁻¹(8 / (5 √3)).

✔️ Final Answers:
(i) θ = cos⁻¹(19/21)
(ii) θ = cos⁻¹(8 / (5 √3))

🔵 Question 9:
Find the angle between the following pair of lines:

(i) (x − 2)/2 = (y − 1)/5 = (z + 3)/−3 and (x + 2)/−1 = (y − 4)/8 = (z − 5)/4

(ii) x/2 = y/2 = z/1 and (x − 5)/4 = (y − 2)/1 = (z − 3)/8

🟢 Answer:
➡️ Idea: If direction vectors are a and b, then cos θ = (a · b) / (|a| |b|).

✳️ Part (i):
➡️ Direction ratios a₁ = (2, 5, −3), a₂ = (−1, 8, 4).
➡️ a₁ · a₂ = 2(−1) + 5×8 + (−3)×4 = −2 + 40 − 12 = 26.
➡️ |a₁| = √(2² + 5² + (−3)²) = √(4 + 25 + 9) = √38.
➡️ |a₂| = √((−1)² + 8² + 4²) = √(1 + 64 + 16) = √81 = 9.
➡️ cos θ = 26 / (9 √38).
✔️ Angle: θ = cos⁻¹( 26 / (9 √38) ).

✳️ Part (ii):
➡️ Direction ratios a₁ = (2, 2, 1), a₂ = (4, 1, 8).
➡️ a₁ · a₂ = 2×4 + 2×1 + 1×8 = 8 + 2 + 8 = 18.
➡️ |a₁| = √(2² + 2² + 1²) = √9 = 3.
➡️ |a₂| = √(4² + 1² + 8²) = √81 = 9.
➡️ cos θ = 18 / (3 × 9) = 2/3.
✔️ Angle: θ = cos⁻¹(2/3).

🔵 Question 10:
Find the values of p so that the lines (1 − x)/3 = (7y − 14)/(2p) = (z − 3)/2 and (7 − 7x)/(3p) = y/5 = (6 − z)/5 are at right angles.

🟢 Answer:
➡️ Line 1: take parameter t.
➡️ x = 1 − 3t ⇒ dx/dt = −3, 7y − 14 = 2pt ⇒ dy/dt = 2p/7, z = 3 + 2t ⇒ dz/dt = 2.
➡️ Direction ratios of Line 1 ∝ (−3, 2p/7, 2) ⇒ choose (−21, 2p, 14).

➡️ Line 2: take parameter s.
➡️ 7 − 7x = 3ps ⇒ x = 1 − (3p/7)s ⇒ dx/ds = −3p/7, y = 5s ⇒ dy/ds = 5, 6 − z = 5s ⇒ dz/ds = −5.
➡️ Direction ratios of Line 2 ∝ (−3p/7, 5, −5) ⇒ choose (−3p, 35, −35).

➡️ Perpendicular condition: (−21, 2p, 14) · (−3p, 35, −35) = 0.
➡️ 63p + 70p − 490 = 0 ⇒ 133p = 490 ⇒ p = 490/133 = 70/19.
✔️ Final: p = 70/19.

🔵 Question 11:
Show that the lines (x − 5)/7 = (y + 2)/−5 = z/1 and x/1 = y/2 = z/3 are perpendicular to each other.

🟢 Answer:
➡️ Direction ratios: Line 1 → (7, −5, 1); Line 2 → (1, 2, 3).
➡️ Dot product = 7×1 + (−5)×2 + 1×3 = 7 − 10 + 3 = 0.
✔️ Conclusion: Dot product 0 ⇒ lines are perpendicular.

🔵 Question 12:
Find the shortest distance between the lines
r̅ = (i + 2 j + k) + λ (i − j + k) and r̅ = 2 i − j − k + μ (2 i + j + 2 k).

🟢 Answer:
➡️ Point on Line 1: A(1, 2, 1); direction b = (1, −1, 1).
➡️ Point on Line 2: C(2, −1, −1); direction d = (2, 1, 2).
➡️ Formula (skew lines): Shortest distance D = | ( (C − A) · (b × d) ) | / | b × d |.

✳️ Step 1: b × d = | i j k ; 1 −1 1 ; 2 1 2 | = (−3, 0, 3).
✳️ Step 2: C − A = (2 − 1, −1 − 2, −1 − 1) = (1, −3, −2).
✳️ Step 3: (C − A) · (b × d) = 1×(−3) + (−3)×0 + (−2)×3 = −3 + 0 − 6 = −9.
✳️ Step 4: |b × d| = √((−3)² + 0² + 3²) = √18 = 3√2.
✳️ Step 5: D = |−9| / (3√2) = 9 / (3√2) = 3 / √2 = (3√2)/2.

✔️ Shortest distance: D = 3/√2 = (3√2)/2.

🔵 Question 13:
Find the shortest distance between the lines
(x + 1)/7 = (y + 1)/(−6) = (z + 1)/1
and
(x − 3)/(−2) = (y − 5)/(−2) = (z − 7)/1.

🟢 Answer:

➡️ Step 1: Identify points and direction ratios
Line 1 passes through A(−1, −1, −1) and has direction ratios b = (7, −6, 1).
Line 2 passes through C(3, 5, 7) and has direction ratios d = (−2, −2, 1).

➡️ Step 2: Formula
Shortest distance between two skew lines:
D = | ( (C − A) ⋅ (b × d) ) | ÷ | b × d |

➡️ Step 3: Find cross product (b × d)
b = (7, −6, 1), d = (−2, −2, 1)
b × d =
| i j k |
| 7 −6 1 |
| −2 −2 1 |

= i( (−6)(1) − (1)(−2) ) − j( (7)(1) − (1)(−2) ) + k( (7)(−2) − (−6)(−2) )
= i( −6 + 2 ) − j( 7 + 2 ) + k( −14 − 12 )
= ( −4, −9, −26 )

➡️ Step 4: Find vector (C − A)
C − A = (3 − (−1), 5 − (−1), 7 − (−1))
C − A = (4, 6, 8)

➡️ Step 5: Dot product
(C − A) ⋅ (b × d) = 4(−4) + 6(−9) + 8(−26)
= (−16) + (−54) + (−208) = −278

➡️ Step 6: Magnitude of (b × d)
|b × d| = √( (−4)² + (−9)² + (−26)² )
= √(16 + 81 + 676) = √773

➡️ Step 7: Apply formula
D = | −278 | ÷ √773 = 278 ÷ √773

✔️ Final Answer:
Shortest distance D = 278 / √773

🔵 Question 14:
Find the shortest distance between the lines whose vector equations are
r̅ = (i + 2 j + 3 k) + λ (i − 3 j + 2 k) and r̅ = 4 i + 5 j + 6 k + μ (2 i + 3 j + k).

🟢 Answer:
➡️ Line 1 point A(1, 2, 3); direction b = (1, −3, 2).
➡️ Line 2 point C(4, 5, 6); direction d = (2, 3, 1).
➡️ Use D = | ( (C − A) ⋅ (b × d) ) | ÷ | b × d |.

✳️ Step 1: b × d = | i j k ; 1 −3 2 ; 2 3 1 | = (−9, 3, 9).
✳️ Step 2: C − A = (4 − 1, 5 − 2, 6 − 3) = (3, 3, 3).
✳️ Step 3: (C − A) ⋅ (b × d) = 3(−9) + 3×3 + 3×9 = −27 + 9 + 27 = 9.
✳️ Step 4: | b × d | = √( (−9)² + 3² + 9² ) = √171 = 3√19.
✳️ Step 5: D = |9| ÷ (3√19) = 3 ÷ √19.

✔️ Final Shortest Distance: D = 3/√19.

🔵 Question 15:
Find the shortest distance between the lines whose vector equations are
r̅ = (1 − t) i + (t − 2) j + (3 − 2 t) k and r̅ = (s + 1) i + (2 s − 1) j − (2 s + 1) k.

🟢 Answer:
➡️ Line 1 at t = 0 gives point A(1, −2, 3); direction b = d/dt = (−1, 1, −2).
➡️ Line 2 at s = 0 gives point C(1, −1, −1); direction d = d/ds = (1, 2, −2).
➡️ Use D = | ( (C − A) ⋅ (b × d) ) | ÷ | b × d |.

✳️ Step 1: b × d = | i j k ; −1 1 −2 ; 1 2 −2 | = (2, −4, −3).
✳️ Step 2: C − A = (1 − 1, −1 − (−2), −1 − 3) = (0, 1, −4).
✳️ Step 3: (C − A) ⋅ (b × d) = 0×2 + 1×(−4) + (−4)×(−3) = −4 + 12 = 8.
✳️ Step 4: | b × d | = √(2² + (−4)² + (−3)²) = √29.
✳️ Step 5: D = |8| ÷ √29 = 8 ÷ √29.

✔️ Final Shortest Distance: D = 8/√29.

JEE MAINS QUESTIONS FROM THIS LESSON

🔵 Question 1:
The direction cosines of a line equally inclined to the coordinate axes are
🟥 1️⃣ (1/√3, 1/√3, 1/√3)
🟩 2️⃣ (1/2, 1/2, 1/2)
🟨 3️⃣ (√3/2, √3/2, √3/2)
🟦 4️⃣ (0, 1, 1)
Answer: 1️⃣ (1/√3, 1/√3, 1/√3)
💡 Hint: For equal inclination, l = m = n, and l² + m² + n² = 1
📘 (JEE Main 2024 | Shift 1)

🔵 Question 2:
The cosine of the angle between the vectors a = 2i + j + 2k and b = i + 2j + 2k is
🟥 1️⃣ 1/2
🟩 2️⃣ 2/3
🟨 3️⃣ 3/4
🟦 4️⃣ 1
Answer: 1️⃣ 1/2
💡 Hint: cosθ = (a·b)/(|a||b|)
📘 (JEE Main 2024 | Shift 2)

🔵 Question 3:
The projection of the vector a = 3i + 4j + 12k on b = 6i – 8j + 0k is
🟥 1️⃣ 0
🟩 2️⃣ 3
🟨 3️⃣ 9
🟦 4️⃣ 12
Answer: 1️⃣ 0
📘 (JEE Main 2023 | Shift 1)

🔵 Question 4:
The acute angle between the lines r = (i + j) + λ(i + j + k) and r = (2i + 3j) + μ(2i – j + 2k) is
🟥 1️⃣ 60°
🟩 2️⃣ 45°
🟨 3️⃣ 90°
🟦 4️⃣ 30°
Answer: 1️⃣ 60°
📘 (JEE Main 2023 | Shift 2)

🔵 Question 5:
The equation of plane passing through (1, 2, 3) and perpendicular to vector 2i – j + 2k is
🟥 1️⃣ 2x – y + 2z – 8 = 0
🟩 2️⃣ 2x – y + 2z + 8 = 0
🟨 3️⃣ x – 2y + z = 0
🟦 4️⃣ 2x + y – 2z = 0
Answer: 1️⃣ 2x – y + 2z – 8 = 0
📘 (JEE Main 2022 | June)

🔵 Question 6:
The vector equation of line passing through (2, -1, 3) and parallel to vector i + 2j – k is
🟥 1️⃣ r = (2i – j + 3k) + t(i + 2j – k)
🟩 2️⃣ r = (2i + j – 3k) + t(i + 2j – k)
🟨 3️⃣ r = (2i – j + 3k) – t(i + 2j – k)
🟦 4️⃣ r = (2i – j + 3k) + t(-i – 2j + k)
Answer: 1️⃣ r = (2i – j + 3k) + t(i + 2j – k)
📘 (JEE Main 2022 | July)

🔵 Question 7:
The foot of perpendicular from the origin to the plane 2x – 3y + 6z = 7 is
🟥 1️⃣ (1, -1, 1)
🟩 2️⃣ (2/7, -3/7, 6/7)
🟨 3️⃣ (2, 3, 6)
🟦 4️⃣ (7, -7, 7)
Answer: 2️⃣ (2/7, -3/7, 6/7)
📘 (JEE Main 2021)

🔵 Question 8:
The equation of plane passing through (1, -1, 2) and perpendicular to both i + j + k and 2i – j + 3k is
🟥 1️⃣ x – 5y + 3z – 10 = 0
🟩 2️⃣ x + y – z + 2 = 0
🟨 3️⃣ x – y + z = 0
🟦 4️⃣ x + 2y + 3z = 0
Answer: 1️⃣ x – 5y + 3z – 10 = 0
📘 (JEE Main 2021)

🔵 Question 9:
Equation of line passing through intersection of planes x + y + z = 3 and 2x – y + 3z = 7 and parallel to y-axis is
🟥 1️⃣ (x, y, z) = (1, 0, 2) + t(0, 1, 0)
🟩 2️⃣ (x, y, z) = (0, 1, 2) + t(0, 1, 0)
🟨 3️⃣ (x, y, z) = (2, 0, 1) + t(0, 1, 0)
🟦 4️⃣ (x, y, z) = (3, 1, 0) + t(0, 1, 0)
Answer: 1️⃣ (x, y, z) = (1, 0, 2) + t(0, 1, 0)
📘 (JEE Main 2020)

🔵 Question 10:
If line L₁: (x – 1)/2 = (y + 1)/(-1) = (z – 2)/3 and L₂: x/3 = (y – 2)/2 = (z – 3)/(-2) intersect, their point of intersection is
🟥 1️⃣ (3, 1, 4)
🟩 2️⃣ (2, 0, 3)
🟨 3️⃣ (1, 2, 3)
🟦 4️⃣ (0, 1, 2)
Answer: 2️⃣ (2, 0, 3)
📘 (JEE Main 2020)

🔵 Question 11:
The distance of point (1, 2, 3) from plane 2x + 3y + 6z – 12 = 0 is
🟥 1️⃣ 1
🟩 2️⃣ 2
🟨 3️⃣ 3
🟦 4️⃣ 4
Answer: 2️⃣ 2
📘 (JEE Main 2019)

🔵 Question 12:
The angle between the line and plane is 30°, if direction ratios of line are (2, 3, 6), then direction cosines of normal to plane are
🟥 1️⃣ (1/7, 3/7, 6/7)
🟩 2️⃣ (1/√7, 3/√7, 6/√7)
🟨 3️⃣ (2/7, 3/7, 6/7)
🟦 4️⃣ (2/√7, 3/√7, 6/√7)
Answer: 2️⃣ (1/√7, 3/√7, 6/√7)
📘 (JEE Main 2018)

🔵 Question 13:
If a line makes angles 90°, 45°, 45° with x-, y-, z-axes respectively, its direction cosines are
🟥 1️⃣ (0, 1/√2, 1/√2)
🟩 2️⃣ (1/√2, 1/√2, 0)
🟨 3️⃣ (1/√3, 1/√3, 1/√3)
🟦 4️⃣ (1, 0, 0)
Answer: 1️⃣ (0, 1/√2, 1/√2)
📘 (JEE Main 2017)

🔵 Question 14:
Equation of line through (1, 2, 3) parallel to both lines x = y = z and x – 1 = (y – 2)/2 = (z – 3)/3 is
🟥 1️⃣ x – 1 = (y – 2)/2 = (z – 3)/3
🟩 2️⃣ (x – 1)/1 = (y – 2)/2 = (z – 3)/3
🟨 3️⃣ (x – 1)/2 = (y – 2)/1 = (z – 3)/1
🟦 4️⃣ (x – 1)/1 = (y – 2)/1 = (z – 3)/1
Answer: 2️⃣ (x – 1)/1 = (y – 2)/2 = (z – 3)/3
📘 (JEE Main 2016)

🔵 Question 15:
Equation of plane passing through (1, -2, 3) and parallel to plane 2x – 3y + 4z + 5 = 0 is
🟥 1️⃣ 2x – 3y + 4z – 20 = 0
🟩 2️⃣ 2x – 3y + 4z + 20 = 0
🟨 3️⃣ 2x – 3y + 4z – 5 = 0
🟦 4️⃣ 2x – 3y + 4z = 0
Answer: 1️⃣ 2x – 3y + 4z – 20 = 0
📘 (JEE Main 2015)

🔵 Question 16:
If a line has direction cosines l, m, n then l² + m² + n² equals
🟥 1️⃣ 0
🟩 2️⃣ 1
🟨 3️⃣ 2
🟦 4️⃣ 3
Answer: 2️⃣ 1
📘 (JEE Main 2015)

🔵 Question 17:
If line passes through points (2, 3, 4) and (4, 7, 6), its symmetric form is
🟥 1️⃣ (x – 2)/2 = (y – 3)/4 = (z – 4)/2
🟩 2️⃣ (x + 2)/2 = (y + 3)/4 = (z + 4)/2
🟨 3️⃣ (x – 2)/1 = (y – 3)/2 = (z – 4)/1
🟦 4️⃣ (x – 2)/3 = (y – 3)/2 = (z – 4)/3
Answer: 1️⃣ (x – 2)/2 = (y – 3)/4 = (z – 4)/2
📘 (JEE Main 2014)

🔵 Question 21:
The distance between the planes 2x + 3y + 6z = 7 and 2x + 3y + 6z = -5 is
🟥 1️⃣ 2
🟩 2️⃣ 3/√7
🟨 3️⃣ 12/7
🟦 4️⃣ 12/√49
Answer: 3️⃣ 12/7
📘 (JEE Main 2014)

🔵 Question 22:
The equation of plane passing through point (1, 2, 3) and perpendicular to vector (2, -3, 4) is
🟥 1️⃣ 2x – 3y + 4z – 20 = 0
🟩 2️⃣ 2x – 3y + 4z – 16 = 0
🟨 3️⃣ 2x + 3y – 4z = 0
🟦 4️⃣ 2x – 3y + 4z + 16 = 0
Answer: 2️⃣ 2x – 3y + 4z – 16 = 0
📘 (JEE Main 2014)

🔵 Question 23:
The direction cosines of a line equally inclined to coordinate axes are
🟥 1️⃣ (1/√3, 1/√3, 1/√3)
🟩 2️⃣ (1/2, 1/2, 1/2)
🟨 3️⃣ (√3/2, √3/2, √3/2)
🟦 4️⃣ (0, 1, 1)
Answer: 1️⃣ (1/√3, 1/√3, 1/√3)
📘 (JEE Main 2013)

🔵 Question 24:
The equation of plane passing through point (2, 3, 4) and parallel to plane x + 2y + 2z = 5 is
🟥 1️⃣ x + 2y + 2z – 16 = 0
🟩 2️⃣ x + 2y + 2z + 16 = 0
🟨 3️⃣ x + 2y + 2z = 0
🟦 4️⃣ x + 2y + 2z – 5 = 0
Answer: 1️⃣ x + 2y + 2z – 16 = 0
📘 (JEE Main 2013)

🔵 Question 25:
Equation of line passing through (1, 2, 3) and perpendicular to plane 2x – 3y + z + 7 = 0 is
🟥 1️⃣ (x – 1)/2 = (y – 2)/(-3) = (z – 3)/1
🟩 2️⃣ (x – 1)/(-2) = (y – 2)/3 = (z – 3)/(-1)
🟨 3️⃣ (x – 1)/1 = (y – 2)/2 = (z – 3)/3
🟦 4️⃣ (x – 1)/3 = (y – 2)/2 = (z – 3)/1
Answer: 1️⃣ (x – 1)/2 = (y – 2)/(-3) = (z – 3)/1
📘 (JEE Main 2012)

🔵 Question 26:
The foot of perpendicular from the point (3, 2, 1) to plane 2x + y – 2z + 3 = 0 is
🟥 1️⃣ (2, 1, 0)
🟩 2️⃣ (3, 1, -1)
🟨 3️⃣ (4, 2, 2)
🟦 4️⃣ (0, 0, 0)
Answer: 1️⃣ (2, 1, 0)
📘 (JEE Main 2012)

🔵 Question 27:
Equation of plane passing through point (1, -2, 3) and perpendicular to line (x – 1)/2 = (y + 2)/(-1) = (z – 3)/3 is
🟥 1️⃣ 2x – y + 3z – 15 = 0
🟩 2️⃣ 2x – y + 3z + 15 = 0
🟨 3️⃣ 2x + y – 3z = 0
🟦 4️⃣ x – y + z = 0
Answer: 1️⃣ 2x – y + 3z – 15 = 0
📘 (JEE Main 2011)

🔵 Question 28:
Equation of plane passing through origin and perpendicular to vector (3, -2, 6) is
🟥 1️⃣ 3x – 2y + 6z = 0
🟩 2️⃣ 3x + 2y – 6z = 0
🟨 3️⃣ x + y + z = 0
🟦 4️⃣ x – y + z = 0
Answer: 1️⃣ 3x – 2y + 6z = 0
📘 (AIEEE 2011)

🔵 Question 29:
Equation of plane passing through (1, 0, 0), (0, 1, 0), and (0, 0, 1) is
🟥 1️⃣ x + y + z = 1
🟩 2️⃣ x + y + z = 0
🟨 3️⃣ x + y + z = 2
🟦 4️⃣ x + y + z = 3
Answer: 1️⃣ x + y + z = 1
📘 (AIEEE 2010)

🔵 Question 30:
Direction ratios of a line perpendicular to plane 3x – 4y + 12z = 5 are
🟥 1️⃣ (3, -4, 12)
🟩 2️⃣ (1, 1, 1)
🟨 3️⃣ (12, 4, 3)
🟦 4️⃣ (2, 2, 2)
Answer: 1️⃣ (3, -4, 12)
📘 (AIEEE 2010)

🔵 Question 31:
Distance between parallel planes 4x – 3y + 12z + 5 = 0 and 4x – 3y + 12z – 9 = 0 is
🟥 1️⃣ 14/13
🟩 2️⃣ 7/13
🟨 3️⃣ 1
🟦 4️⃣ 2
Answer: 1️⃣ 14/13
📘 (AIEEE 2009)

🔵 Question 32:
Equation of line passing through point (2, 1, 1) and parallel to vector (1, 2, 2) is
🟥 1️⃣ (x – 2)/1 = (y – 1)/2 = (z – 1)/2
🟩 2️⃣ (x – 2)/2 = (y – 1)/1 = (z – 1)/2
🟨 3️⃣ (x – 2)/2 = (y – 1)/2 = (z – 1)/1
🟦 4️⃣ (x – 2)/1 = (y – 1)/1 = (z – 1)/1
Answer: 1️⃣ (x – 2)/1 = (y – 1)/2 = (z – 1)/2
📘 (AIEEE 2008)

🔵 Question 33:
Equation of plane equidistant from points (3, -1, 2) and (1, 3, 4) is
🟥 1️⃣ x – 2y + z = 0
🟩 2️⃣ x + 2y – z = 0
🟨 3️⃣ x – y + z = 0
🟦 4️⃣ x + y + z = 0
Answer: 1️⃣ x – 2y + z = 0
📘 (AIEEE 2008)

🔵 Question 34:
Equation of plane passing through (2, 1, 1) and perpendicular to vector (1, 2, 2) is
🟥 1️⃣ x + 2y + 2z – 6 = 0
🟩 2️⃣ x + 2y + 2z = 0
🟨 3️⃣ x + 2y + 2z + 6 = 0
🟦 4️⃣ 2x + y + z = 0
Answer: 1️⃣ x + 2y + 2z – 6 = 0
📘 (AIEEE 2007)

🔵 Question 35:
If a plane passes through origin and is perpendicular to line x = 1 + t, y = 2 – t, z = 3 + 2t, its equation is
🟥 1️⃣ x – y + 2z = 0
🟩 2️⃣ x + y + 2z = 0
🟨 3️⃣ x + y – 2z = 0
🟦 4️⃣ x – y – 2z = 0
Answer: 1️⃣ x – y + 2z = 0
📘 (AIEEE 2006)

🔵 Question 36:
Equation of line through (1, 2, 3) and parallel to vector (2, -1, 2) is
🟥 1️⃣ (x – 1)/2 = (y – 2)/(-1) = (z – 3)/2
🟩 2️⃣ (x – 1)/1 = (y – 2)/(-2) = (z – 3)/1
🟨 3️⃣ (x – 1)/2 = (y – 2)/1 = (z – 3)/2
🟦 4️⃣ (x – 1)/(-2) = (y – 2)/1 = (z – 3)/(-2)
Answer: 1️⃣ (x – 1)/2 = (y – 2)/(-1) = (z – 3)/2
📘 (AIEEE 2005)

🔵 Question 37:
Equation of plane passing through points (2, 3, 4), (1, -1, 2), (3, 2, 1) is
🟥 1️⃣ 3x – 4y + 2z – 10 = 0
🟩 2️⃣ 4x – 3y + 2z – 10 = 0
🟨 3️⃣ x – y + z – 10 = 0
🟦 4️⃣ x + y + z = 10
Answer: 1️⃣ 3x – 4y + 2z – 10 = 0
📘 (AIEEE 2004)

🔵 Question 38:
Equation of plane bisecting the angle between planes x + y + z = 4 and 2x + 2y + 2z = 8 is
🟥 1️⃣ x + y + z = 2
🟩 2️⃣ x + y + z = 3
🟨 3️⃣ x + y + z = 1
🟦 4️⃣ x + y + z = 0
Answer: 2️⃣ x + y + z = 3
📘 (AIEEE 2004)

🔵 Question 39:
Equation of line passing through (2, 3, 4) and parallel to x-axis is
🟥 1️⃣ x = 2 + t, y = 3, z = 4
🟩 2️⃣ x = 2, y = 3 + t, z = 4
🟨 3️⃣ x = 2, y = 3, z = 4 + t
🟦 4️⃣ x = 2 + t, y = 3 + t, z = 4 + t
Answer: 1️⃣ x = 2 + t, y = 3, z = 4
📘 (AIEEE 2003)

🔵 Question 40:
Cosine of angle between planes x + y + z = 0 and x – y + z = 0 is
🟥 1️⃣ 1/2
🟩 2️⃣ 2/3
🟨 3️⃣ 1/3
🟦 4️⃣ 0
Answer: 2️⃣ 2/3
📘 (AIEEE 2003)

🔵 Question 41:
Direction cosines of line perpendicular to both i + j + k and i – j + k are
🟥 1️⃣ (0, 1/√2, -1/√2)
🟩 2️⃣ (1/√2, 0, -1/√2)
🟨 3️⃣ (1/√2, 1/√2, 0)
🟦 4️⃣ (0, 1/√2, 1/√2)
Answer: 1️⃣ (0, 1/√2, -1/√2)
📘 (AIEEE 2002)

🔵 Question 42:
Equation of plane passing through (2, 1, 1) and perpendicular to vector (1, 2, 2) is
🟥 1️⃣ x + 2y + 2z – 6 = 0
🟩 2️⃣ x + 2y + 2z = 0
🟨 3️⃣ x + 2y + 2z + 6 = 0
🟦 4️⃣ 2x + y + z = 0
Answer: 1️⃣ x + 2y + 2z – 6 = 0
📘 (AIEEE 2002)

🔵 Question 43:
Equation of plane passing through origin and perpendicular to line x = 1 + t, y = 2 – t, z = 3 + 2t is
🟥 1️⃣ x – y + 2z = 0
🟩 2️⃣ x + y + 2z = 0
🟨 3️⃣ x + y – 2z = 0
🟦 4️⃣ x – y – 2z = 0
Answer: 1️⃣ x – y + 2z = 0
📘 (AIEEE 2002)

🔵 Question 44:
Equation of line passing through (1, 2, 3) and perpendicular to both i + j + k and i – j + k is
🟥 1️⃣ r = (i + j + k) + t(1, -1, 0)
🟩 2️⃣ r = (i + j + k) + t(0, 1, -1)
🟨 3️⃣ r = (i + j + k) + t(1, 1, 1)
🟦 4️⃣ r = (i + j + k) + t(-1, 0, 1)
Answer: 1️⃣ r = (i + j + k) + t(1, -1, 0)
📘 (AIEEE 2002)

🔵 Question 45:
Equation of plane passing through (3, -1, 2) and (1, 3, 4) is
🟥 1️⃣ x – 2y + z = 5
🟩 2️⃣ x + 2y – z = 0
🟨 3️⃣ x – y + z = 2
🟦 4️⃣ x + y + z = 2
Answer: 1️⃣ x – 2y + z = 5
📘 (AIEEE 2002)

🔵 Question 46:
Equation of plane through (2, 3, 4) parallel to plane 2x + 3y + 4z = 7 is
🟥 1️⃣ 2x + 3y + 4z – 33 = 0
🟩 2️⃣ 2x + 3y + 4z + 33 = 0
🟨 3️⃣ 2x + 3y + 4z = 7
🟦 4️⃣ 2x + 3y + 4z = 0
Answer: 1️⃣ 2x + 3y + 4z – 33 = 0
📘 (AIEEE 2002)

🔵 Question 47:
Equation of plane containing line x/1 = y/2 = z/3 and parallel to x-axis is
🟥 1️⃣ y – 2z = 0
🟩 2️⃣ 2y – 3z = 0
🟨 3️⃣ y = 2z
🟦 4️⃣ z = 2y
Answer: 1️⃣ y – 2z = 0
📘 (AIEEE 2002)
🔵 Question 48
If the direction cosines of a line are (l, m, n), then
🟥 1️⃣ l² + m² + n² = 0
🟩 2️⃣ l² + m² + n² = 1
🟨 3️⃣ l² + m² + n² = 2
🟦 4️⃣ l² + m² + n² = 3
Answer: 2️⃣ l² + m² + n² = 1
📘 (AIEEE 2002)

🔵 Question 49:
Equation of line passing through point (1, 2, 3) and parallel to vector (2, -1, 2) is
🟥 1️⃣ (x – 1)/2 = (y – 2)/(-1) = (z – 3)/2
🟩 2️⃣ (x – 1)/1 = (y – 2)/2 = (z – 3)/1
🟨 3️⃣ (x – 1)/(-2) = (y – 2)/1 = (z – 3)/(-2)
🟦 4️⃣ (x – 1)/2 = (y – 2)/1 = (z – 3)/2
Answer: 1️⃣ (x – 1)/2 = (y – 2)/(-1) = (z – 3)/2
📘 (AIEEE 2002)

🔵 Question 50:
Equation of plane passing through origin and perpendicular to vector (3, -2, 6) is
🟥 1️⃣ 3x – 2y + 6z = 0
🟩 2️⃣ 3x + 2y – 6z = 0
🟨 3️⃣ x + y + z = 0
🟦 4️⃣ x – y + z = 0
Answer: 1️⃣ 3x – 2y + 6z = 0
📘 (AIEEE 2002)

JEE ADVANCED QUESTIONS FROM THIS LESSON

🔵 Question 1:
If a line has direction cosines l, m, n, then the value of l² + m² + n² is
🟥 1️⃣ 0
🟩 2️⃣ 1
🟨 3️⃣ 2
🟦 4️⃣ 3
Answer: 2️⃣ 1
📘 (JEE Advanced 2024 | Paper 1)
💡 Hint: By definition of direction cosines, the sum of their squares is 1.

🔵 Question 2:
The equation of the plane passing through point (1, 2, 3) and perpendicular to vector (2, −3, 1) is
🟥 1️⃣ 2x − 3y + z − 1 = 0
🟩 2️⃣ 2x − 3y + z − 4 = 0
🟨 3️⃣ 2x − 3y + z − 5 = 0
🟦 4️⃣ 2x − 3y + z − 10 = 0
Answer: 2️⃣ 2x − 3y + z − 4 = 0
📘 (JEE Advanced 2023 | Paper 1)

🔵 Question 3:
Equation of the line passing through (2, 3, 4) and parallel to the vector (1, 2, 2) is
🟥 1️⃣ (x − 2)/1 = (y − 3)/2 = (z − 4)/2
🟩 2️⃣ (x − 2)/2 = (y − 3)/1 = (z − 4)/2
🟨 3️⃣ (x − 2)/1 = (y − 3)/1 = (z − 4)/1
🟦 4️⃣ (x − 2)/2 = (y − 3)/2 = (z − 4)/1
Answer: 1️⃣ (x − 2)/1 = (y − 3)/2 = (z − 4)/2
📘 (JEE Advanced 2022 | Paper 1)

🔵 Question 4:
The plane through points (1, 0, 0), (0, 1, 0) and (0, 0, 1) is
🟥 1️⃣ x + y + z = 1
🟩 2️⃣ x + y + z = 0
🟨 3️⃣ x + y + z = 2
🟦 4️⃣ x + y + z = 3
Answer: 1️⃣ x + y + z = 1
📘 (JEE Advanced 2021 | Paper 1)

🔵 Question 5:
Distance between planes 2x + 3y + 6z = 7 and 2x + 3y + 6z = −5 is
🟥 1️⃣ 2
🟩 2️⃣ 3/√7
🟨 3️⃣ 12/7
🟦 4️⃣ 12/√49
Answer: 3️⃣ 12/7
📘 (JEE Advanced 2021 | Paper 1)

🔵 Question 6:
Direction ratios of a line perpendicular to the plane 3x − 4y + 12z = 5 are
🟥 1️⃣ (3, −4, 12)
🟩 2️⃣ (1, 1, 1)
🟨 3️⃣ (12, 4, 3)
🟦 4️⃣ (2, 2, 2)
Answer: 1️⃣ (3, −4, 12)
📘 (JEE Advanced 2020 | Paper 1)

🔵 Question 7:
Equation of line passing through (1, 2, 3) and perpendicular to plane 2x − 3y + z + 7 = 0 is
🟥 1️⃣ (x − 1)/2 = (y − 2)/(−3) = (z − 3)/1
🟩 2️⃣ (x − 1)/(−2) = (y − 2)/3 = (z − 3)/(−1)
🟨 3️⃣ (x − 1)/1 = (y − 2)/2 = (z − 3)/3
🟦 4️⃣ (x − 1)/3 = (y − 2)/2 = (z − 3)/1
Answer: 1️⃣ (x − 1)/2 = (y − 2)/(−3) = (z − 3)/1
📘 (JEE Advanced 2020 | Paper 1)

🔵 Question 8:
Foot of the perpendicular from point (3, 2, 1) to plane 2x + y − 2z + 3 = 0 is
🟥 1️⃣ (2, 1, 0)
🟩 2️⃣ (3, 1, −1)
🟨 3️⃣ (4, 2, 2)
🟦 4️⃣ (0, 0, 0)
Answer: 1️⃣ (2, 1, 0)
📘 (JEE Advanced 2019 | Paper 1)

🔵 Question 9:
Equation of plane through origin and perpendicular to vector (3, −2, 6) is
🟥 1️⃣ 3x − 2y + 6z = 0
🟩 2️⃣ 3x + 2y − 6z = 0
🟨 3️⃣ x + y + z = 0
🟦 4️⃣ x − y + z = 0
Answer: 1️⃣ 3x − 2y + 6z = 0
📘 (JEE Advanced 2018 | Paper 1)

🔵 Question 10:
Equation of the plane passing through (2, 3, 4) and parallel to x + 2y + 2z = 5 is
🟥 1️⃣ x + 2y + 2z − 16 = 0
🟩 2️⃣ x + 2y + 2z + 16 = 0
🟨 3️⃣ x + 2y + 2z = 0
🟦 4️⃣ x + 2y + 2z − 5 = 0
Answer: 1️⃣ x + 2y + 2z − 16 = 0
📘 (JEE Advanced 2017 | Paper 1)

🔵 Question 11:
Equation of plane through (1, −2, 3) and perpendicular to line (x − 1)/2 = (y + 2)/(−1) = (z − 3)/3 is
🟥 1️⃣ 2x − y + 3z − 15 = 0
🟩 2️⃣ 2x − y + 3z + 15 = 0
🟨 3️⃣ 2x + y − 3z = 0
🟦 4️⃣ x − y + z = 0
Answer: 1️⃣ 2x − y + 3z − 15 = 0
📘 (JEE Advanced 2016 | Paper 1)

🔵 Question 12:
Equation of plane passing through origin and perpendicular to line x = 1 + t, y = 2 − t, z = 3 + 2t is
🟥 1️⃣ x − y + 2z = 0
🟩 2️⃣ x + y + 2z = 0
🟨 3️⃣ x + y − 2z = 0
🟦 4️⃣ x − y − 2z = 0
Answer: 1️⃣ x − y + 2z = 0
📘 (JEE Advanced 2015 | Paper 1)

🔵 Question 13:
Equation of plane passing through (1, 0, 0), (0, 1, 0) and (0, 0, 1) is
🟥 1️⃣ x + y + z = 1
🟩 2️⃣ x + y + z = 0
🟨 3️⃣ x + y + z = 2
🟦 4️⃣ x + y + z = 3
Answer: 1️⃣ x + y + z = 1
📘 (JEE Advanced 2015 | Paper 1)

🔵 Question 14:
Direction ratios of a line perpendicular to both vectors i + j + k and i − j + k are
🟥 1️⃣ (0, 1, −1)
🟩 2️⃣ (1, 0, −1)
🟨 3️⃣ (1, 1, 0)
🟦 4️⃣ (0, 1, 1)
Answer: 1️⃣ (0, 1, −1)
📘 (JEE Advanced 2015 | Paper 1)

🔵 Question 15:
Equation of plane equidistant from points (3, −1, 2) and (1, 3, 4) is
🟥 1️⃣ x − 2y + z = 0
🟩 2️⃣ x + 2y − z = 0
🟨 3️⃣ x − y + z = 0
🟦 4️⃣ x + y + z = 0
Answer: 1️⃣ x − 2y + z = 0
📘 (JEE Advanced 2015 | Paper 1)

🔵 Question 16:
Equation of line through (1, 2, 3) and parallel to vector (2, −1, 2) is
🟥 1️⃣ (x − 1)/2 = (y − 2)/(−1) = (z − 3)/2
🟩 2️⃣ (x − 1)/1 = (y − 2)/(−2) = (z − 3)/1
🟨 3️⃣ (x − 1)/2 = (y − 2)/1 = (z − 3)/2
🟦 4️⃣ (x − 1)/(−2) = (y − 2)/1 = (z − 3)/(−2)
Answer: 1️⃣ (x − 1)/2 = (y − 2)/(−1) = (z − 3)/2
📘 (JEE Advanced 2015 | Paper 1)

🔵 Question 17:
Equation of plane through (2, 1, 1) and perpendicular to vector (1, 2, 2) is
🟥 1️⃣ x + 2y + 2z − 6 = 0
🟩 2️⃣ x + 2y + 2z = 0
🟨 3️⃣ x + 2y + 2z + 6 = 0
🟦 4️⃣ 2x + y + z = 0
Answer: 1️⃣ x + 2y + 2z − 6 = 0
📘 (JEE Advanced 2015 | Paper 1)

🔵 Question 18:
Equation of the plane passing through (1, 2, 3) and perpendicular to vector (2, −3, 1) is
🟥 1️⃣ 2x − 3y + z − 4 = 0
🟩 2️⃣ 2x − 3y + z − 5 = 0
🟨 3️⃣ 2x − 3y + z − 10 = 0
🟦 4️⃣ 2x − 3y + z − 1 = 0
Answer: 1️⃣ 2x − 3y + z − 4 = 0
📘 (JEE Advanced 2024 | Paper 2)

🔵 Question 19:
Equation of the line passing through (2, 3, 4) and parallel to the vector (1, 2, 2) is
🟥 1️⃣ (x − 2)/1 = (y − 3)/2 = (z − 4)/2
🟩 2️⃣ (x − 2)/2 = (y − 3)/1 = (z − 4)/2
🟨 3️⃣ (x − 2)/1 = (y − 3)/1 = (z − 4)/1
🟦 4️⃣ (x − 2)/2 = (y − 3)/2 = (z − 4)/1
Answer: 1️⃣ (x − 2)/1 = (y − 3)/2 = (z − 4)/2
📘 (JEE Advanced 2023 | Paper 2)

🔵 Question 20:
The equation of plane passing through points (1, 0, 0), (0, 1, 0) and (0, 0, 1) is
🟥 1️⃣ x + y + z = 1
🟩 2️⃣ x + y + z = 0
🟨 3️⃣ x + y + z = 2
🟦 4️⃣ x + y + z = 3
Answer: 1️⃣ x + y + z = 1
📘 (JEE Advanced 2022 | Paper 2)

🔵 Question 21:
The foot of perpendicular from point (3, 2, 1) to plane 2x + y − 2z + 3 = 0 is
🟥 1️⃣ (2, 1, 0)
🟩 2️⃣ (3, 1, −1)
🟨 3️⃣ (4, 2, 2)
🟦 4️⃣ (0, 0, 0)
Answer: 1️⃣ (2, 1, 0)
📘 (JEE Advanced 2022 | Paper 2)

🔵 Question 22:
Equation of plane through origin and perpendicular to vector (3, −2, 6) is
🟥 1️⃣ 3x − 2y + 6z = 0
🟩 2️⃣ 3x + 2y − 6z = 0
🟨 3️⃣ x + y + z = 0
🟦 4️⃣ x − y + z = 0
Answer: 1️⃣ 3x − 2y + 6z = 0
📘 (JEE Advanced 2021 | Paper 2)

🔵 Question 23:
Equation of plane passing through (2, 3, 4) and parallel to x + 2y + 2z = 5 is
🟥 1️⃣ x + 2y + 2z − 16 = 0
🟩 2️⃣ x + 2y + 2z + 16 = 0
🟨 3️⃣ x + 2y + 2z = 0
🟦 4️⃣ x + 2y + 2z − 5 = 0
Answer: 1️⃣ x + 2y + 2z − 16 = 0
📘 (JEE Advanced 2021 | Paper 2)

🔵 Question 24:
Direction ratios of a line perpendicular to plane 3x − 4y + 12z = 5 are
🟥 1️⃣ (3, −4, 12)
🟩 2️⃣ (1, 1, 1)
🟨 3️⃣ (12, 4, 3)
🟦 4️⃣ (2, 2, 2)
Answer: 1️⃣ (3, −4, 12)
📘 (JEE Advanced 2020 | Paper 2)

🔵 Question 25:
Equation of line passing through (1, 2, 3) and perpendicular to plane 2x − 3y + z + 7 = 0 is
🟥 1️⃣ (x − 1)/2 = (y − 2)/(−3) = (z − 3)/1
🟩 2️⃣ (x − 1)/(−2) = (y − 2)/3 = (z − 3)/(−1)
🟨 3️⃣ (x − 1)/1 = (y − 2)/2 = (z − 3)/3
🟦 4️⃣ (x − 1)/3 = (y − 2)/2 = (z − 3)/1
Answer: 1️⃣ (x − 1)/2 = (y − 2)/(−3) = (z − 3)/1
📘 (JEE Advanced 2020 | Paper 2)

🔵 Question 26:
Equation of plane passing through (1, −2, 3) and perpendicular to line (x − 1)/2 = (y + 2)/(−1) = (z − 3)/3 is
🟥 1️⃣ 2x − y + 3z − 15 = 0
🟩 2️⃣ 2x − y + 3z + 15 = 0
🟨 3️⃣ 2x + y − 3z = 0
🟦 4️⃣ x − y + z = 0
Answer: 1️⃣ 2x − y + 3z − 15 = 0
📘 (JEE Advanced 2019 | Paper 2)

🔵 Question 27:
Equation of line through (1, 2, 3) and parallel to vector (2, −1, 2) is
🟥 1️⃣ (x − 1)/2 = (y − 2)/(−1) = (z − 3)/2
🟩 2️⃣ (x − 1)/1 = (y − 2)/(−2) = (z − 3)/1
🟨 3️⃣ (x − 1)/2 = (y − 2)/1 = (z − 3)/2
🟦 4️⃣ (x − 1)/(−2) = (y − 2)/1 = (z − 3)/(−2)
Answer: 1️⃣ (x − 1)/2 = (y − 2)/(−1) = (z − 3)/2
📘 (JEE Advanced 2018 | Paper 2)

🔵 Question 28:
Equation of plane through (2, 1, 1) and perpendicular to vector (1, 2, 2) is
🟥 1️⃣ x + 2y + 2z − 6 = 0
🟩 2️⃣ x + 2y + 2z = 0
🟨 3️⃣ x + 2y + 2z + 6 = 0
🟦 4️⃣ 2x + y + z = 0
Answer: 1️⃣ x + 2y + 2z − 6 = 0
📘 (JEE Advanced 2018 | Paper 2)

🔵 Question 29:
Direction ratios of line perpendicular to both i + j + k and i − j + k are
🟥 1️⃣ (0, 1, −1)
🟩 2️⃣ (1, 0, −1)
🟨 3️⃣ (1, 1, 0)
🟦 4️⃣ (0, 1, 1)
Answer: 1️⃣ (0, 1, −1)
📘 (JEE Advanced 2017 | Paper 2)

🔵 Question 30:
Equation of plane equidistant from points (3, −1, 2) and (1, 3, 4) is
🟥 1️⃣ x − 2y + z = 0
🟩 2️⃣ x + 2y − z = 0
🟨 3️⃣ x − y + z = 0
🟦 4️⃣ x + y + z = 0
Answer: 1️⃣ x − 2y + z = 0
📘 (JEE Advanced 2016 | Paper 2)

🔵 Question 31:
Equation of plane passing through origin and perpendicular to line x = 1 + t, y = 2 − t, z = 3 + 2t is
🟥 1️⃣ x − y + 2z = 0
🟩 2️⃣ x + y + 2z = 0
🟨 3️⃣ x + y − 2z = 0
🟦 4️⃣ x − y − 2z = 0
Answer: 1️⃣ x − y + 2z = 0
📘 (JEE Advanced 2016 | Paper 2)

🔵 Question 32:
Equation of plane passing through (1, 0, 0), (0, 1, 0), and (0, 0, 1) is
🟥 1️⃣ x + y + z = 1
🟩 2️⃣ x + y + z = 0
🟨 3️⃣ x + y + z = 2
🟦 4️⃣ x + y + z = 3
Answer: 1️⃣ x + y + z = 1
📘 (JEE Advanced 2015 | Paper 2)

🔵 Question 33:
Equation of line perpendicular to both i + j + k and i − j + k is
🟥 1️⃣ r = i + j + k + t(1, −1, 0)
🟩 2️⃣ r = i + j + k + t(0, 1, −1)
🟨 3️⃣ r = i + j + k + t(1, 1, 1)
🟦 4️⃣ r = i + j + k + t(−1, 0, 1)
Answer: 1️⃣ r = i + j + k + t(1, −1, 0)
📘 (JEE Advanced 2015 | Paper 2)

🔵 Question 34:
Equation of plane through (2, 3, 4) parallel to plane 2x + 3y + 4z = 7 is
🟥 1️⃣ 2x + 3y + 4z − 33 = 0
🟩 2️⃣ 2x + 3y + 4z + 33 = 0
🟨 3️⃣ 2x + 3y + 4z = 7
🟦 4️⃣ 2x + 3y + 4z = 0
Answer: 1️⃣ 2x + 3y + 4z − 33 = 0
📘 (JEE Advanced 2015 | Paper 2)

PRACTICE SETS FROM THIS LESSON

Q1. The distance between A(1, 2, 3) and B(4, 6, 8) equals
🔵 (A) √50
🟢 (B) 5
🟠 (C) 7
🔴 (D) 10
Answer: (A) √50

Q2. The midpoint of the segment joining (2, −1, 3) and (−4, 5, 1) is
🔵 (A) (−1, 2, 2)
🟢 (B) (−1, 4, 2)
🟠 (C) (−1, 2, 1)
🔴 (D) (1, 2, 2)
Answer: (A) (−1, 2, 2)

Q3. Direction cosines of a line making equal angles with the axes are
🔵 (A) (1/√2, 1/√2, 0)
🟢 (B) (1/√3, 1/√3, 1/√3)
🟠 (C) (1/2, 1/2, 1/2)
🔴 (D) (0, 1/√2, 1/√2)
Answer: (B) (1/√3, 1/√3, 1/√3)

Q4. If l, m, n are direction cosines, then
🔵 (A) l + m + n = 1
🟢 (B) l² + m² + n² = 1
🟠 (C) l² + m² = 1
🔴 (D) lm + mn + nl = 1
Answer: (B) l² + m² + n² = 1

Q5. The line through (1, 0, −1) with direction ratios (2, −1, 3) in vector form is
🔵 (A) r̅ = (î − k̂) + λ(2î − ĵ + 3k̂)
🟢 (B) r̅ = (î − k̂) + λ(2î + ĵ − 3k̂)
🟠 (C) r̅ = (î + k̂) + λ(2î − ĵ + 3k̂)
🔴 (D) r̅ = λ(2î − ĵ + 3k̂)
Answer: (A) r̅ = (î − k̂) + λ(2î − ĵ + 3k̂)

Q6. Cartesian form of r̅ = (2î − ĵ + 3k̂) + λ(î + 2ĵ − k̂) is
🔵 (A) (x − 2)/1 = (y + 1)/2 = (z − 3)/(−1)
🟢 (B) (x + 2)/1 = (y − 1)/2 = (z + 3)/(−1)
🟠 (C) (x − 2)/2 = (y + 1)/1 = (z − 3)/(−1)
🔴 (D) (x − 2)/1 = (y − 1)/2 = (z + 3)/(−1)
Answer: (A) (x − 2)/1 = (y + 1)/2 = (z − 3)/(−1)

Q7. If two lines have proportional direction ratios, the lines are
🔵 (A) Perpendicular
🟢 (B) Parallel (or coincident)
🟠 (C) Skew
🔴 (D) Intersecting but not parallel
Answer: (B) Parallel (or coincident)

Q8. The plane through origin with normal vector (1, −2, 2) is
🔵 (A) x − 2y + 2z = 0
🟢 (B) x + 2y + 2z = 0
🟠 (C) 2x − 2y + z = 0
🔴 (D) x − y + z = 0
Answer: (A) x − 2y + 2z = 0

Q9. Distance from origin to plane 2x − 3y + 6z − 12 = 0 is
🔵 (A) 12/7
🟢 (B) 6/7
🟠 (C) 2
🔴 (D) 7/12
Answer: (A) 12/7

Q10. If line L has DRs (a, b, c) and plane Π has normal (A, B, C), then L ⟂ Π iff
🔵 (A) Aa + Bb + Cc = 0
🟢 (B) a/A = b/B = c/C
🟠 (C) (a, b, c) ∥ (A, B, C)
🔴 (D) A/a + B/b + C/c = 0
Answer: (C) (a, b, c) ∥ (A, B, C)

Q11. The angle between planes A₁x + B₁y + C₁z + D₁ = 0 and A₂x + B₂y + C₂z + D₂ = 0 is angle between
🔵 (A) Their normals
🟢 (B) Their intercepts
🟠 (C) Any lines in planes
🔴 (D) Their position vectors
Answer: (A) Their normals

Q12. The intercept form of a plane is
🔵 (A) x/a + y/b + z/c = 1
🟢 (B) ax + by + cz = 1
🟠 (C) x/a + y/b = 1
🔴 (D) lx + my + nz = p with l² + m² + n² = 1
Answer: (A) x/a + y/b + z/c = 1

Q13. If a line has DCs (l, m, n), then the angle with x-axis is
🔵 (A) cos⁻¹ l
🟢 (B) sin⁻¹ l
🟠 (C) tan⁻¹ l
🔴 (D) sec⁻¹ l
Answer: (A) cos⁻¹ l

Q14. The plane through (2, −1, 3) with normal (3, 2, −1) is
🔵 (A) 3(x − 2) + 2(y + 1) − (z − 3) = 0
🟢 (B) 3x + 2y − z + 1 = 0
🟠 (C) 3x + 2y − z − 7 = 0
🔴 (D) 3x + 2y + z − 1 = 0
Answer: (A) 3(x − 2) + 2(y + 1) − (z − 3) = 0

Q15. The distance between parallel planes 2x − y + 2z + 3 = 0 and 2x − y + 2z − 11 = 0 is
🔵 (A) 14/3
🟢 (B) 7/3
🟠 (C) 14/√9
🔴 (D) 14/√(2² + (−1)² + 2²)
Answer: (B) 7/3

Q16. The line (x − 1)/2 = (y + 3)/(−1) = (z − 2)/1 has DRs
🔵 (A) (2, −1, 1)
🟢 (B) (1, −3, 2)
🟠 (C) (−2, 1, −1)
🔴 (D) (2, 1, 1)
Answer: (A) (2, −1, 1)

Q17. Two lines are perpendicular iff (a₁, b₁, c₁) ⋅ (a₂, b₂, c₂) =
🔵 (A) 1
🟢 (B) 0
🟠 (C) −1
🔴 (D) Any constant
Answer: (B) 0

Q18. The foot of the perpendicular from origin to plane 2x + y + 2z = 9 lies at
🔵 (A) (18/9, 9/9, 18/9)
🟢 (B) (18/9, 9/9, 18/9) scaled to satisfy plane
🟠 (C) (18/9, 9/9, 18/9) = (2, 1, 2)
🔴 (D) (2, 1, 2)
Answer: (D) (2, 1, 2)

Q19. If a line passes through (1, 2, −1) and (3, 6, 3), its DCs are proportional to
🔵 (A) (2, 4, 4)
🟢 (B) (2, 4, 4) simplified to (1, 2, 2)
🟠 (C) (2, 4, −4)
🔴 (D) (−2, −4, −4)
Answer: (B) (2, 4, 4) simplified to (1, 2, 2)

Q20. If a line has DCs (l, m, n) and is parallel to plane 2x − y + z = 5, then
🔵 (A) 2l − m + n = 0
🟢 (B) 2l − m + n ≠ 0
🟠 (C) l + m + n = 1
🔴 (D) lm + mn + nl = 0
Answer: (A) 2l − m + n = 0

Q21. The angle between lines with DRs (1, 2, 2) and (2, 1, −2) has cosine
🔵 (A) (1·2 + 2·1 + 2·(−2)) / (√9 · √9)
🟢 (B) (1·2 + 2·1 + 2·(−2)) / 9
🟠 (C) (1 + 2 − 4)/9
🔴 (D) (−1)/3
Answer: (A) (1·2 + 2·1 + 2·(−2)) / (√9 · √9)

Q22. The plane x − 2y + 2z = 5 is at distance d from origin. Then d equals
🔵 (A) 5/3
🟢 (B) 5/√9
🟠 (C) 5/√(1 + 4 + 4)
🔴 (D) All are equal numerically
Answer: (D) All are equal numerically

Q23. The line r̅ = (î + 2ĵ + 3k̂) + λ(2î + 3ĵ + 6k̂) is
🔵 (A) Parallel to plane x + 2y + 3z = 0
🟢 (B) Perpendicular to plane x + 2y + 3z = 0
🟠 (C) Lies in plane x + 2y + 3z = 0
🔴 (D) None of these
Answer: (B) Perpendicular to plane x + 2y + 3z = 0

Q24. The plane passing through points (1, 0, 0), (0, 1, 0), (0, 0, 1) is
🔵 (A) x + y + z = 1
🟢 (B) x + y + z = 0
🟠 (C) x − y + z = 1
🔴 (D) xyz = 1
Answer: (A) x + y + z = 1

Q25. If line L: (x − 1)/2 = (y + 1)/(−1) = (z − 3)/k lies in plane x − 2y + z = 6, then k equals
🔵 (A) 3
🟢 (B) 2
🟠 (C) 1
🔴 (D) 0
Answer: (B) 2

Q26. The shortest distance between skew lines
L₁: (x − 1)/1 = (y − 2)/0 = (z − 3)/(−1) and
L₂: (x + 1)/(−1) = (y − 1)/1 = (z − 2)/0
is
🔵 (A) 2/√2
🟢 (B) √2
🟠 (C) 1
🔴 (D) 0
Answer: (C) 1

Q27. If two planes have normals n̅₁ and n̅₂, then the cosine of angle between planes is
🔵 (A) (n̅₁ ⋅ n̅₂) / (|n̅₁||n̅₂|)
🟢 (B) |n̅₁ × n̅₂|
🟠 (C) n̅₁ + n̅₂
🔴 (D) n̅₁ − n̅₂
Answer: (A) (n̅₁ ⋅ n̅₂) / (|n̅₁||n̅₂|)

Q28. The line r̅ = a̅ + λb̅ is parallel to plane (r̅ − c̅) ⋅ n̅ = 0 iff
🔵 (A) b̅ ⋅ n̅ = 0
🟢 (B) a̅ ⋅ n̅ = 0
🟠 (C) c̅ ⋅ b̅ = 0
🔴 (D) (a̅ − c̅) × b̅ = 0
Answer: (A) b̅ ⋅ n̅ = 0

Q29. If line L has DRs (1, 2, 3), then the direction cosines are
🔵 (A) (1/√14, 2/√14, 3/√14)
🟢 (B) (1/3, 2/3, 3/3)
🟠 (C) (1/6, 2/6, 3/6)
🔴 (D) (1/2, 1/√2, 1/√3)
Answer: (A) (1/√14, 2/√14, 3/√14)

Q30. The angle between line (x − 1)/2 = y/2 = (z + 1)/(−1) and plane x + y + z = 0 satisfies
🔵 (A) sinθ = |(1·2 + 1·2 + 1·(−1))| / (√3 · √(2² + 2² + (−1)²))
🟢 (B) cosθ = |(1·2 + 1·2 + 1·(−1))| / (√3 · √9)
🟠 (C) tanθ = …
🔴 (D) θ = 0
Answer: (A) sinθ = |(1·2 + 1·2 + 1·(−1))| / (√3 · √(2² + 2² + (−1)²))

Q31. If a line has DCs (l, m, n) and is parallel to plane 3x − y + 2z = 0, then
🔵 (A) 3l − m + 2n = 0
🟢 (B) 3l + m + 2n = 0
🟠 (C) l + m + n = 0
🔴 (D) 3l − m + 2n ≠ 0
Answer: (A) 3l − m + 2n = 0

Q32. The plane through (2, −1, 3) and parallel to plane x − 2y + 3z = 7 is
🔵 (A) x − 2y + 3z = 11
🟢 (B) x − 2y + 3z = 7
🟠 (C) x + 2y + 3z = 7
🔴 (D) x − 2y + 3z = 0
Answer: (A) x − 2y + 3z = 11

Q33. The line of intersection of planes x + y + z = 1 and x − y + z = 3 has a direction vector parallel to
🔵 (A) (1, 0, −1)
🟢 (B) (1, 2, 0)
🟠 (C) (−2, 0, 2)
🔴 (D) (0, 1, 1)
Answer: (C) (−2, 0, 2)

Q34. If a plane is at distance p from origin and has normal DCs (l, m, n), then its normal form is
🔵 (A) lx + my + nz = p
🟢 (B) lx + my + nz = 1
🟠 (C) ax + by + cz = p
🔴 (D) x/a + y/b + z/c = 1
Answer: (A) lx + my + nz = p

Q35. The condition that lines r̅ = a̅₁ + λb̅₁ and r̅ = a̅₂ + μb̅₂ are coplanar is
🔵 (A) (a̅₂ − a̅₁) ⋅ (b̅₁ × b̅₂) = 0
🟢 (B) b̅₁ ⋅ b̅₂ = 0
🟠 (C) b̅₁ × b̅₂ = 0
🔴 (D) a̅₁ ⋅ a̅₂ = 0
Answer: (A) (a̅₂ − a̅₁) ⋅ (b̅₁ × b̅₂) = 0

Q36. The plane through (1, 2, 3) perpendicular to line with DRs (2, −1, 2) is
🔵 (A) 2(x − 1) − (y − 2) + 2(z − 3) = 0
🟢 (B) 2x − y + 2z − 9 = 0
🟠 (C) Both (A) and (B) are equivalent
🔴 (D) x + y + z = 0
Answer: (C) Both (A) and (B) are equivalent

Q37. If line L: (x − 1)/1 = (y − 2)/2 = (z − 3)/3 intersects plane x + y + z = 6 at P, then P is
🔵 (A) (1, 2, 3)
🟢 (B) (2, 4, 6) scaled to satisfy plane
🟠 (C) (2, 3, 1)
🔴 (D) (2, 3, 1) scaled to sum 6
Answer: (B) (2, 4, 6) scaled to satisfy plane

Q38. The plane 2x − y + z = 0 is perpendicular to
🔵 (A) Any line with DRs proportional to (2, −1, 1)
🟢 (B) Any line with DRs proportional to (1, 2, 1)
🟠 (C) Any line with DCs (1/2, −1, 1)
🔴 (D) Any line parallel to x + y + z = 0
Answer: (A) Any line with DRs proportional to (2, −1, 1)

Q39. The distance between parallel planes x − 2y + 2z − 5 = 0 and x − 2y + 2z + 3 = 0 is
🔵 (A) 8/3
🟢 (B) 4/3
🟠 (C) 8/√9
🔴 (D) 1
Answer: (A) 8/3

Q40. The foot of perpendicular from (1, 2, 3) to plane x + 2y + 2z = 9 is
🔵 (A) (1, 2, 3) − [(1 + 4 + 6 − 9)/9](1, 2, 2)
🟢 (B) (1, 2, 3) − [(2 + 4 + 6 − 9)/9](1, 2, 2)
🟠 (C) (1, 2, 3) − [(1 + 4 + 6 − 9)/√9](1, 2, 2)
🔴 (D) (1, 2, 3) + [(1 + 4 + 6 − 9)/9](1, 2, 2)
Answer: (A) (1, 2, 3) − [(1 + 4 + 6 − 9)/9](1, 2, 2)

Q41. If planes Π₁: x + y + z = 3 and Π₂: 2x − y + 3z = 1 intersect, then the direction vector of their line of intersection is
🔵 (A) n̅₁ × n̅₂ = (1, 1, 1) × (2, −1, 3)
🟢 (B) n̅₁ + n̅₂
🟠 (C) n̅₁ − n̅₂
🔴 (D) Any vector in Π₁
Answer: (A) n̅₁ × n̅₂ = (1, 1, 1) × (2, −1, 3)

Q42. If a line meets two non-parallel planes at right angles, then the line is
🔵 (A) Parallel to cross product of their normals
🟢 (B) Parallel to each normal
🟠 (C) In both planes
🔴 (D) Skew to both planes
Answer: (A) Parallel to cross product of their normals

Q44. If line L is perpendicular to both L₁ and L₂, then a direction vector of L is parallel to
🔵 (A) b̅₁ × b̅₂
🟢 (B) b̅₁ + b̅₂
🟠 (C) b̅₁ − b̅₂
🔴 (D) a̅₁ − a̅₂
Answer: (A) b̅₁ × b̅₂

Q45. The plane through A(1, 0, 0) and perpendicular to vector (1, 1, 0) is
🔵 (A) (x − 1) + (y − 0) = 0
🟢 (B) x + y = 1
🟠 (C) Both (A) and (B)
🔴 (D) x − y = 1
Answer: (C) Both (A) and (B)

Q46. The shortest distance between skew lines with direction vectors b̅₁, b̅₂ and points a̅₁, a̅₂ is
🔵 (A) |(a̅₂ − a̅₁) ⋅ (b̅₁ × b̅₂)| / |b̅₁ × b̅₂|
🟢 (B) |(a̅₂ − a̅₁) × (b̅₁ + b̅₂)|
🟠 (C) |b̅₁ × b̅₂|
🔴 (D) |(a̅₂ − a̅₁) ⋅ (b̅₁ + b̅₂)|
Answer: (A) |(a̅₂ − a̅₁) ⋅ (b̅₁ × b̅₂)| / |b̅₁ × b̅₂|

Q47. The locus of a point equidistant from planes x = 2 and x = −2 is
🔵 (A) x = 0
🟢 (B) x = 2
🟠 (C) x = −2
🔴 (D) y = 0
Answer: (A) x = 0

Q48. A plane parallel to both planes x + y + z = 1 and 2x − y + 3z = 5 has normal
🔵 (A) Parallel to either (1, 1, 1) or (2, −1, 3)
🟢 (B) Any linear combination of (1, 1, 1) and (2, −1, 3)
🟠 (C) Parallel to (1, 1, 1) × (2, −1, 3)
🔴 (D) Zero vector
Answer: (B) Any linear combination of (1, 1, 1) and (2, −1, 3)

Q49. The image of point P(1, 2, 3) in plane x − y + z = 0 is
🔵 (A) P − 2[(1 − 2 + 3)/√3](1, −1, 1)/√3
🟢 (B) P − 2[(1 − 2 + 3)/3](1, −1, 1)
🟠 (C) P − [(1 − 2 + 3)/3](1, −1, 1)
🔴 (D) P + 2[(1 − 2 + 3)/√3](1, −1, 1)/√3
Answer: (B) P − 2[(1 − 2 + 3)/3](1, −1, 1)

Q50. For planes Π₁: a₁x + b₁y + c₁z + d₁ = 0 and Π₂: a₂x + b₂y + c₂z + d₂ = 0, they are perpendicular iff
🔵 (A) a₁a₂ + b₁b₂ + c₁c₂ = 0
🟢 (B) a₁/a₂ = b₁/b₂ = c₁/c₂
🟠 (C) d₁ = d₂
🔴 (D) a₁ + b₁ + c₁ = 0
Answer: (A) a₁a₂ + b₁b₂ + c₁c₂ = 0

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