JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Angle between two planes in a family of planes

The acute angle between the planes P_1 and P_2, when P_1 and P_2 are the planes passing through the intersection of the planes 5 x+8 y+13 z-29=0 and 8 x-7 y+z-20=0 and the points (2,1,3) and (0,1,2), respectively, is

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Aπ/3Bπ/4Cπ/6Dπ/12

Step-by-step solutionView

Correct answer

The acute angle between the two planes is $\frac{π}{3}$ .

Option analysis

Why each option works or fails

A · π/3

The normal vectors are determined to be ${\vec{n}}_{1} = (1, 2, 2)$ and ${\vec{n}}_{2} = (2, - 1, 2)$ , giving $\cos θ = \frac{∣ 2 - 2 + 4 ∣}{\sqrt{9} \sqrt{9}} = \frac{4}{9}$ ? Wait, let's recalculate the planes carefully. Correctly find the planes using the family of planes equation $P_{1} + λ P_{2} = 0$ passing through the respective points, then evaluate the dot product of their normal vectors to find $\cos θ = \frac{1}{2}$ , giving $θ = \frac{π}{3}$ .

B · π/4

The student miscalculates the dot product or magnitudes of the normal vectors, obtaining $\cos θ = \frac{1}{\sqrt{2}}$ instead of $\frac{1}{2}$ . Ensure normal vectors are simplified properly and compute $\cos θ = \frac{∣ {\vec{n}}_{1} \cdot {\vec{n}}_{2} ∣}{∣ {\vec{n}}_{1} ∣ ∣ {\vec{n}}_{2} ∣}$ carefully without arithmetic errors.

C · π/6

The student confuses the sine and cosine relations for the angle between planes, identifying $\sin θ = \frac{1}{2}$ as leading to $θ = \frac{π}{6}$ rather than evaluating $\cos θ$ . Remember that the angle $θ$ between two planes with normal vectors ${\vec{n}}_{1}$ and ${\vec{n}}_{2}$ satisfies $\cos θ = \frac{∣ {\vec{n}}_{1} \cdot {\vec{n}}_{2} ∣}{∣ {\vec{n}}_{1} ∣ ∣ {\vec{n}}_{2} ∣}$ , not $\sin θ$ .

D · π/12

The student makes an algebraic sign error when solving for $λ_{1}$ or $λ_{2}$ , leading to an erroneous angle calculation. Substitute the coordinates into $(5 x + 8 y + 13 z - 29) + λ (8 x - 7 y + z - 20) = 0$ carefully to find the correct values of $λ$ .

Reviewed route

Solution

StepWorking

01Given

Given two base planes $Π_{1} : 5 x + 8 y + 13 z - 29 = 0$ and $Π_{2} : 8 x - 7 y + z - 20 = 0$ . Plane $P_{1}$ passes through the intersection of $Π_{1}, Π_{2}$ and $(2, 1, 3)$ . Plane $P_{2}$ passes through the intersection of $Π_{1}, Π_{2}$ and $(0, 1, 2)$ .

02Goal

Find the acute angle $θ$ between the planes $P_{1}$ and $P_{2}$ .

03Approach

Use the family of planes $Π_{1} + λ Π_{2} = 0$ . Substitute the respective passing points to determine $λ_{1}$ and $λ_{2}$ , obtain the normal vectors ${\vec{n}}_{1}$ and ${\vec{n}}_{2}$ , and compute $\cos θ = \frac{∣ {\vec{n}}_{1} \cdot {\vec{n}}_{2} ∣}{∣ {\vec{n}}_{1} ∣ ∣ {\vec{n}}_{2} ∣}$ .

04Execute

For $P_{1}$ : $(5 x + 8 y + 13 z - 29) + λ_{1} (8 x - 7 y + z - 20) = 0$ . Plugging in $(2, 1, 3)$ : $[5 (2) + 8 (1) + 13 (3) - 29] + λ_{1} [8 (2) - 7 (1) + 3 - 20] = 0 ⟹ 28 + λ_{1} (- 8) = 0 ⟹ λ_{1} = \frac{7}{2}$ . Substituting $λ_{1}$ back gives: $2 (5 x + 8 y + 13 z - 29) + 7 (8 x - 7 y + z - 20) = 0 ⟹ 66 x - 33 y + 33 z - 198 = 0 ⟹ 2 x - y + z = 6$ . The normal vector is ${\vec{n}}_{1} = ⟨ 2, - 1, 1 ⟩$ .

05Execute

For $P_{2}$ : plugging $(0, 1, 2)$ into $Π_{1} + λ_{2} Π_{2} = 0$ : $[5 (0) + 8 (1) + 13 (2) - 29] + λ_{2} [8 (0) - 7 (1) + 2 - 20] = 0 ⟹ 5 + λ_{2} (- 25) = 0 ⟹ λ_{2} = \frac{1}{5}$ . Substituting $λ_{2}$ : $5 (5 x + 8 y + 13 z - 29) + 1 (8 x - 7 y + z - 20) = 0 ⟹ 33 x + 33 y + 66 z - 165 = 0 ⟹ x + y + 2 z = 5$ . The normal vector is ${\vec{n}}_{2} = ⟨ 1, 1, 2 ⟩$ .

06Execute

Compute $\cos θ = \frac{∣ {\vec{n}}_{1} \cdot {\vec{n}}_{2} ∣}{∣ {\vec{n}}_{1} ∣ ∣ {\vec{n}}_{2} ∣} = \frac{∣ (2) (1) + (- 1) (1) + (1) (2) ∣}{\sqrt{2^{2} + (- 1)^{2} + 1^{2}} \sqrt{1^{2} + 1^{2} + 2^{2}}} = \frac{∣ 2 - 1 + 2 ∣}{\sqrt{6} \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$ . Thus, $θ = \frac{π}{3}$ .

✓Verify

Check point satisfaction: $P_{1} : 2 (2) - 1 (1) + 1 (3) = 6$ (holds). $P_{2} : 1 (0) + 1 (1) + 2 (2) = 5$ (holds). Magnitudes $∣ {\vec{n}}_{1} ∣ = \sqrt{6}$ , $∣ {\vec{n}}_{2} ∣ = \sqrt{6}$ , dot product is 3, leading unambiguously to $\cos θ = 1 / 2$ , so $θ = π / 3$ .

Hints that build this answer step by step

How can any plane passing through the line of intersection of $5 x + 8 y + 13 z - 29 = 0$ and $8 x - 7 y + z - 20 = 0$ be represented?
$(5 x + 8 y + 13 z - 29) + λ (8 x - 7 y + z - 20) = 0$
What are the values of parameter $λ$ for the planes $P_{1}$ passing through $(2, 1, 3)$ and $P_{2}$ passing through $(0, 1, 2)$ ?
$λ_{1} = - 2$ and $λ_{2} = 1$
Using $P_{1} : 7 x - 2 y + 5 z - 17 = 0$ (or normal ${\vec{n}}_{1}$ ) and $P_{2}$ with its corresponding normal ${\vec{n}}_{2}$ , what is the cosine of the acute angle $θ$ between them?
$\cos θ = \frac{1}{2}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

Quick checks

Students also ask

Why do we use the family of planes

P_{1} + λ P_{2} = 0

Any plane passing through the line of intersection of two planes $P_{1} = 0$ and $P_{2} = 0$ can be represented as $P_{1} + λ P_{2} = 0$ (or $P_{2} = 0$ ). This guarantees the plane contains their line of intersection.

Answer

The acute angle between the two planes is $\frac{π}{3}$ .

Why each option works or fails

A · correct: π/3 - The normal vectors are determined to be ${\vec{n}}_{1} = (1, 2, 2)$ and ${\vec{n}}_{2} = (2, - 1, 2)$ , giving $\cos θ = \frac{∣ 2 - 2 + 4 ∣}{\sqrt{9} \sqrt{9}} = \frac{4}{9}$ ? Wait, let's recalculate the planes carefully. Correctly find the planes using the family of planes equation $P_{1} + λ P_{2} = 0$ passing through the respective points, then evaluate the dot product of their normal vectors to find $\cos θ = \frac{1}{2}$ , giving $θ = \frac{π}{3}$ .
B: π/4 - The student miscalculates the dot product or magnitudes of the normal vectors, obtaining $\cos θ = \frac{1}{\sqrt{2}}$ instead of $\frac{1}{2}$ . Ensure normal vectors are simplified properly and compute $\cos θ = \frac{∣ {\vec{n}}_{1} \cdot {\vec{n}}_{2} ∣}{∣ {\vec{n}}_{1} ∣ ∣ {\vec{n}}_{2} ∣}$ carefully without arithmetic errors.
C: π/6 - The student confuses the sine and cosine relations for the angle between planes, identifying $\sin θ = \frac{1}{2}$ as leading to $θ = \frac{π}{6}$ rather than evaluating $\cos θ$ . Remember that the angle $θ$ between two planes with normal vectors ${\vec{n}}_{1}$ and ${\vec{n}}_{2}$ satisfies $\cos θ = \frac{∣ {\vec{n}}_{1} \cdot {\vec{n}}_{2} ∣}{∣ {\vec{n}}_{1} ∣ ∣ {\vec{n}}_{2} ∣}$ , not $\sin θ$ .
D: π/12 - The student makes an algebraic sign error when solving for $λ_{1}$ or $λ_{2}$ , leading to an erroneous angle calculation. Substitute the coordinates into $(5 x + 8 y + 13 z - 29) + λ (8 x - 7 y + z - 20) = 0$ carefully to find the correct values of $λ$ .

Step-by-step solution

Given: Given two base planes $Π_{1} : 5 x + 8 y + 13 z - 29 = 0$ and $Π_{2} : 8 x - 7 y + z - 20 = 0$ . Plane $P_{1}$ passes through the intersection of $Π_{1}, Π_{2}$ and $(2, 1, 3)$ . Plane $P_{2}$ passes through the intersection of $Π_{1}, Π_{2}$ and $(0, 1, 2)$ .
Goal: Find the acute angle $θ$ between the planes $P_{1}$ and $P_{2}$ .
Approach: Use the family of planes $Π_{1} + λ Π_{2} = 0$ . Substitute the respective passing points to determine $λ_{1}$ and $λ_{2}$ , obtain the normal vectors ${\vec{n}}_{1}$ and ${\vec{n}}_{2}$ , and compute $\cos θ = \frac{∣ {\vec{n}}_{1} \cdot {\vec{n}}_{2} ∣}{∣ {\vec{n}}_{1} ∣ ∣ {\vec{n}}_{2} ∣}$ .
Execute: For $P_{1}$ : $(5 x + 8 y + 13 z - 29) + λ_{1} (8 x - 7 y + z - 20) = 0$ . Plugging in $(2, 1, 3)$ : $[5 (2) + 8 (1) + 13 (3) - 29] + λ_{1} [8 (2) - 7 (1) + 3 - 20] = 0 ⟹ 28 + λ_{1} (- 8) = 0 ⟹ λ_{1} = \frac{7}{2}$ . Substituting $λ_{1}$ back gives: $2 (5 x + 8 y + 13 z - 29) + 7 (8 x - 7 y + z - 20) = 0 ⟹ 66 x - 33 y + 33 z - 198 = 0 ⟹ 2 x - y + z = 6$ . The normal vector is ${\vec{n}}_{1} = ⟨ 2, - 1, 1 ⟩$ .
Execute: For $P_{2}$ : plugging $(0, 1, 2)$ into $Π_{1} + λ_{2} Π_{2} = 0$ : $[5 (0) + 8 (1) + 13 (2) - 29] + λ_{2} [8 (0) - 7 (1) + 2 - 20] = 0 ⟹ 5 + λ_{2} (- 25) = 0 ⟹ λ_{2} = \frac{1}{5}$ . Substituting $λ_{2}$ : $5 (5 x + 8 y + 13 z - 29) + 1 (8 x - 7 y + z - 20) = 0 ⟹ 33 x + 33 y + 66 z - 165 = 0 ⟹ x + y + 2 z = 5$ . The normal vector is ${\vec{n}}_{2} = ⟨ 1, 1, 2 ⟩$ .
Execute: Compute $\cos θ = \frac{∣ {\vec{n}}_{1} \cdot {\vec{n}}_{2} ∣}{∣ {\vec{n}}_{1} ∣ ∣ {\vec{n}}_{2} ∣} = \frac{∣ (2) (1) + (- 1) (1) + (1) (2) ∣}{\sqrt{2^{2} + (- 1)^{2} + 1^{2}} \sqrt{1^{2} + 1^{2} + 2^{2}}} = \frac{∣ 2 - 1 + 2 ∣}{\sqrt{6} \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$ . Thus, $θ = \frac{π}{3}$ .
Verify: Check point satisfaction: $P_{1} : 2 (2) - 1 (1) + 1 (3) = 6$ (holds). $P_{2} : 1 (0) + 1 (1) + 2 (2) = 5$ (holds). Magnitudes $∣ {\vec{n}}_{1} ∣ = \sqrt{6}$ , $∣ {\vec{n}}_{2} ∣ = \sqrt{6}$ , dot product is 3, leading unambiguously to $\cos θ = 1 / 2$ , so $θ = π / 3$ .

Angle between two planes in a family of planes

What feels right?

The acute angle between the two planes is $\frac{π}{3}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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