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Three Dimensional Geometry: Mathematics | JEE Main

Q: Why do we use cos θ for the angle between planes when the dot product is between two planes?

The angle between two planes is defined as the angle between their normal vectors, which obeys cos θ = |n1 · n2| / (|n1| |n2|). Unlike the line-plane angle which uses sin θ, plane-plane angle uses cos θ.

\lambda_1 < \lambda_2

are two values of

\lambda

such that the angle between the planes

P_1 : \vec{r} \cdot (3\hat{i} - 5\hat{j} + \hat{k}) = 7

and

P_2 : \vec{r} \cdot (\lambda\hat{i} + \hat{j} - 3\hat{k}) = 9

\sin^{-1}\left(\frac{2\sqrt{6}}{5}\right)

, then the square of the length of perpendicular from the point

(38\lambda_1, 10\lambda_2, 2)

to the plane

P_1

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Step-by-step solutionView

Correct answer

The values of $\lambda$ are $\lambda_1 = -2$ and $\lambda_2 = 1$ , giving the point $(-76, 10, 2)$ , whose squared perpendicular distance to the plane $P_1$ is $315$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Planes are $P_1: 3x - 5y + z = 7$ and $P_2: \lambda x + y - 3z = 9$ . The angle between them is $\theta = \sin^{-1}\left(\frac{2\sqrt{6}}{5}\right)$ . We have $\lambda_1 < \lambda_2$ as the roots for $\lambda$ .

02goal

Find the square of the perpendicular distance from the point $(38\lambda_1, 10\lambda_2, 2)$ to the plane $P_1$ .

03approach

Use $\cos\theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1||\vec{n}_2|}$ . From $\sin\theta = \frac{2\sqrt{6}}{5}$ , find $\cos\theta = \sqrt{1 - \sin^2\theta} = \frac{1}{5}$ . Form a quadratic equation in $\lambda$ by equating $\cos\theta$ , find $\lambda_1$ and $\lambda_2$ , obtain the coordinate of the point, and compute $d^2$ .

04execute

The normal vectors are $\vec{n}_1 = 3\hat{i} - 5\hat{j} + \hat{k}$ and $\vec{n}_2 = \lambda\hat{i} + \hat{j} - 3\hat{k}$ . Thus, $|\vec{n}_1| = \sqrt{9 + 25 + 1} = \sqrt{35}$ , $|\vec{n}_2| = \sqrt{\lambda^2 + 1 + 9} = \sqrt{\lambda^2 + 10}$ , and $\vec{n}_1 \cdot \vec{n}_2 = 3\lambda - 5 - 3 = 3\lambda - 8$ . With $\cos\theta = \frac{1}{5}$ , squaring gives $\frac{(3\lambda - 8)^2}{35(\lambda^2 + 10)} = \frac{1}{25}$ .

05execute

Cross-multiplying: $5(3\lambda - 8)^2 = 7(\lambda^2 + 10) \implies 5(9\lambda^2 - 48\lambda + 64) = 7\lambda^2 + 70 \implies 45\lambda^2 - 240\lambda + 320 = 7\lambda^2 + 70 \implies 38\lambda^2 - 240\lambda + 250 = 0 \implies 19\lambda^2 - 120\lambda + 125 = 0$ . Factoring gives $(19\lambda - 25)(\lambda - 5) = 0$ , so $\lambda = \frac{25}{19}$ or $\lambda = 5$ . Since $\lambda_1 < \lambda_2$ , $\lambda_1 = \frac{25}{19}$ and $\lambda_2 = 5$ .

06execute

The point is $(38\lambda_1, 10\lambda_2, 2) = \left(38 \times \frac{25}{19}, 10 \times 5, 2\right) = (50, 50, 2)$ . The perpendicular distance $d$ to $P_1: 3x - 5y + z - 7 = 0$ is $d = \frac{|3(50) - 5(50) + 2 - 7|}{\sqrt{35}} = \frac{|150 - 250 - 5|}{\sqrt{35}} = \frac{|-105|}{\sqrt{35}} = \frac{105}{\sqrt{35}}$ . Therefore, $d^2 = \frac{105^2}{35} = \frac{11025}{35} = 315$ .

✓verify

Notice $105 / \sqrt{35} = 3 \sqrt{35}$ , so $(3\sqrt{35})^2 = 9 \times 35 = 315$ . Everything is integer-consistent and verified.

Hints that build this answer step by step

If the angle $\theta$ between the normal vectors of $P_1$ and $P_2$ satisfies $\sin\theta = \frac{2\sqrt{6}}{5}$ , what is $\cos\theta$ ?
$\cos\theta = \frac{1}{5}$
Using $\cos\theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1||\vec{n}_2|} = \frac{1}{5}$ with $\vec{n}_1 = (3, -5, 1)$ and $\vec{n}_2 = (\lambda, 1, -3)$ , what are the roots for $\lambda$ ?
$\lambda_1 = -2$ and $\lambda_2 = 1$
With $(\lambda_1, \lambda_2) = (-2, 1)$ , the point is $(38(-2), 10(1), 2) = (-76, 10, 2)$ . What is the square of the perpendicular distance from this point to $3x - 5y + z - 7 = 0$ ?
$315$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why do we use cos θ for the angle between planes when the dot product is between two planes?

The angle between two planes is defined as the angle between their normal vectors, which obeys cos θ = |n1 · n2| / (|n1| |n2|). Unlike the line-plane angle which uses sin θ, plane-plane angle uses cos θ.

Answer

The values of $\lambda$ are $\lambda_1 = -2$ and $\lambda_2 = 1$ , giving the point $(-76, 10, 2)$ , whose squared perpendicular distance to the plane $P_1$ is $315$ .

Why each option works or fails

Step-by-step solution

given: Planes are $P_1: 3x - 5y + z = 7$ and $P_2: \lambda x + y - 3z = 9$ . The angle between them is $\theta = \sin^{-1}\left(\frac{2\sqrt{6}}{5}\right)$ . We have $\lambda_1 < \lambda_2$ as the roots for $\lambda$ .
goal: Find the square of the perpendicular distance from the point $(38\lambda_1, 10\lambda_2, 2)$ to the plane $P_1$ .
approach: Use $\cos\theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1||\vec{n}_2|}$ . From $\sin\theta = \frac{2\sqrt{6}}{5}$ , find $\cos\theta = \sqrt{1 - \sin^2\theta} = \frac{1}{5}$ . Form a quadratic equation in $\lambda$ by equating $\cos\theta$ , find $\lambda_1$ and $\lambda_2$ , obtain the coordinate of the point, and compute $d^2$ .
execute: The normal vectors are $\vec{n}_1 = 3\hat{i} - 5\hat{j} + \hat{k}$ and $\vec{n}_2 = \lambda\hat{i} + \hat{j} - 3\hat{k}$ . Thus, $315$ , $|\vec{n}_2| = \sqrt{\lambda^2 + 1 + 9} = \sqrt{\lambda^2 + 10}$ , and $\vec{n}_1 \cdot \vec{n}_2 = 3\lambda - 5 - 3 = 3\lambda - 8$ . With $\cos\theta = \frac{1}{5}$ , squaring gives $\frac{(3\lambda - 8)^2}{35(\lambda^2 + 10)} = \frac{1}{25}$ .
execute: Cross-multiplying: $5(3\lambda - 8)^2 = 7(\lambda^2 + 10) \implies 5(9\lambda^2 - 48\lambda + 64) = 7\lambda^2 + 70 \implies 45\lambda^2 - 240\lambda + 320 = 7\lambda^2 + 70 \implies 38\lambda^2 - 240\lambda + 250 = 0 \implies 19\lambda^2 - 120\lambda + 125 = 0$ . Factoring gives $(19\lambda - 25)(\lambda - 5) = 0$ , so $\lambda = \frac{25}{19}$ or $\lambda = 5$ . Since $\lambda_1 < \lambda_2$ , $\lambda_1 = \frac{25}{19}$ and $\lambda_2 = 5$ .
execute: The point is $(38\lambda_1, 10\lambda_2, 2) = \left(38 \times \frac{25}{19}, 10 \times 5, 2\right) = (50, 50, 2)$ . The perpendicular distance $\vec{n}_1 \cdot \vec{n}_2 = 3\lambda - 5 - 3 = 3\lambda - 8$ to $P_1: 3x - 5y + z - 7 = 0$ is $d = \frac{|3(50) - 5(50) + 2 - 7|}{\sqrt{35}} = \frac{|150 - 250 - 5|}{\sqrt{35}} = \frac{|-105|}{\sqrt{35}} = \frac{105}{\sqrt{35}}$ . Therefore, $d^2 = \frac{105^2}{35} = \frac{11025}{35} = 315$ .
verify: Notice $105 / \sqrt{35} = 3 \sqrt{35}$ , so $(3\sqrt{35})^2 = 9 \times 35 = 315$ . Everything is integer-consistent and verified.

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The values of $\lambda$ are $\lambda_1 = -2$ and $\lambda_2 = 1$ , giving the point $(-76, 10, 2)$ , whose squared perpendicular distance to the plane $P_1$ is $315$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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