JEE MainMathematics

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Angle Between Planes and Lines in Three Dimensions

Q: Why was the point (4, -2, 5) given if it's not used?

The point specifies where line L intersects P_2 to show L is well-defined in space, but the angle between a line and a plane depends only on their directional vectors, not on the position of intersection.

Let θ be the angle between the planes P_1:r vector·(i hat+j hat+2k hat)=9 and P_2:r vector·(2i hat-j hat+k hat)=15. Let L be the line that meets P_2 at the point (4,-2,5) and makes an angle θ with the normal of P_2. If α is the angle between L and P_2, then (tan^2θ)(cot^2α) is equal to

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Correct answer

The value of $(\tan^{2} θ) (\cot^{2} α)$ is 9.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

Planes $P_{1} : \vec{r} \cdot (\hat{i} + \hat{j} + 2 \hat{k}) = 9$ and $P_{2} : \vec{r} \cdot (2 \hat{i} - \hat{j} + \hat{k}) = 15$ . The angle between $P_{1}$ and $P_{2}$ is $θ$ . A line $L$ meets $P_{2}$ at $(4, - 2, 5)$ and makes an angle $θ$ with the normal to $P_{2}$ . $α$ is the angle between $L$ and the plane $P_{2}$ .

02Goal

Find the value of $(\tan^{2} θ) (\cot^{2} α)$ .

03Approach

First, calculate $\cos θ$ via the dot product of the normal vectors of $P_{1}$ and $P_{2}$ : ${\vec{n}}_{1} = \hat{i} + \hat{j} + 2 \hat{k}$ and ${\vec{n}}_{2} = 2 \hat{i} - \hat{j} + \hat{k}$ . Second, use the geometric relation between a line and a plane: the angle $α$ between line $L$ and plane $P_{2}$ is complementary to the angle between $L$ and the normal vector of $P_{2}$ , which is given as $θ$ . Thus, $α = 90^{\circ} - θ$ . Finally, evaluate $(\tan^{2} θ) (\cot^{2} α)$ .

04Execute

Find $\cos θ$ using ${\vec{n}}_{1} \cdot {\vec{n}}_{2} / (∣ {\vec{n}}_{1} ∣ ∣ {\vec{n}}_{2} ∣)$ : $∣ {\vec{n}}_{1} ∣ = \sqrt{1^{2} + 1^{2} + 2^{2}} = \sqrt{6}$ $∣ {\vec{n}}_{2} ∣ = \sqrt{2^{2} + (- 1)^{2} + 1^{2}} = \sqrt{6}$ ${\vec{n}}_{1} \cdot {\vec{n}}_{2} = (1) (2) + (1) (- 1) + (2) (1) = 2 - 1 + 2 = 3$ . Therefore, $\cos θ = \frac{3}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$ . Since $θ$ is an acute angle between planes, $θ = 60^{\circ} = \frac{π}{3}$ .

05Execute

Since line $L$ makes an angle $θ$ with the normal to $P_{2}$ , the angle between $L$ and plane $P_{2}$ is $α = 90^{\circ} - θ = 90^{\circ} - 60^{\circ} = 30^{\circ} = \frac{π}{6}$ . Now evaluate $(\tan^{2} θ) (\cot^{2} α)$ : $\tan θ = \tan 60^{\circ} = \sqrt{3} ⟹ \tan^{2} θ = 3$ . $\cot α = \cot 30^{\circ} = \sqrt{3} ⟹ \cot^{2} α = 3$ . Product $= 3 \times 3 = 9$ .

✓Verify

Notice that $α = 90^{\circ} - θ ⟹ \cot α = \cot (90^{\circ} - θ) = \tan θ$ . Thus, $(\tan^{2} θ) (\cot^{2} α) = \tan^{4} θ = (\sqrt{3})^{4} = 9$ . The point $(4, - 2, 5)$ is extraneous information and does not affect the angle calculation.

Hints that build this answer step by step

What is the cosine of the angle $θ$ between the planes $P_{1}$ and $P_{2}$ , using their normal vectors ${\vec{n}}_{1} = \hat{i} + \hat{j} + 2 \hat{k}$ and ${\vec{n}}_{2} = 2 \hat{i} - \hat{j} + \hat{k}$ ?
$\cos θ = \frac{∣ {\vec{n}}_{1} \cdot {\vec{n}}_{2} ∣}{∣ {\vec{n}}_{1} ∣ ∣ {\vec{n}}_{2} ∣} = \frac{3}{\sqrt{6} \sqrt{6}} = \frac{1}{2}$
Line $L$ makes an angle $θ$ with the normal to the plane $P_{2}$ . What is the relationship between $θ$ and the angle $α$ that $L$ makes with the plane $P_{2}$ ?
$α = \frac{π}{2} - θ$
Using $θ = \frac{π}{3}$ and $α = \frac{π}{6}$ , what is the value of $(\tan^{2} θ) (\cot^{2} α)$ ?
9

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why was the point

(4, - 2, 5)

given if it's not used?

The point specifies where line $L$ intersects $P_{2}$ to show $L$ is well-defined in space, but the angle between a line and a plane depends only on their directional vectors, not on the position of intersection.

Why is

α = 90^{\circ} - θ

The angle between a line and a plane is measured relative to the plane itself. It is the angle between the line and its orthogonal projection on the plane. The normal vector is perpendicular to the plane ( $90^{\circ}$ ). Therefore, the angle with the normal and the angle with the plane always sum to $90^{\circ}$ .

Answer

Accepted answer: 9

The value of $(\tan^{2} θ) (\cot^{2} α)$ is 9.

Why each option works or fails

Step-by-step solution

Given: Planes $P_{1} : \vec{r} \cdot (\hat{i} + \hat{j} + 2 \hat{k}) = 9$ and $P_{2} : \vec{r} \cdot (2 \hat{i} - \hat{j} + \hat{k}) = 15$ . The angle between $P_{1}$ and $P_{2}$ is $θ$ . A line $L$ meets $P_{2}$ at $(4, - 2, 5)$ and makes an angle $θ$ with the normal to $P_{2}$ . $α$ is the angle between $L$ and the plane $P_{2}$ .
Goal: Find the value of $(\tan^{2} θ) (\cot^{2} α)$ .
Approach: First, calculate $\cos θ$ via the dot product of the normal vectors of $P_{1}$ and $P_{2}$ : ${\vec{n}}_{1} = \hat{i} + \hat{j} + 2 \hat{k}$ and ${\vec{n}}_{2} = 2 \hat{i} - \hat{j} + \hat{k}$ . Second, use the geometric relation between a line and a plane: the angle $α$ between line $L$ and plane $P_{2}$ is complementary to the angle between $L$ and the normal vector of $P_{2}$ , which is given as $θ$ . Thus, $α = 90^{\circ} - θ$ . Finally, evaluate $(\tan^{2} θ) (\cot^{2} α)$ .
Execute: Find $\cos θ$ using ${\vec{n}}_{1} \cdot {\vec{n}}_{2} / (∣ {\vec{n}}_{1} ∣ ∣ {\vec{n}}_{2} ∣)$ : $∣ {\vec{n}}_{1} ∣ = \sqrt{1^{2} + 1^{2} + 2^{2}} = \sqrt{6}$ $∣ {\vec{n}}_{2} ∣ = \sqrt{2^{2} + (- 1)^{2} + 1^{2}} = \sqrt{6}$ ${\vec{n}}_{1} \cdot {\vec{n}}_{2} = (1) (2) + (1) (- 1) + (2) (1) = 2 - 1 + 2 = 3$ . Therefore, $\cos θ = \frac{3}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$ . Since $θ$ is an acute angle between planes, $θ = 60^{\circ} = \frac{π}{3}$ .
Execute: Since line $L$ makes an angle $θ$ with the normal to $P_{2}$ , the angle between $L$ and plane $P_{2}$ is $α = 90^{\circ} - θ = 90^{\circ} - 60^{\circ} = 30^{\circ} = \frac{π}{6}$ . Now evaluate $(\tan^{2} θ) (\cot^{2} α)$ : $\tan θ = \tan 60^{\circ} = \sqrt{3} ⟹ \tan^{2} θ = 3$ . $\cot α = \cot 30^{\circ} = \sqrt{3} ⟹ \cot^{2} α = 3$ . Product $= 3 \times 3 = 9$ .
Verify: Notice that $α = 90^{\circ} - θ ⟹ \cot α = \cot (90^{\circ} - θ) = \tan θ$ . Thus, $(\tan^{2} θ) (\cot^{2} α) = \tan^{4} θ = (\sqrt{3})^{4} = 9$ . The point $(4, - 2, 5)$ is extraneous information and does not affect the angle calculation.

Angle Between Planes and Lines in Three Dimensions

What feels right?

The value of $(\tan^{2} θ) (\cot^{2} α)$ is 9.

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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