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JEE MainMathematics

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Three Dimensional Geometry: Mathematics | JEE Main

Let a unit vector

\vec{OP}

make angles

\alpha, \beta, \gamma

with the positive directions of the co-ordinate axes OX, OY, OZ respectively, where

\beta \in \left(0, \frac{\pi}{2}\right)

. If

\vec{OP}

is perpendicular to the plane through points

(1,2,3)

,

(2,3,4)

and

(1,5,7)

, then which one of the following is true ?

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A

\alpha \in \left(0, \frac{\pi}{2}\right)

and

\gamma \in \left(0, \frac{\pi}{2}\right)

B

\alpha \in \left(0, \frac{\pi}{2}\right)

and

\gamma \in \left(\frac{\pi}{2}, \pi\right)

C

\alpha \in \left(\frac{\pi}{2}, \pi\right)

and

\gamma \in \left(\frac{\pi}{2}, \pi\right)

D

\alpha \in \left(\frac{\pi}{2}, \pi\right)

and

\gamma \in \left(0, \frac{\pi}{2}\right)

Step-by-step solutionView

Correct answer

Because the normal vector to the plane with a positive y-component has negative x- and z-components, both $\alpha$ and $\gamma$ lie in the interval $\left(\frac{\pi}{2}, \pi\right)$ .

Option analysis

Why each option works or fails

A ·

\alpha \in \left(0, \frac{\pi}{2}\right)

and

\gamma \in \left(0, \frac{\pi}{2}\right)

The student calculates the normal vector via cross product but flips or ignores the signs of the components, assuming all direction angles are acute. Check the sign of each direction cosine: $\cos\theta > 0 \implies \theta \in \left(0, \frac{\pi}{2}\right)$ and $\cos\theta < 0 \implies \theta \in \left(\frac{\pi}{2}, \pi\right)$ .

B ·

\alpha \in \left(0, \frac{\pi}{2}\right)

and

\gamma \in \left(\frac{\pi}{2}, \pi\right)

The student makes a sign error in the cross-product determinant expansion specifically for the x-component, obtaining $\cos\alpha > 0$ . Carefully compute $\vec{u} \times \vec{v} = (u_y v_z - u_z v_y)\hat{i} - (u_x v_z - u_z v_x)\hat{j} + (u_x v_y - u_y v_x)\hat{k}$ and adjust by multiplying by $-1$ so the $\hat{j}$ coefficient is positive.

C ·

\alpha \in \left(\frac{\pi}{2}, \pi\right)

and

\gamma \in \left(\frac{\pi}{2}, \pi\right)

None. The student correctly finds that the normal vector with $\cos\beta > 0$ has $\cos\alpha < 0$ and $\cos\gamma < 0$ , placing both angles in $\left(\frac{\pi}{2}, \pi\right)$ . This option is correct.

D ·

\alpha \in \left(\frac{\pi}{2}, \pi\right)

and

\gamma \in \left(0, \frac{\pi}{2}\right)

The student makes a sign error in the cross-product expansion for the z-component, arriving at $\cos\gamma > 0$ while having $\cos\alpha < 0$ . Re-evaluate the determinant expansion for the $\hat{k}$ component: $(1)(3) - (1)(0) = 3$ , which becomes $-3$ after ensuring the $\hat{j}$ component is positive.

Reviewed route

Solution

StepWorking

01given

Let points on the plane be $A(1, 2, 3)$ , $B(2, 3, 4)$ , and $C(1, 5, 7)$ . The unit vector $\vec{OP}$ is perpendicular to this plane, with direction angles $\alpha, \beta, \gamma$ such that $\beta \in \left(0, \frac{\pi}{2}\right)$ .

02goal

Determine the intervals containing the angles $\alpha$ and $\gamma$ .

03approach

Find two vectors lying in the plane, $\vec{AB}$ and $\vec{AC}$ . Compute their cross product to get a normal vector $\vec{n}$ . Since $\vec{OP}$ is normal to the plane, $\vec{OP} = \pm \frac{\vec{n}}{|\vec{n}|}$ . Use the condition $\beta \in (0, \pi/2) \implies \cos\beta > 0$ to fix the unique sign for $\vec{OP}$ , which gives the signs of $\cos\alpha$ and $\cos\gamma$ .

04execute

Form in-plane vectors: $\vec{AB} = (2-1)\hat{i} + (3-2)\hat{j} + (4-3)\hat{k} = \hat{i} + \hat{j} + \hat{k}$ $\vec{AC} = (1-1)\hat{i} + (5-2)\hat{j} + (7-3)\hat{k} = 0\hat{i} + 3\hat{j} + 4\hat{k}$

05execute

Find the cross product $\vec{n} = \vec{AB} \times \vec{AC}$ : $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 3 & 4 \end{vmatrix} = \hat{i}(4 - 3) - \hat{j}(4 - 0) + \hat{k}(3 - 0) = \hat{i} - 4\hat{j} + 3\hat{k}$

06execute

The unit normal vector $\vec{OP}$ is proportional to $\vec{n}$ : $\vec{OP} = \lambda (\hat{i} - 4\hat{j} + 3\hat{k})$ where $\lambda = \pm \frac{1}{\sqrt{1^2 + (-4)^2 + 3^2}} = \pm \frac{1}{\sqrt{26}}$ . The direction cosines are $\cos\alpha = \lambda$ , $\cos\beta = -4\lambda$ , $\cos\gamma = 3\lambda$ . Since $\beta \in \left(0, \frac{\pi}{2}\right)$ , we have $\cos\beta > 0 \implies -4\lambda > 0 \implies \lambda < 0$ . Therefore, $\lambda = -\frac{1}{\sqrt{26}}$ . This gives: $\cos\alpha = -\frac{1}{\sqrt{26}} < 0 \implies \alpha \in \left(\frac{\pi}{2}, \pi\right)$ $\cos\gamma = -\frac{3}{\sqrt{26}} < 0 \implies \gamma \in \left(\frac{\pi}{2}, \pi\right)$ .

✓verify

Check scalar products: $\vec{OP} \cdot \vec{AB} \propto -1(1) + 4(1) - 3(1) = 0$ and $\vec{OP} \cdot \vec{AC} \propto -1(0) + 4(3) - 3(4) = 0$ . The condition $\cos\beta = 4/\sqrt{26} > 0$ holds. Thus both $\cos\alpha < 0$ and $\cos\gamma < 0$ , placing $\alpha, \gamma \in (\pi/2, \pi)$ .

Hints that build this answer step by step

Let $A=(1,2,3)$ , $B=(2,3,4)$ , and $C=(1,5,7)$ . What are the vectors $\vec{AB}$ and $\vec{AC}$ lying in the plane?
$\vec{AB} = \langle 1, 1, 1 \rangle$ and $\vec{AC} = \langle 0, 3, 4 \rangle$
What is the cross product $\vec{n} = \vec{AB} \times \vec{AC}$ ?
$\vec{n} = \langle 1, -4, 3 \rangle$
Given that $\vec{OP}$ is a unit normal vector along $\vec{n}$ with $\beta \in \left(0, \frac{\pi}{2}\right)$ , what are the signs of $\cos\alpha$ and $\cos\gamma$ ?
$\cos\beta > 0$ requires choosing $-\vec{n} = \langle -1, 4, -3 \rangle$ , so $\cos\alpha < 0$ and $\cos\gamma < 0$ .

✓ Source and academic review

Question type: Single correct
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 can't

\alpha

or

\gamma

be in

(\pi, 2\pi)

?

Direction angles of a vector with coordinate axes are defined strictly in the range $[0, \pi]$ . A negative direction cosine directly implies the angle is in $(\pi/2, \pi)$ .

Answer

Because the normal vector to the plane with a positive y-component has negative x- and z-components, both $\alpha$ and $\gamma$ lie in the interval $\left(\frac{\pi}{2}, \pi\right)$ .

Why each option works or fails

A: $\alpha \in \left(0, \frac{\pi}{2}\right)$ and $\gamma \in \left(0, \frac{\pi}{2}\right)$ - The student calculates the normal vector via cross product but flips or ignores the signs of the components, assuming all direction angles are acute. Check the sign of each direction cosine: $\cos\theta > 0 \implies \theta \in \left(0, \frac{\pi}{2}\right)$ and $\cos\theta < 0 \implies \theta \in \left(\frac{\pi}{2}, \pi\right)$ .
B: $\alpha \in \left(0, \frac{\pi}{2}\right)$ and $\gamma \in \left(\frac{\pi}{2}, \pi\right)$ - The student makes a sign error in the cross-product determinant expansion specifically for the x-component, obtaining $\cos\alpha > 0$ . Carefully compute $\vec{u} \times \vec{v} = (u_y v_z - u_z v_y)\hat{i} - (u_x v_z - u_z v_x)\hat{j} + (u_x v_y - u_y v_x)\hat{k}$ and adjust by multiplying by $-1$ so the $\hat{j}$ coefficient is positive.
C · correct: $\alpha \in \left(\frac{\pi}{2}, \pi\right)$ and $\gamma \in \left(\frac{\pi}{2}, \pi\right)$ - None. The student correctly finds that the normal vector with $\cos\beta > 0$ has $\cos\alpha < 0$ and $\cos\gamma < 0$ , placing both angles in $\left(\frac{\pi}{2}, \pi\right)$ . This option is correct.
D: $\alpha \in \left(\frac{\pi}{2}, \pi\right)$ and $\gamma \in \left(0, \frac{\pi}{2}\right)$ - The student makes a sign error in the cross-product expansion for the z-component, arriving at $\cos\gamma > 0$ while having $\cos\alpha < 0$ . Re-evaluate the determinant expansion for the $\hat{k}$ component: $(1)(3) - (1)(0) = 3$ , which becomes $-3$ after ensuring the $\hat{j}$ component is positive.

Step-by-step solution

given: Let points on the plane be $A(1, 2, 3)$ , $\cos\theta > 0 \implies \theta \in \left(0, \frac{\pi}{2}\right)$ , and $C(1, 5, 7)$ . The unit vector $\vec{OP}$ is perpendicular to this plane, with direction angles $\alpha, \beta, \gamma$ such that $\beta \in \left(0, \frac{\pi}{2}\right)$ .
goal: Determine the intervals containing the angles $\alpha$ and $\gamma$ .
approach: Find two vectors lying in the plane, $\vec{AB}$ and $\vec{AC}$ . Compute their cross product to get a normal vector $\vec{n}$ . Since $\vec{OP}$ is normal to the plane, $\vec{OP} = \pm \frac{\vec{n}}{|\vec{n}|}$ . Use the condition $\beta \in (0, \pi/2) \implies \cos\beta > 0$ to fix the unique sign for $\vec{OP}$ , which gives the signs of $\cos\alpha$ and $\cos\gamma$ .
execute: Form in-plane vectors: $\vec{AB} = (2-1)\hat{i} + (3-2)\hat{j} + (4-3)\hat{k} = \hat{i} + \hat{j} + \hat{k}$ $\vec{AC} = (1-1)\hat{i} + (5-2)\hat{j} + (7-3)\hat{k} = 0\hat{i} + 3\hat{j} + 4\hat{k}$
execute: Find the cross product $\vec{n} = \vec{AB} \times \vec{AC}$ : $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 3 & 4 \end{vmatrix} = \hat{i}(4 - 3) - \hat{j}(4 - 0) + \hat{k}(3 - 0) = \hat{i} - 4\hat{j} + 3\hat{k}$
execute: The unit normal vector $\vec{OP}$ is proportional to $\vec{n}$ : $\vec{OP} = \lambda (\hat{i} - 4\hat{j} + 3\hat{k})$ where $\lambda = \pm \frac{1}{\sqrt{1^2 + (-4)^2 + 3^2}} = \pm \frac{1}{\sqrt{26}}$ . The direction cosines are $\cos\alpha = \lambda$ , $\cos\beta = -4\lambda$ , $\cos\gamma = 3\lambda$ . Since $\beta \in \left(0, \frac{\pi}{2}\right)$ , we have $\cos\beta > 0 \implies -4\lambda > 0 \implies \lambda < 0$ . Therefore, $\lambda = -\frac{1}{\sqrt{26}}$ . This gives: $\cos\alpha = -\frac{1}{\sqrt{26}} < 0 \implies \alpha \in \left(\frac{\pi}{2}, \pi\right)$ $\cos\gamma = -\frac{3}{\sqrt{26}} < 0 \implies \gamma \in \left(\frac{\pi}{2}, \pi\right)$ .
verify: Check scalar products: $\vec{OP} \cdot \vec{AB} \propto -1(1) + 4(1) - 3(1) = 0$ and $\vec{OP} \cdot \vec{AC} \propto -1(0) + 4(3) - 3(4) = 0$ . The condition $\cos\beta = 4/\sqrt{26} > 0$ holds. Thus both $\cos\alpha < 0$ and $\cos\gamma < 0$ , placing $\alpha, \gamma \in (\pi/2, \pi)$ .