The problem specifies that is in the first octant and makes an acute angle with the z-axis, which means .

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Three Dimensional Geometry: Mathematics | JEE Main

A vector

\vec{\text{v}}

in the first octant is inclined to the x-axis at

60^\circ

, to the y-axis at

45

and to the z-axis at an acute angle. If a plane passing through the points

(\sqrt{2},-1,1)

and

(\text{a},\text{b},\text{c})

, is normal to

\vec{\text{v}}

, then

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

A

\sqrt{2}\text{a} + \text{b} + \text{c} = 1

B

\text{a} + \sqrt{2}\text{b} + \text{c} = 1

C

\text{a} + \text{b} + \sqrt{2}\text{c} = 1

D

\sqrt{2}\text{a} - \text{b} + \text{c} = 1

Step-by-step solutionView

Correct answer

The normal vector has direction ratios proportional to $1 : \sqrt{2} : 1$ , leading to the plane equation $a + \sqrt{2}b + c = 1$ .

Option analysis

Why each option works or fails

A ·

\sqrt{2}\text{a} + \text{b} + \text{c} = 1

Swapping the inclination angles between the x-axis and y-axis, assigning $45^\circ$ to the x-axis instead of the y-axis. Ensure direction angles correspond correctly: $\alpha = 60^\circ$ gives the x-component $\cos 60^\circ = \frac{1}{2}$ , while $\beta = 45^\circ$ gives the y-component $\cos 45^\circ = \frac{1}{\sqrt{2}}$ .

B ·

\text{a} + \sqrt{2}\text{b} + \text{c} = 1

Correct option. Using $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$ , we find $\cos\gamma = \frac{1}{2}$ , giving normal vector ratios $(1, \sqrt{2}, 1)$ . Substituting $(\sqrt{2}, -1, 1)$ into $x + \sqrt{2}y + z = d$ gives $d = 1$ , so $(a, b, c)$ satisfies $a + \sqrt{2}b + c = 1$ .

C ·

\text{a} + \text{b} + \sqrt{2}\text{c} = 1

Swapping the inclination angles between the y-axis and z-axis, mistakenly setting $\gamma = 45^\circ$ instead of $\beta = 45^\circ$ . Read the given angles carefully: the angle with the y-axis is $45^\circ$ , so the coefficient of $y$ takes the $\sqrt{2}$ factor.

D ·

\sqrt{2}\text{a} - \text{b} + \text{c} = 1

Making a sign error when computing the dot product with the normal vector using the point $(\sqrt{2}, -1, 1)$ . The vector components are all positive because $\vec{\text{v}}$ lies in the first octant; the minus sign comes from the y-coordinate of the point, not the normal vector coefficients.

Reviewed route

Solution

StepWorking

01given

Vector $\vec{v}$ is inclined to the x-axis at $\alpha = 60^\circ$ , to the y-axis at $\beta = 45^\circ$ , and to the z-axis at an acute angle $\gamma$ . The plane is normal to $\vec{v}$ and passes through $(\sqrt{2}, -1, 1)$ and $(a, b, c)$ .

02approach

Use the direction cosine identity $l^2 + m^2 + n^2 = 1$ to determine the direction ratios of the normal vector $\vec{v}$ . Then write the plane equation $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$ and substitute $(a, b, c)$ .

03execute

Compute direction cosines: $l = \cos 60^\circ = \frac{1}{2}$ , $m = \cos 45^\circ = \frac{1}{\sqrt{2}}$ . Since $l^2 + m^2 + n^2 = 1$ , we have $\frac{1}{4} + \frac{1}{2} + n^2 = 1 \implies n^2 = \frac{1}{4}$ . Since $\gamma$ is acute, $n = \frac{1}{2}$ .

04execute

The normal vector can be taken as proportional to $(l, m, n) = \left(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}\right)$ , or multiplying by 2: $(1, \sqrt{2}, 1)$ . The plane passing through $(\sqrt{2}, -1, 1)$ has equation: $1\cdot(x - \sqrt{2}) + \sqrt{2}(y + 1) + 1\cdot(z - 1) = 0 \implies x + \sqrt{2}y + z = \sqrt{2} - \sqrt{2} + 1 = 1$ .

05execute

Since $(a, b, c)$ lies on the plane, its coordinates must satisfy the plane equation: $a + \sqrt{2}b + c = 1$ .

✓verify

Check that the normal vector coefficients match: the coefficient of $y$ must be $\sqrt{2}$ times that of $x$ and $z$ because $\frac{m}{l} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$ . Only Option (1) has coefficients in ratio $1 : \sqrt{2} : 1$ .

Hints that build this answer step by step

If the direction angles of $\vec{\text{v}}$ are $\alpha = 60^\circ$ , $\beta = 45^\circ$ , and an acute angle $\gamma$ , what is $\cos\gamma$ ?
$\cos\gamma = \frac{1}{2}$
What are the direction ratios of the normal vector $\vec{\text{v}}$ in simplest integer/radical form?
$\langle 1, \sqrt{2}, 1 \rangle$
Using the normal vector $\vec{\text{n}} = \langle 1, \sqrt{2}, 1 \rangle$ and the point $(\sqrt{2}, -1, 1)$ , what is the Cartesian equation of the plane?
$x + \sqrt{2}y + z = 1$

✓ 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 do we discard

n = -1/2

?

The problem specifies that $\vec{v}$ is in the first octant and makes an acute angle with the z-axis, which means $\cos\gamma > 0$ .

Answer

The normal vector has direction ratios proportional to $1 : \sqrt{2} : 1$ , leading to the plane equation $a + \sqrt{2}b + c = 1$ .

Why each option works or fails

A: $\sqrt{2}\text{a} + \text{b} + \text{c} = 1$ - Swapping the inclination angles between the x-axis and y-axis, assigning $45^\circ$ to the x-axis instead of the y-axis. Ensure direction angles correspond correctly: $\alpha = 60^\circ$ gives the x-component $\cos 60^\circ = \frac{1}{2}$ , while $\beta = 45^\circ$ gives the y-component $\cos 45^\circ = \frac{1}{\sqrt{2}}$ .
B · correct: $\text{a} + \sqrt{2}\text{b} + \text{c} = 1$ - Correct option. Using $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$ , we find $\cos\gamma = \frac{1}{2}$ , giving normal vector ratios $(1, \sqrt{2}, 1)$ . Substituting $(\sqrt{2}, -1, 1)$ into $x + \sqrt{2}y + z = d$ gives $d = 1$ , so $(a, b, c)$ satisfies $a + \sqrt{2}b + c = 1$ .
C: $\text{a} + \text{b} + \sqrt{2}\text{c} = 1$ - Swapping the inclination angles between the y-axis and z-axis, mistakenly setting $\gamma = 45^\circ$ instead of $\beta = 45^\circ$ . Read the given angles carefully: the angle with the y-axis is $45^\circ$ , so the coefficient of $y$ takes the $\sqrt{2}$ factor.
D: $\sqrt{2}\text{a} - \text{b} + \text{c} = 1$ - Making a sign error when computing the dot product with the normal vector using the point $(\sqrt{2}, -1, 1)$ . The vector components are all positive because $\vec{\text{v}}$ lies in the first octant; the minus sign comes from the y-coordinate of the point, not the normal vector coefficients.

Step-by-step solution

given: Vector $\vec{v}$ is inclined to the x-axis at $\alpha = 60^\circ$ , to the y-axis at $\beta = 45^\circ$ , and to the z-axis at an acute angle $\gamma$ . The plane is normal to $\vec{v}$ and passes through $(\sqrt{2}, -1, 1)$ and $(a, b, c)$ .
approach: Use the direction cosine identity $l^2 + m^2 + n^2 = 1$ to determine the direction ratios of the normal vector $\vec{v}$ . Then write the plane equation $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$ and substitute $(a, b, c)$ .
execute: Compute direction cosines: $l = \cos 60^\circ = \frac{1}{2}$ , $m = \cos 45^\circ = \frac{1}{\sqrt{2}}$ . Since $l^2 + m^2 + n^2 = 1$ , we have $\frac{1}{4} + \frac{1}{2} + n^2 = 1 \implies n^2 = \frac{1}{4}$ . Since $\gamma$ is acute, $\vec{\text{v}}$ .
execute: The normal vector can be taken as proportional to $(l, m, n) = \left(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2}\right)$ , or multiplying by 2: $(1, \sqrt{2}, 1)$ . The plane passing through $(\sqrt{2}, -1, 1)$ has equation: $1\cdot(x - \sqrt{2}) + \sqrt{2}(y + 1) + 1\cdot(z - 1) = 0 \implies x + \sqrt{2}y + z = \sqrt{2} - \sqrt{2} + 1 = 1$ .
execute: Since $(a, b, c)$ lies on the plane, its coordinates must satisfy the plane equation: $a + \sqrt{2}b + c = 1$ .
verify: Check that the normal vector coefficients match: the coefficient of $y$ must be $\sqrt{2}$ times that of $x$ and $z$ because $\frac{m}{l} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$ . Only Option (1) has coefficients in ratio $1 : \sqrt{2} : 1$ .

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The normal vector has direction ratios proportional to $1 : \sqrt{2} : 1$ , leading to the plane equation $a + \sqrt{2}b + c = 1$ .

Why each option works or fails

Solution

01given

02approach

03execute

04execute

05execute

✓verify

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