JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

A vector $\vec{v}$ in the first octant is inclined to the $x$ -axis at $60^\circ$ , to the $y$ -axis at $45^\circ$ and to the $z$ -axis at an acute angle. If a plane passing through the points $(\sqrt{2}, -1, 1)$ and $(a, b, c)$ , is normal to $\vec{v}$ , then:

Choose the correct answer:

Explanation

Calculation

1. Direction Cosines ( $l, m, n$ ):

\cos^2 60^\circ + \cos^2 45^\circ + \cos^2 \gamma = 1

\left(\frac{1}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \cos^2 \gamma = 1

\frac{1}{4} + \frac{1}{2} + \cos^2 \gamma = 1

\cos^2 \gamma = 1 - \frac{3}{4} = \frac{1}{4} \implies \cos \gamma = \frac{1}{2}

2. Vector $\vec{v}$ (Normal to the plane):

\vec{v} = \frac{1}{2}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{2}\hat{k}

Direction Ratios ( $a, b, c$ ) of normal: $(1, \sqrt{2}, 1)$

3. Equation of Plane:

1(x - \sqrt{2}) + \sqrt{2}(y + 1) + 1(z - 1) = 0

x - \sqrt{2} + \sqrt{2}y + \sqrt{2} + z - 1 = 0

x + \sqrt{2}y + z = 1

4. Putting point $(a, b, c)$ in the equation:

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

Sahi Option: (2)

Choose the correct answer:

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