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Q.

A vector $n$ is inclined to x-axis at $45^{0}$ , to y-axis at $60^{0}$ and at an acute angle to z-axis. If $n$ is a normal to plane passing through the point $(\sqrt{2}, - 1, 1)$ , then the equation of plane is :

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a

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

b

$3 \sqrt{2} x - 4 y - 3 z = 7$

c

$4 \sqrt{2} x + 7 y + z = 2$

d

$\sqrt{2} x - y - z = 2$

answer is B.

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Detailed Solution

$α = 45^{0} β = 60^{0} \cos^{2} α + \cos^{2} β + \cos^{2} r = 1 \Rightarrow \cos^{2} r = 1 - \frac{1}{2} - \frac{1}{4} = \frac{4 - 2 - 1}{4} = \frac{1}{4} r = 60^{0} n = (\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}) = (\sqrt{2}, 1, 1) eq of plane is \sqrt{2} (x - \sqrt{2}) + 1 (y + 1) + 1 (2 - 1) = 0$

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