First slide

Planes in 3D

Question

If $θ$ denotes the acute angle between the line $r = (i + 2 j - k) + A (i - j + k)$ and the plane $r . (2 i - j + k) = 4$ , then sin $θ + \sqrt{2} \cos θ =$

Moderate

A

$1 / \sqrt{2}$
B

1
C

$\sqrt{2}$
D

$1 + \sqrt{2}$

Solution

Direction ratios of the line are 1, - 1, 1 and that of the normal to the plane are 2, - 1, 1,
so $\sin θ = \frac{2 + 1 + 1}{\sqrt{3} \sqrt{6}} = \frac{4}{\sqrt{18}}, \cos θ = \frac{\sqrt{2}}{\sqrt{18}}$ .
$\sin θ + \sqrt{2} \cos θ = \frac{4}{\sqrt{18}} + \frac{2}{\sqrt{18}} = \frac{6}{\sqrt{18}} = \sqrt{2}$

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