The angle between the line $6 x = 4 y = 3 z$ and the plane $3 x + 2 y - 3 z = 4$ is

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$π / 6$

$π / 3$

$π / 2$

answer is A.

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Detailed Solution

The equation of the given line is

$6 x = 4 y = 3 z$

which is written in symmetric form as

$\frac{x - 0}{1 / 6} = \frac{y - 0}{1 / 4} = \frac{z - 0}{1 / 3}$

Direction ratios of this line are $(\frac{1}{6}, \frac{1}{4}, \frac{1}{3})$

and equation of the plane is

$3 x + 2 y - 3 z - 4 = 0$

If $θ$ be the angle between line and plane then direction ratios of the normal to this plane is (3, 2, -3)

$\sin θ = |\frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}|$

$= |\frac{\frac{1}{6} \times 3 + \frac{1}{4} \times 2 + \frac{1}{3} (- 3)}{\sqrt{\frac{1}{36} + \frac{1}{16} + \frac{1}{9}} \sqrt{9 + 4 + 9}}| = |\frac{1 - 1}{\sqrt{\frac{1}{36} + \frac{1}{16} + \frac{1}{9} \sqrt{22}}}| = 0 \Rightarrow θ = 0^{\circ}$