Question

The sine of the angle between the straight line $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ and the plane $2x-2y+z=5$ is
(a) $\frac{10}{6\sqrt{5}}$

(b) $\frac{4}{5\sqrt{2}}$

(c) $\frac{2\sqrt{3}}{5}$

(d) $\frac{\sqrt{2}}{10}$

Answer 1

For given line, $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$

Direction ratios are given by (3, 4, 5) = (a, b, c) say and for plane, 2x − 2y + z = 5,
Normal vector is given by (2, −2, 1) = (A, B, C)
∴ Since of angle between the straight line and plane is
$\begin{array}{rcl}\sin \theta & =& \frac{\left|aA+bB+cC\right|}{\sqrt{a^2+b^2+c^2}\sqrt{A^2+B^2+C^2}}\\ & =& \frac{\left|3\times 2+4\times \left(-2\right)+5\times 1\right|}{\sqrt{3^2+4^2+5^2}\sqrt{2^2+2^2+1}}=\frac{\left|6-8+5\right|}{\sqrt{50}\sqrt{9}}=\frac{3}{15\sqrt{2}}\\ \mathrm{i}.\mathrm{e}.\sin \theta & =& \frac{3}{15\times \sqrt{2}}\\ & =& \frac{1}{5\sqrt{2}}\\ & =& \frac{1\times \sqrt{2}}{5\sqrt{2}\times \sqrt{2}}\\ & =& \frac{\sqrt{2}}{5\times 2}\\ \mathrm{i}.\mathrm{e}.\sin \theta & =& \frac{\sqrt{2}}{10}\end{array}$

Hence, the correct answer is option D.