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# The sine of the angle between the straight line $\large \frac{x\;-2}{3}=\frac{y\;-3}{4}=\frac{z\;-4}{5}\;$ and the plane $2x-2y+z=5$ is

$(A)\;\frac{10}{6\sqrt 5}\qquad(B)\;\frac{4}{5\sqrt 2}\qquad(C)\;\frac{2\sqrt 3}{5}\qquad(D)\;\frac{\sqrt 2}{10}$

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• $sin\theta = \large\frac{\overrightarrow n.\overrightarrow b}{|\overrightarrow n||\overrightarrow b|}$
• $\overrightarrow n = (2, -2, 1) \: \: \overrightarrow b = (3, 4, 5)$
• Where $\overrightarrow n$ is the normal vector to the plane.
The given line is $\large\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$
The direction ratios of $\overrightarrow n$ are $(2,-2,1)$
and direction ratios of $\overrightarrow b$ are $(3,4,5)$
Now the angle between a line and the plane is
$sin\theta = \large\frac{\overrightarrow n.\overrightarrow b}{|\overrightarrow n||\overrightarrow b|}$
$|\overrightarrow b|=\sqrt {3^2+4^2+5^2}$
$\qquad=\sqrt {50}=5 \sqrt 2$
$|\overrightarrow a|=\sqrt {2^2+(-2)^2+1^2}$
$\qquad=\sqrt {9}=3$
Substituting the respective values we get,
$\sin \theta=\large\frac{(2 \hat i-2 \hat j+\hat k).(3 \hat i+4 \hat j+5 \hat k)}{5 \sqrt 2 \times 3}$
$\qquad= \large\frac{6-8+5}{15 \sqrt {2}}$
$\qquad= \large\frac{3}{15 \sqrt {2}}$
$\qquad= \large\frac{1}{5 \sqrt {2}}$
Multiplying the denominator $\large\frac{1}{\sqrt {10}}$ we get,
$\large\frac{\sqrt 2}{5 \times 2}=\frac{\sqrt 2}{10}$
Hence the correct option is $D$
answered Jun 12, 2013 by