# Find the coordinates of the point where the line $\frac{x-8}{4}= \frac{y-1}{1}= \frac{z-3}{8}$ intersects the plane $2x+2y+z= 3.$ Also find the angle between the line and the plane.

given equation of line is
$\frac{x-8}{4}= \frac{y-1}{1}= \frac{z-3}{8}= \lambda$
$x= 4\lambda +8\: \: y= \lambda +1 , \: z= 8\lambda +3$
Then $\left [ 4\lambda +8,\left ( \lambda +1 \right ),\left ( 8\lambda +3 \right ) \right ]$ be any point on the given line
The points lie on the plane $2x+2y+z= 3$
$2\left ( 4\lambda +8 \right )+2\left ( \lambda +1 \right )+\left ( 8\lambda +3 \right )= 3$
$\therefore 8\lambda +16+2\lambda +2+8\lambda +3= 3$
$\therefore 18\lambda +18=0$
$\lambda = -1$
$\therefore$ point of intersection of the lines and the plane
$= \left [ 4\left ( -1 \right )+8,-1+1,8\left ( -1 \right ) +3\right ]$
$=\left ( 4,0,-6 \right )$
Let $\theta$ be the angle between line and plane
Then
$\sin \theta = \frac{al+bm+cn}{\sqrt{a^{2}+b^{2}+c^{2}}\sqrt{l^{2}+m^{2}+n^{2}}}$
Here a = 4, b = 1,c = 8, l = 2, m = 2,n=1
$\therefore \sin \theta = \frac{4\left ( 2 \right )+1\left ( 2 \right )+8\left ( 1 \right )}{\sqrt{4^{2}+1^{2}+8^{2}}\sqrt{\left ( 2 \right )^{2}+\left ( 2 \right )^{2}+\left ( 1 \right )^{2}}}$
$\therefore \sin \theta =\frac{8+2+8}{\sqrt{81}\sqrt{9}}$
$= \frac{18}{9\times 3}= \frac{2}{3}$
$\theta = \sin^{-1}\left ( \frac{2}{3} \right )$  which is required angle

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