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.

 

 

 

 
 
 
 
 

Answers (1)

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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