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Question
Find the coordinates of the point, where the line `(x-2)/3=(y+1)/4=(z-2)/2` intersects the plane x − y + z − 5 = 0. Also find the angle between the line and the plane.
Solution
The equation of the given line is `(x-2)/3=(y+1)/4=(z-2)/2 ...(1)`
Any point on the given line is `(3lambda+2,4lambda-1,2lambda+2)`
If this point lies on the given plane x − y + z − 5 = 0, then `3lambda+2-(4lambda-1)+2lambda+2-5=0`
`lambda=0`
Putting `lambda =0` in `(3lambda+2,4lambda-1,2lambda+2)`,we get `(3lambda+2,4lambda-1,2lambda+2)=(2,-1,2)`
So, the point of intersection of the given line and the plane is (2,-1,2)
Let θ be the angle between the given line and the plane.
`therefore sin theta=(veca.vecb)/(|veca|.|vecb|)=((3hati+4hatj+2hatk)(hati-hatj+hatk))/(sqrt(3^2+4^2+2^2)sqrt(1^2_(-1)^2+1^2))`
`=(3xx14xx(-1)+2xx1)/(sqrt(29)sqrt(3))=1/sqrt87`
`theta=sin^(-1)(1/sqrt87)`
Thus, the angle between the given line and the plane is `sin ^(-1)(1/sqrt87).`
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