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Question
Find the distance of the point (2, 12, 5) from the point of intersection of the line
`vecr=2hati-4hat+2hatk+lambda(3hati+4hatj+2hatk) `
Solution
Let the point of intersection of the line and the plane be(x1,y1,z1)
Equation of line is `(x−2)/3=(y+4)/4=(z−2)/2` Point (x1,y1,z1) will satisfy the equation of the line.
`(x_1−2)/3=(y_1+4)/4=(z_1−2)/k`
`⇒x_1=3k+2; y_1=4k−4; z_1=2k+2 ..........(1)`
Equation of plane is x-2y+z=0.
Point (x1,y1,z1) will also satisfy the equation of plane.
`⇒x_1−2y_1+z_1=0 `
Substituting equation (1) in equation (2), we get:
3k+2−2(4k−4)+2k+2=0
⇒k=4
Hence,
`x_1=3(4)+2;`
` y_1=4(4)−4;`
` z_1=2(4)+2`
` x_1=14;`
`y_1=12; `
`z_1=10 `
Thus, the point of intersection is (14,12,10).
Distance between the point (14,12,10) and (2,12,5)`=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)`
`=sqrt((14-2)^2+(12-12)^2+(10-5)^2)`
`=sqrt(144+0+25)`
`=sqrt169`
`=13 units`
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