Find the distance of the point (−1, −5, −10) from the point of intersection of the line `vecr=2hati-hatj+2hatk+lambda(3hati+4hatj+2hatk) ` and the plane `vec r (hati-hatj+hatk)=5`
Solution
Let us suppose that the given line and plane intersect at the point P(x,y,z).
∴ The position vector of P is `vecr=xhati+yhatj+zhatk`
Thus, the given equations of the line and the plane can be rewritten as
`xhati+yhatj+zhatk=(2+3λ)hati−(1+4λ)hayj+(2+2λ)hatk and (xhati+yhatj+zhatk).(hati−hatj+hatk) = 5,` respectively.
On simplifying `xhati+yhatj+zhatk=(2+3λ)hati−(1+4λ)hatj+(2+2λ)hatk and (xhati+yhatj+zhatk).(hati−hatj+hatk) = 5` , we get:
x=(2+3λ)y =−(1+4λ)z = (2+2λ)
Also, x−y+z = 5
On putting the values of x, y and z in the equation x−y+z = 5, we get:
2+3λ+1+4λ+2+2λ = 5⇒9λ+5 = 5⇒λ = 0
∴ x=2, y =−1 and z =2
Hence, the distance between the points (−1, −5, −10) and (2, −1, 2) is
`sqrt((2+1)^2+(−1+5)^2+(2+10)^2)=sqrt(9+16+144)=sqrt169=13 units`
