Question

Find the distance of the point (-1,-5,-10) from the point of intersection of the line $\hat r= 2\hat i\hat -j+2\hat k+λ(3\hat i+4\hat j+2\hat k)$ and the plane $\vec r.(\hat i-\hat j+\hat k)=5$.

Answer 1

The equation of the given line is

$\hat r= 2\hat i\hat -j+2\hat k+λ(3\hat i+4\hat j+2\hat k)$ ...(1)

The equation of the given plane is

$\vec r.(\hat i-\hat j+\hat k)=5$ ...(2)

Substituting the value of $\vec r$ from equation (1) in equation (2), we obtain

$[2\hat i\hat -j+2\hat k+λ(3\hat i+4\hat j+2\hat k)]$.$(\hat i-\hat j+\hat k)=5$

$⇒[(3λ+2)\hat i+(4λ-1)\hat j+(2λ+2)\hat k].(\hat i-\hat j+\hat k)=5$

$⇒(3λ+2)-(4λ-1)+(2λ+2)=5$

$⇒λ=0$

Substituting this value in equation (1), we obtain the equation of the line as

$\vec r=(2\hat i-\hat j+2\hat k$

This means that the position vector of the point of intersection of the line and the plane is

$\vec r=(2\hat i-\hat j+2\hat k$

This shows that the point of intersection of the given line and plane is given by the coordinates (2, -1, 2).

The point is (-1,5,-10).

The distance d between the points,(2,-1,2)and(-1,5,-10),is

$d =\sqrt {(-1-2)^2+(-5+1)^2+(-10-2)^2}$
$d =\sqrt {9+16+144}$
$d =\sqrt {169}$
$d =13$.