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

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