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 \(\vec r.(\hat i-\hat j+\hat k)=5\).
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 \(\vec r.(\hat i-\hat j+\hat k)=5\).
Solution and Explanation
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
\(d =\sqrt {(-1-2)^2+(-5+1)^2+(-10-2)^2}\)
\(d =\sqrt {9+16+144}\)
\(d =\sqrt {169}\)
\(d =13\).
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