Question

Consider the plane $(x,y,z)=(0,1,1)+\lambda (1,-1,1)+\mu (2,-1,0)$. The distance of this plane from the origin is:

• A. $\frac{1}{3}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\sqrt{\frac{3}{2}}$
• D. $\frac{2}{\sqrt{3}}$

Answer 1

Answer: (C)

$\sqrt{\frac{3}{2}}$

Given $(x,y,z)=(0,1,1)+\lambda (1,-1,1)+\mu (2,-1,0)$
$\Rightarrow xi+yj+zk=j+k+\lambda (i-j+k)+\mu (2i-j)$
Comparing coefficients of $i,j$ and $k$,
$\lambda +2\mu =x..\left(1\right)$
$-\lambda -\mu =y-1..\left(2\right)$
$-2\mu =y+z-2..\left(3\right)$
Eliminating $\lambda$ and $\mu$ from (1), (2) and (3), we get
$x+2y+z+3=0$
Hence, distance of this plane from origin is
$=\left|\frac{3}{\sqrt{1^2+2^2+1^2}}\right|=\sqrt{\frac{3}{2}}$