The equation of the plane perpendicular to the line `(x-1)/1=(y-2)/(-1)=(z+1)/2` and passing through the point (2,3,1), is

Options

A. `r*(hat(i)+hat(j)+2hat(k))=1`
B. `r*(hat(i)-hat(j)+2hat(k))=1`
C. `r*(hat(i)-hat(j)+2hat(k))=7`
D. `r*(hat(i)+hat(j)-2hat(k))=10`

Answer

B

Explanation

The given line is parallel to the vector `n=hat(i)-hat(j)+2hat(k)`. The required plane passing through the point (2,3,1) i.e. `2hat(i)+3hat(j)+hat(k)` and is perpendicular to the vector `n=hat(i)-hat(j)+2hat(k)`
is `[r-(2hat(i)+3hat(j)+hat(k))]*(hat(i)-hat(j)+2hat(k))=0`
`rArr" "r*(hat(i)-hat(j)2hat(k))=(2hat(i)+3hat(j)+hat(k))*(hat(i)-hat(j)+2hat(k))`
`rArr" "r*(hat(i)-hat(j)+2hat(k))=(2hat(i)+3hat(j)+hat(k))*(hat(i)-hat(j)+2hat(k))`
`rArr" "r*(hat(i)-hat(j)+2hat(k))=1`