Question

The projection of the line segment joining the points $(1,2,3)$ and $(4,5,6)$ on the plane $2x+y+z=1$ is

• A. $1$
• B. $\sqrt{3}$
• C. $5$
• D. $6$

Answer 1

The correct option is B $\sqrt{3}$

Points $A(1,2,3)$ and $B(4,5,6)$ , Plane :$2x+y+z=1$

length of projection is distance between foot of perpendicular of $A$ & $B$ on Plane.

Directions of normal to plane $\Rightarrow 2,1,1$

let $(2r+1,r+2,r+3)$ is foot of $\perp$ from $A(1,2,3)$.

$(2r+1)2+(r+2)+(r+3)=1\Rightarrow r=-1$

foot of $\perp$ from $A=(-1,1,2)$

Similarly,If $(2k+4,k+5,k+6)$ is foot of $\perp$ from $B(4,5,6)$

$\therefore (2k+1)2+(k+5)+(k+6)=1\Rightarrow k=-3$

foot of $\perp$ from $B=(-2,2,3)$

$\therefore$ distance between foot of $\perp$ from $A$ & foot of $\perp$ from $B$.

$\Rightarrow \sqrt{{(-2-(-1\left)\right)}^2+{(2-1)}^2+{(3-2)}^2}=\sqrt{3}$