Question

If the length of the projection of the line segment with points $(1,0,-1)$ and $(-1,2,2)$ to the plane $x+3y-5z=6$ is $d$, then the value of $\left[d/2\right],$ is
where [.] represent greatest interger function

Answer 1

Let $A(1,0,-1),B(-1,2,2)$
Direction ratios of AB are $(2,-2,-3)$
Let $\theta$ be the angle between the line and normal to plane; then
$\cos \theta =\frac{|2.1+3(-2)-5(-3)|}{\sqrt{1+9+25}\sqrt{4+4+9}}$
$=\frac{11}{\sqrt{17}\sqrt{35}}=\frac{11}{\sqrt{595}}$

Length of projection,
$d=\left(AB\right)\sin \theta$
$=\sqrt{(2{)}^2+(2{)}^2+(3{)}^2}\times \sqrt{1-\frac{121}{595}}$
$=\sqrt{\frac{474}{35}}$

$\therefore \left[d/2\right]=1$