Question

Let 'k' is distance of the plane through (1, 1, 1) and perpendicular to the line $\frac{x-1}{3}=\frac{y-1}{0}=\frac{z-1}{4}$ from the origin, then value of 5 k is __

Answer 1

equation of plane perpendicular to line $\frac{x-1}{3}=\frac{y-1}{0}=\frac{z-1}{4}$

is $3x+0\left(y\right)+4z=P$

Given that plane passes through $(1,1,1)$

$3\left(1\right)+4\left(1\right)=P$

$\Rightarrow P=7$

$\therefore$ equation of the plane $\Rightarrow 3x+4z=7$

perpendicular distance from origin to plane $=k$

$\Rightarrow K=\frac{|3\left(0\right)+4\left(0\right)-7|}{\sqrt{3^2+4^2}}=\frac{7}{\sqrt{25}}=\frac{7}{5}$

$\Rightarrow K=\frac{7}{5}$

$\therefore 5K=7$

$\therefore$ value of $5K$ is $7$