Question

If the points (1, 1, p) and (-3, 0, 1) be equidistant from the plane $r.(3\hat{i}+4\hat{j}-12\hat{k})+13=0$, then find value of p.

Answer 1

The distance of point (1, 1, p) from the plane $r.(3\hat{i}+4\hat{j}-12\hat{k})+13=0$ or in Cartesian from 3x + 4y - 12z + 13 = 0, is
$d_1=\left|\frac{3\times 1+4\times 1-12\times p+13}{\sqrt{3^2+4^2+(-12{)}^2}}\right|=\left|\frac{3+4-12p+13}{\sqrt{169}}\right|=\left|\frac{20-12p}{13}\right|\cdots \left(i\right)$
The distance of the point (-3, 0, 1) from the plane 3x + 4y - 12z + 13 = 0 is
$d_2=\left|\frac{3\times (-3)+4\times 0-12\times 1+13}{\sqrt{3^2+4^2+(-12{)}^2}}\right|=\left|\frac{-9+0-12+13}{\sqrt{169}}\right|=\left|\frac{-8}{13}\right|=\frac{8}{13}$
According to the given condition,
$d_1=d_2\Rightarrow \left|\frac{20-12p}{13}\right|=\frac{8}{13}\Rightarrow \frac{20-12p}{13}=\pm \frac{8}{13}$
Taking +ve sign, we get
$\frac{20-12p}{13}=\frac{8}{13}\Rightarrow 20-12p=8\Rightarrow 12p=12\Rightarrow p=1$
Taking -ve sign, we get $\frac{20-12p}{13}=\frac{-8}{13}\Rightarrow 20-12p=-8$
$\Rightarrow 12p=28\Rightarrow p=\frac{28}{12}=\frac{7}{3}$