Question

The shortest distance of a point $(1,2,-3)$ from a plane making intercepts $1,2$ and $3$ units on position $x,y$ and $z$-axes respectively, is

• A. $2$
• B. $0$
• C. $\frac{13}{12}$
• D. $\frac{12}{7}$

Answer 1

The correct option is B $0$

$e{q}^n$ of plane making intercept $1,2,3$ on $x,y,z$ axis Respectively

$\Rightarrow \frac{x}{1}+\frac{y}{2}+\frac{x}{3}=1$

$\Rightarrow \frac{6x+3y+2z}{6}=1$

$\Rightarrow 6x+3y+2z-6=0$ the shortest distance from $(1,2,3)$ is $\left|\frac{6\left(1\right)+3\left(2\right)+2(-3)-6}{\sqrt{36+9+4}}\right|$

$=\left|\frac{6+6-6-6}{7}\right|=0$

here we can say the shortest distance of point $(x_1,y_1,z_1)$ from $ax+vy+cz+d=0$ is $\left|\frac{a{x}_1+b{y}_1+c{z}_1+d}{\sqrt{a^2+b^2+c^2}}\right|$