Question

The points $(1,1,k)$ and $(-3,0,1)$ are equidistant from the plane $3x+4y-12z+13=0$, if $k=$

• A. $0$
• B. $1$
• C. $2$
• D. $\frac{7}{3}$

Answer 1

Answer: (B), (D)

$1$
$\frac{7}{3}$

Distance $\left(d\right)$ of a point $(x_1,y_1,z_1)$ from perpendicular plane $ax+by+cz+d=0$-

$d=\frac{|a{x}_1+b{y}_1+c{z}_1+d|}{\sqrt{a^2+b^2+c^2}}$

Therefore, distance $\left(d_1\right)$ of the point $(1,1,k)$ from the plane $3x+4y-12z+13=0$ is-

$d_1=\frac{|3\times 1+4\times 1-12\times k+13|}{\sqrt{3^2+4^2+{(-12)}^2}}$

$d_1=\frac{|20-12k|}{\sqrt{169}}$

$\Rightarrow d_1=\frac{|20-12k|}{13}$

Similarly, distance $\left(d_1\right)$ of the point $(-3,0,1)$ from the plane $3x+4y-12z+13=0$ is-

$d_1=\frac{|3\times (-3)+4\times 0-12\times 1+13|}{\sqrt{3^2+4^2+{(-12)}^2}}$

$d_1=\frac{|-8|}{\sqrt{169}}$

$\Rightarrow d_1=\frac{8}{13}$

Given that the points are equidistant from the plane, i.e.,

$d_1=d_2$

$\frac{|20-12k|}{13}=\frac{8}{13}$

$|20-12k|=8$

Case I:-

$20-12k=8$

$1k=12$

$\Rightarrow k=1$

Case II:-

$20-12k=-8$

$12k=28$

$\Rightarrow k=\frac{28}{12}=\frac{7}{3}$

Hence the value of $k$ will be $1,\frac{7}{3}$.