Question

The equation of sphere with centre $(1, - 2, 3)$ and touching the plane $6x-3y+2z-4 = 0$ is

• A. $\left|\vec{r}\left(\hat{i}-2 \hat{j}+3 \hat{k}\right)\right|=2$
• B. $\left|\vec{r}+\left(\hat{i}-2 \hat{j}+3 \hat{k}\right)\right|=2$
• C. $\left|\vec{r}-\left(\hat{i}-2 \hat{j}+3 \hat{k}\right)\right|=\sqrt{10}$
• D. $\left|\vec{r}-\left(\hat{i}-2 \hat{j}+3 \hat{k}\right)\right|=\sqrt{14}$

Answer 1

Answer: (A)

Centre $( 1 , - 2 , 3)$
Radius of sphere = length of perpendicular drawn from $(1, -2, 3)$ to the plane
$=\frac{\left|6\left(1\right)-3\left(-2\right)+2\left(3\right)-4\right|}{\sqrt{6^{2}+\left(-3\right)^{2}+2^{2}}}=2$
$\therefore $ Required equation of sphere is
$\left|\vec{r}-\left(\hat{i}-2 \hat{j}+3 \hat{k}\right)\right|$ = radius of sphere = 2