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
$=\frac{|6\left(1\right)-3\left(-2\right)+2\left(3\right)-4|}{\sqrt{6^2+{\left(-3\right)}^2+2^2}}=2$
$\therefore$ Required equation of sphere is
$\left|\vec{r}-\left(\hat{i}-2\hat{j}+3\hat{k}\right)\right|$ = radius of sphere = 2