Question

If the distance of points $2\hat{i}+3\hat{j}+\lambda \hat{k}$ from the plane $\overline{r}\cdot (3\hat{i}+2\hat{j}+6\hat{k})=13$ is $5$ units then $\lambda =$

• A. $6,-\frac{17}{3}$
• B. $6,\frac{17}{3}$
• C. $-6,-\frac{17}{3}$
• D. $-6,\frac{17}{3}$

Answer 1

Answer: (A)

$6,-\frac{17}{3}$

Equation of plane $\overline{r}.(3\hat{i}+2\hat{j}+6\hat{k})=13$
i.e. $3x+2y+6z-13=0$

Given point $(2,3,\lambda )$
distance of plane from the points $=\left|\frac{a{x}_1+b{y}_1+c{z}_1+d}{\sqrt{a^2+b^2+c^2}}\right|$

$5=\left|\frac{3\left(2\right)+2\left(3\right)+6\lambda -13}{\sqrt{9+4+36}}\right|$
$\therefore 5=\left|\frac{6\lambda -1}{7}\right|$

$\Rightarrow 6\lambda -1=\pm 35$

$\Rightarrow 6\lambda =36,6\lambda =-34$

$\therefore \lambda =6,\lambda =-\frac{17}{3}$