Question

The largest value of a, for which the perpendicular distance of the plane containing the lines
$\vec{r} (\hat{i}+\hat{j})+λ(\hat{i}+a\hat{j}−\hat{k})\ and\ \vec{r}=(\hat{i}+\hat{j})+μ(−\hat{i}+\hat{j}−a\hat{k})$
from the point (2, 1, 4) is √3, is _____________.

Answer 1

The correct answer is 2
Normal to plane
$=\left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\\ 1 & a & -1 \\\ -1 & 1 & -a \end{matrix} \right|$
$=\hat{i}(1−a^2)−\hat{j}(−a−1)+\hat{k}(1+a)$
$=(1−a)\hat{i}+\hat{j}+\hat{k}$
∴ Plane (1 – a) (x – 1) + (y – 1) + z = 0
Distance from(2,1,4) is $\sqrt3$
i.e
$⇒ |\frac{(1-a)+0+4}{\sqrt{(1-a)^2+1+1}}| = \sqrt3$
$⇒ 25+a^2-10a = 3a^2-6a+9$
$⇒ 2a^2+4a-16 = 0$
$⇒ a^2+2a-8 = 0$
a=2 or -4
Therefore ,$a_{max} = 2$