Mathematics - Vector Question with Solution | TestHub

Line is ${\perp }^{\mathrm{r}}$ to 2 line $\Rightarrow$ line will be parallel to
$(i+a\hat{j}+b\hat{k})\times (-b\hat{i}+a\hat{j}+5\hat{k})$
Parallel vector along the required line is
$\hat{\mathrm{i}}(5\mathrm{a}-\mathrm{ab})-\hat{\mathrm{j}}\left({\mathrm{b}}^2+5\right)+\hat{\mathrm{k}}(\mathrm{a}+\mathrm{ab})$
Dr's of required line $\alpha (5a-ab),-\left(b^2+5\right),(a+ab)$
Also Dr's of required line $\alpha -2,\mathrm{d},-4$

Also point $\left(0,\frac{-1}{2},0\right)$ will lie on $\frac{\mathrm{x}-1}{-2}=\frac{\mathrm{y}+4}{\mathrm{d}}=\frac{\mathrm{z}-\mathrm{c}}{-4}$
$\frac{0-1}{-2}=\frac{\frac{-1}{2}+4}{d}=\frac{0-c}{-4}\Rightarrow d=7,c=2$
$\begin{array}{rl}& \text{ From (1) }\frac{5\mathrm{a}-\mathrm{ab}}{-2}=\frac{-{\mathrm{b}}^2-5}{7}=\frac{\mathrm{a}+\mathrm{ab}}{-4}\\ & \frac{5\mathrm{a}-\mathrm{ab}}{-2}=\frac{\mathrm{a}+\mathrm{ab}}{-4};\frac{-{\mathrm{b}}^2-5}{7}=\frac{\mathrm{a}+\mathrm{ab}}{-4}\end{array}$
\begin{array}{c|c}-20 a+4 a b=-2 a-2 a b & 4 b^2+20=70+7 a b \\18 a=6 a b & 36+20=70+21 a \\b=3 & 56=28 a \Rightarrow a=2\end{array}
$a+b+c+d=2+3+2+7=14$