If the equation of the line passing through the point $\left(0,-\frac{1}{2}, 0\right)$ and perpendicular to the lines $\vec{r}=\lambda(\hat{i}+a \hat{j}+b \hat{k})$ and $\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}-6 \hat{\mathrm{k}})+\mu(-b \hat{\mathrm{i}}+\mathrm{a}…

Question

If the equation of the line passing through the point $\left(0,-\frac{1}{2}, 0\right)$ and perpendicular to the lines $\vec{r}=\lambda(\hat{i}+a \hat{j}+b \hat{k})$ and
$\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}-6 \hat{\mathrm{k}})+\mu(-b \hat{\mathrm{i}}+\mathrm{a} \hat{\mathrm{j}}+5 \hat{\mathrm{k}})$
is $\frac{\mathrm{x}-1}{-2}=\frac{\mathrm{y}+4}{\mathrm{~d}}=\frac{\mathrm{z}-\mathrm{c}}{-4}$, then $\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}$ is equal to :

Aicharya · Accepted Answer

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 $\alpha(5 a-a b),-\left(b^2+5\right),(a+a b)$
Also Dr's of required line $\alpha-2, \mathrm{~d},-4$

Also point $\left(0, \frac{-1}{2}, 0\right)$ will lie on $\frac{\mathrm{x}-1}{-2}=\frac{\mathrm{y}+4}{\mathrm{~d}}=\frac{\mathrm{z}-\mathrm{c}}{-4}$
$\frac{0-1}{-2}=\frac{\frac{-1}{2}+4}{d}=\frac{0-c}{-4} \Rightarrow d=7, c=2$
$\begin{aligned}
&\text { From (1) } \frac{5 \mathrm{a}-\mathrm{ab}}{-2}=\frac{-\mathrm{b}^2-5}{7}=\frac{\mathrm{a}+\mathrm{ab}}{-4}\\ &\frac{5 \mathrm{a}-\mathrm{ab}}{-2}=\frac{\mathrm{a}+\mathrm{ab}}{-4} ; \frac{-\mathrm{b}^2-5}{7}=\frac{\mathrm{a}+\mathrm{ab}}{-4}
\end{aligned}$
\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}
$a+b+c+d=2+3+2+7=14$ *

If the equation of the line passing through the point $\left(0,-\frac{1}{2}, 0\right)$ and perpendicular to…

Solution