Question

The equation of the line passing through the point $(1,2,-4)$ and perpendicular to the two lines $\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}$ and $\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$ will be

• A. $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$
• B. $\frac{x-1}{-2}=\frac{y-2}{3}=\frac{z+4}{8}$
• C. $\frac{x-1}{3}=\frac{y-2}{2}=\frac{z+4}{8}$
• D. None of the above

Answer 1

Answer: (A)

Given lines are
$\frac{x-8}{3} =\frac{y+19}{-16}=\frac{z-10}{7} $ ...(i)
$\text { and } \frac{x-15}{3} =\frac{y-29}{8}=\frac{z-5}{-5}$...(ii)
Both lines are in the form
$ \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$
$ \therefore a_1=3 b_1=-16 c_1=7$
$ a_2=3, b_2=8, c_2=-5 $
$ \text { Now, } b_1=a_1 \hat{j}+b_1 \hat{j}+c_1 \hat{k}=3 \hat{i}-16 \hat{j}+7 \hat{k}$
$ \text { and } b_2=a_2 \hat{i}+b_2 \hat{j}+c_2 \hat{k}=3 \hat{i}+8 \hat{j}-5 \hat{k} $
The vector perpendicular to both Eqs. (i) and (ii) is given by $b=b_1 \times b_2$
$=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5\end{vmatrix}$
$ -\hat{i}(80-56)-\hat{j}(-15-21)+\hat{k}(24+48)$
$b_1 \times b_2 =24 \hat{i}+36 \hat{j}+72 \hat{k}$
Hence, the required equation of line is
$\frac{x-1}{24}=\frac{y-2}{36}=\frac{z+4}{72} \text { i.e., } \frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$