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
$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$
$\therefore a_1=3b_1=-16c_1=7$
$a_2=3,b_2=8,c_2=-5$
$\text{ Now, }{b}_1=a_1\hat{j}+b_1\hat{j}+c_1\hat{k}=3\hat{i}-16\hat{j}+7\hat{k}$
$\text{ and }{b}_2=a_2\hat{i}+b_2\hat{j}+c_2\hat{k}=3\hat{i}+8\hat{j}-5\hat{k}$
The vector perpendicular to both Eqs. (i) and (ii) is given by $b=b_1\times b_2$
$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -16& 7\\ 3& 8& -5\end{array}\right|$
$-\hat{i}(80-56)-\hat{j}(-15-21)+\hat{k}(24+48)$
$b_1\times b_2=24\hat{i}+36\hat{j}+72\hat{k}$
Hence, the required equation of line is
$\frac{x-1}{24}=\frac{y-2}{36}=\frac{z+4}{72}\text{ i.e., }\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$