Question

The equation of the line passing through the point $(5,3,2)$ and perpendicular to the lines $\frac{x-2}{1}=\frac{y-3}{-1}=\frac{z-4}{1}$ and $\frac{x-1}{2}=\frac{y-1}{1}=\frac{z+1}{0}$ is

• A. $\frac{x-5}{-1}=\frac{y-3}{2}=\frac{z-2}{3}$
• B. $\frac{x+1}{5}=\frac{y-2}{3}=\frac{z-3}{2}$
• C. $\frac{x-5}{1}=\frac{y-3}{-1}=\frac{z-2}{1}$
• D. $\frac{x-5}{2}=\frac{y-3}{1}=\frac{z-2}{0}$

Answer 1

The correct option is A $\frac{x-5}{-1}=\frac{y-3}{2}=\frac{z-2}{3}$

We have to find the equation of the line passing through the point $(5,3,2)$ and perpendicular to the lines $\frac{x-2}{1}=\frac{y-3}{-1}=\frac{z-4}{1}$ and $\frac{x-1}{2}=\frac{y-1}{1}=\frac{z+1}{0}$.

Consider $\frac{x-2}{1}=\frac{y-3}{-1}=\frac{z-4}{1}$

Thus the directional ratios are $(1,-1,1)$.

Now consider $\frac{x-1}{2}=\frac{y-1}{1}=\frac{z+1}{0}$

The directional ratios are $(2,1,0)$.

To find the vector which is perpendicular to the given lines we need to find the cross product of $(1,-1,1)$ and $(2,1,0)$

$\left|\begin{array}{ccc}i& j& k\\ 1& -1& 1\\ 2& 1& 0\end{array}\right|=i(0-1)-j(0-2)+k(1+2)=-i+2j+3k=li+mj+nk\left(say\right)$

Thus the equation of the required line is given by $\frac{x-a}{l}=\frac{y-b}{m}=\frac{z-c}{n}$ where $(a,b,c)=(5,3,2)$ and $(l,m,n)=(-1,2,3)$

$\frac{x-5}{-1}=\frac{y-3}{2}=\frac{z-2}{3}$ is the required equation of the line.