Question

Let a line having direction ratios $1,-4,2$ intersect the lines $\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}$ and $\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}$ at the points $A$ and $B$. Then $(AB{)}^2$ is equal to

• A. 84.00
• B. 84.0
• C. 84

Answer 1

Answer: (A), (B), (C)

Given lines:
$\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1}=\lambda$
Therefore, the general points are $A(3\lambda +7,-\lambda +1,\lambda -2))$
and $\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1}=\mu$
Therefore, the general points are $B(2\mu ,3\mu +7,\mu )$

Direction ratios of $AB$ are $3\lambda -2\mu +7,-(\lambda +3\mu +6),\lambda -\mu -2$

Clearly, $\frac{3\lambda -2\mu +7}{1}=\frac{\lambda +3\mu +6}{4}=\frac{\lambda -\mu -2}{2}$
Taking first two fractions, we have
$\lambda -\mu +2=0\cdots \left(1\right)$
again taking last two fractions, we have
$\lambda -5\mu -10=0\cdots \left(2\right)$
Solving equation $\left(1\right)$ and $\left(2\right),$ we get
$\lambda =-5,\mu =-3$
So, $A(-8,6,-7)$ and $B(-6,-2,-3)$
Now, $AB=\sqrt{4+64+16}=\sqrt{84}$
$\therefore (AB{)}^2=84$