Question

The line $\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$. Then which of the following is not the side of the triangle?

• A. $\frac{x-7}{2}=\frac{y-2}{-3}=\frac{z-4}{6}$
• B. $\frac{x-7}{3}=\frac{y-2}{6}=\frac{z-4}{2}$
• C. $\frac{x-7}{3}=\frac{y-2}{5}=\frac{z-4}{-1}$
• D. none of these

Answer 1

Answer: (A)

$\frac{x-7}{2}=\frac{y-2}{-3}=\frac{z-4}{6}$

Sides AB and AC make an angle of ${45}^{\circ }$ with the given line.

Let $B=(-6+5t,-10+3t,-14+8t)$

$\overrightarrow{AB}=(5t-13)\hat{i}+(3t-12)\hat{j}+(8t-18)\hat{k}$

$\cos {45}^{\circ }=\frac{\left(\right(5t-13)\hat{i}+(3t-12)\hat{j}+(8t-18\left)\hat{k}\right)(5\hat{i}+3\hat{j}+8\hat{k})}{\sqrt{(5t-13{)}^2+(3t-12{)}^2+(8t-18{)}^2}\sqrt{(5^2+3^2+8^2)}}$

Squaring and simplifying $t^2-5t+6=0\Rightarrow t=2or3$

Taking 2 for B and 3 for C

$\overrightarrow{AB}=-3\hat{i}-6\hat{j}-2\hat{k}$

$\overrightarrow{AC}=2\hat{i}-3\hat{j}+6\hat{k}$

Equation of AB is $\frac{x-7}{3}=\frac{y-2}{6}=\frac{z-4}{2}$

Equation of AC is $\frac{x-7}{2}=\frac{y-2}{-3}=\frac{z-4}{6}$