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: (C)

Given are vertex $A(7,2,4)$ and line
$\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}$.
General point on above line $B\equiv (5\lambda -6,3\lambda -10,8\lambda -14)$
Direction ratios of line $AB$ are $⟨5\lambda -13,3\lambda -12,8\lambda -18>$
Direction ratios of line $BC$ are $<5,3,8>$
Since angle between $AB$ and $BC$ is $\pi /4$, we have
$\cos \frac{\pi }{4}=\frac{(5\lambda -3)5+3(3\lambda -12)+8(8\lambda -18)}{\sqrt{5^2+3^2+8^2}\cdot \sqrt{(5\lambda -13{)}^2+(3\lambda -12{)}^2+(8\lambda -18{)}^2}}$
Squaring and solving, we get $\lambda =3,2$
Hence, equation of lines are $\frac{x-7}{2}=\frac{y-2}{-3}=\frac{z-4}{6}$
and $\frac{x-7}{3}=\frac{y-2}{6}=\frac{z-4}{2}$.