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

The correct option is C $\frac{x-7}{3}=\frac{y-2}{5}=\frac{z-4}{-1}$

Let opposite vertex be $A(7,2,4).$
$\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}=\lambda$

Hence, any point on above line be $B\equiv (5\lambda -6,3\lambda -10,8\lambda -14)$

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, the angle between $AB$ and $BC$ is $\pi /4.$
$\therefore \cos \frac{\pi }{4}=\frac{(5\lambda -3)5+3(3\lambda -12)+8(8\lambda -18)}{\sqrt{5^2+3^2+8^2}.\sqrt{(5\lambda -13{)}^2+(3\lambda -12{)}^2+(8\lambda -18{)}^2}}$

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