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 $(7,2,4)$.Find the equation of the remaining sides.

• 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. $\frac{x-7}{2}=\frac{y+2}{3}=\frac{z-4}{6}$
• D. $\frac{x-7}{3}=\frac{y-2}{6}=\frac{z-4}{2}$

Answer 1

Answer: (A), (D)

The correct options are
A

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

D

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

$\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}=\lambda$(say)

General point on above line is $(5\lambda -6,3\lambda -10,8\lambda -14)$

Let $B\equiv (5\lambda -6,3\lambda -10,8\lambda -14)$

d.r's of line $AB$ are

$((5\lambda -6-7),(3\lambda -10-2),(8\lambda -14-4))\Rightarrow (5\lambda -13,3\lambda -12,8\lambda -18)$

Also d.r.'s of line $BC$ are $(5,3,8)$

Since angle between $AB$ and $BC$ is $\frac{\pi }{4}$

$\Rightarrow \cos \frac{\pi }{4}=\frac{|(5\lambda -13)5+(3\lambda -12)3+(8\lambda -18)8|}{\sqrt{5^2+3^2+8^2}.\sqrt{{(5\lambda -13)}^2+{(3\lambda -12)}^2+{(8\lambda -18)}^2}}$

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{|25\lambda +9\lambda +64\lambda -65-36-144|}{\sqrt{98}.\sqrt{{(5\lambda -13)}^2+{(3\lambda -12)}^2+{(8\lambda -18)}^2}}$

Solving above equation, we get $\lambda =3,2$

Hence, equation of lines are

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