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 $(7,2,4)$

The line $\dfrac{x+6}{5}=\dfrac{y+10}{3}=\dfrac{z+14}{8}$ is the hypotenuse of an isosceles right angled triangle whose opposite vertex is $(7,2,4)$.Find the equation of the remaining sides.

My Attempt:
Let the right angled triangle be $ABC$,right angled at $B(7,2,4).$The remaining two sides $AB$ and $BC$ of the right angled triangle will pass through the vertex $B(7,2,4)$.
Let the equation of $AB$ be $\dfrac{x-7}{a_1}=\dfrac{y-2}{b_1}=\dfrac{z-4}{c_1}$
and the equation of $BC$ be $\dfrac{x-7}{a_2}=\dfrac{y-2}{b_2}=\dfrac{z-4}{c_2}$
As $AB$ and $BC$ are perpendicular to each other. So $a_1a_2+b_1b_2+c_1c_2=0.....(1)$
Also as the triangle is a isosceles right triangle, so angle between $AC$ and $BC$ will be same as the angle between $CA$ and $BA$. Let it be $\theta$.

So $\cos\theta=\dfrac{5a_1+3b_1+8c_1}{\sqrt{5^2+3^2+8^2}\sqrt{a_1^2+b_1^2+c_1^2}}=\dfrac{5a_2+3b_2+8c_2}{\sqrt{5^2+3^2+8^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
Squaring both sides we get
$\dfrac{(5a_1+3b_1+8c_1)^2}{a_1^2+b_1^2+c_1^2}=\dfrac{(5a_2+3b_2+8c_2)^2}{a_2^2+b_2^2+c_2^2}.............(2)$

But I am stuck here.I don't know how to find $a_1,b_1,c_1,a_2,b_2$ and $c_2$. Please help me. Thanks.

There is a much simpler and efficient way to solve this problem.

First, find the foot of the perpendicular from the point B to the hypotenuse, and let's call it $D$. Now calculate the length of the perpendicular $BD$ and let's call it $d$.

Now since the triangle is isosceles and right-angled it should be obvious that the points $A$ and $C$ will be obtained by traversing a distance $d$ along the given line.

So if the direction cosines of the line are $l,m,n$ and the point $D$ is at $(x_0,y_0,z_0)$ then clearly we have the coordinates of $A$ and $C$ as $(x_0\pm ld,y_0\pm md,z_0\pm nd)$

• This is a very good approach and i got the answer using this ,but the calculations and simplifications are lengthy.Thank you @G-man – Vinod Kumar Punia Nov 20 '15 at 16:27

Efficient way is take angle b/w line 45 as given triangle is 90 with isosceles.Take parametric point on given line as (5¥-6,3¥-10,8¥-18) and make vector with (7,2,4) and then then take cos45 of this vector and (5,3,8).You will get a simple quadratic I.e. ¥^2-5¥+6=0. Imply ¥=2,3.means you get the value of ¥ due to other 2 vertex of triangle.

Sides AB and AC make an angle of 45∘ with the given line.

Let B=(−6+5t,−10+3t,−14+8t)

AB→=(5t−13)i^+(3t−12)j^+(8t−18)k^

cos45∘=((5t−13)i^+(3t−12)j^+(8t−18)k^).(5i^+3j^+8k^)(5t−13)2+(3t−12)2+(8t−18)2−−−−−−−−−−−−−−−−−−−−−−−−−−−√52+32+82−−−−−−−−−−√

Squaring and simplifying t2–5t+6=0⟹t=2or3

Taking 2 for B and 3 for C.

AB→=−3i^−6j^−2k^

AC→=2i^−3j^+6k^

Equation of AB is x−73=y−26=z−42

Equation of AC is x−72=y−2−3=z−46