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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.

My approach $\frac{x+6}{5}=\frac{y+10}{3}=\frac{z+14}{8}=t$

$x=-6+5t\\y=-10+3t\\z=-14+8t$

Vector drawn from opposite to hypotenuse$\langle 13-5t,12-3t,18-8t\rangle$

Vector of hypotenuse $\langle 5,3,8\rangle$

Vector product should be zero: $65-25t+36-9t+144-64t=0$, I could get the value of $t$.

Is there any way I could find end point of the hypotenuse by getting the value of $t$ for end points. I don't want to find the value by distance but by angle between line viz 45°.

Samar Imam Zaidi
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2 Answers2

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you must compute $$\vec{AB}\cdot\vec{AC}=0$$ (dot product) and the Points $B,C$ are situated on the given line

Dr. Sonnhard Graubner
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If you don’t want to use lengths directly, here are a couple of ways to proceed.

Let $t_0$ be the value of $t$ that you’ve computed for the foot of the altitude. By symmetry, the endpoints of the hypotenuse are at $t_0\pm\Delta t$ and the dot product of the vectors from $(7,2,4)$ to these points must be $0$. This will give you a quadratic equation in $\Delta t$.

Instead of computing the end points of the hypotenuse, however, once you’ve found the foot of the altitude, you can take direct advantage of the fact that this is an isosceles right triangle. The two missing sides will be parallel to the angle bisectors of the line through the hypotenuse and the altitude from $(7,2,4)$. If the respective direction vectors are $\mathbf v$ and $\mathbf w$, the directions of the angle bisectors are $\|\mathbf w\|\mathbf v\pm\|\mathbf v\|\mathbf w$, from which you can construct the two equations of the lines.

amd
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