Question

Prove that $\overline{a}=(4,-3,1),\overline{b}=(2,-4,5)$ and $\overline{c}=(1,-1,0)$ are vertices of a right angle.

Answer 1

$\overrightarrow{a}=(4,-3,1)$

$\overrightarrow{b}=(2,-4,5)$

$\overrightarrow{c}=(1,-1,0)$

$\overrightarrow{ab}=\overrightarrow{b}-\overrightarrow{a}=(2-4)\hat{i}+(-4+3)\hat{j}+(5-1)\hat{k}$

$=2\hat{i}-\hat{j}+4\hat{k}$

$\overrightarrow{bc}=\overrightarrow{c}-\overrightarrow{b}=(1-2)\hat{i}+(-1+4)\hat{j}+(0-5)\hat{k}$

$=-\hat{i}+3\hat{j}-5\hat{k}$

$\overrightarrow{ca}=\overrightarrow{a}-\overrightarrow{c}=(4-1)\hat{i}+(-3+1)\hat{j}+(1-0)\hat{k}$

$=3\hat{i}-2\hat{j}+\hat{k}$

now $\left|\overrightarrow{ab}\right|=\sqrt{4+1+16}=\sqrt{21}$ units

$\left|\overrightarrow{bc}\right|=\sqrt{25+9+1}=\sqrt{35}$ units

$\left|\overrightarrow{ca}\right|=\sqrt{9+4+1}=\sqrt{14}$ units

we can see that

${\left|\overrightarrow{ab}\right|}^2+{\left|\overrightarrow{ca}\right|}^2={\left|\overrightarrow{bc}\right|}^2$

$\because$ $({\sqrt{21}}^2+{\sqrt{14}}^2={\left(\sqrt{35}\right)}^2)$

So we have the square of the two sides of a triangles is equal to the square of the third which is basically the pythagoras theorem, Hence $\overrightarrow{a},\overrightarrow{b}\&\overrightarrow{c}$ are the coordinates of a right angle triangle, right angled at $\overrightarrow{a}$.