Question

If $A(-1,3,2),B(2,3,5)$ and $C(3,5,-2)$ are vertices of a $△ABC$, then angles of $△ABC$ are

• A. $A=90^{\circ },\angle B=30^{\circ },\angle C=60^{\circ }$
• B. $\angle A=\angle B=\angle C=60^{\circ }$
• C. $\angle A=\angle B=45^{\circ },\angle C=90^{\circ }$
• D. None of the above

Answer 1

Answer: (D)

Given vertices of a $△ABC$ are $A(-1,3,2),B(2,3,5)$ and $C(3,5,-2)$
Now DR's of
$AB=(2+1,3-3,5-2)=(3,0,3)$
DR's of
$BC=(3-2,5-3,-2-5)=(1,2,-7)$
and DR's of $CA=(-1-3,3-5,2+2)$ $=(-4,-2,4)$
Now, the angle between $AB$ and $BC$,
$\cos B=\frac{|3\times 1+0\times 2+3\times (-7)|}{\sqrt{3^2+0^2+3^2}\sqrt{1^2+2^2+(-7{)}^2}}$
$=\frac{|3+0-21|}{\sqrt{9+0+9}\sqrt{1+4+49}}$
$=\frac{18}{3\sqrt{2}\times 3\sqrt{6}}=\frac{2}{2\sqrt{3}}=\frac{1}{\sqrt{3}}$
angle between $BC$ and $CA$,
$\cos C=\frac{|1\times (-4)+2(-2)+(-7)(4)|}{\sqrt{1^2+2^2+(-7{)}^2}\sqrt{(-4{)}^2+(-2{)}^2+(4{)}^2}}$
$=\frac{|-4-4-28|}{\sqrt{1+4+49}\sqrt{16+4+16}}$
$=\frac{36}{\sqrt{54}\sqrt{36}}=\frac{36}{3\sqrt{6}\times 6}$
$=\frac{2}{\sqrt{2}\sqrt{3}}=\frac{\sqrt{2}}{\sqrt{3}}$
and angle between $AC$ and $AB$,
$\cos A=\frac{|-4\times 3+(-2)\times 0+4\times 3|}{\sqrt{(-4{)}^2+(-2{)}^2+(4{)}^2}\sqrt{3^2+0^2+3^2}}$
$=|0|$
$\Rightarrow A=90^{\circ }$