Question

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