Question

If the coordinate of the vertices of a triangle $ABC$ be $A(-1,3,2),$ $B(2,3,5)$ and $C(3,5,-2)$ then $\angle A$ is equal to:

• A. $45^{\circ }$
• B. $60^{\circ }$
• C. $90^{\circ }$
• D. $30^{\circ }$

Answer 1

Answer: (C)

Given that coordinates of the vertices of a triangle $ABC$ are
$A(-1,3,2),B(2,3,5)$ and $C(3,5,-2)$.
$D{R}^{\prime }s$ of a line $AB$ is $(3,0,3)$
and $CA$ is $(-4,-2,4)$ .
$\therefore \cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
$=\frac{3(-4)+0(-2)+3(4)}{\sqrt{9+0+9}\sqrt{16+4+16}}=0$
Alternative Solution:
Given vertices of a triangle are
$A(-1,3,2),B(2,3,5)$ and $C(3,5,-2)$.
$\therefore AB=\sqrt{(2+1{)}^2+(3-3{)}^2+(5-2{)}^2}$
$=\sqrt{9+0+9}=3\sqrt{2}$
$BC=\sqrt{(3-2{)}^2+(5-3{)}^2+(-2-5{)}^2}$
$=\sqrt{1+4+49}=3\sqrt{6}$
$CA=\sqrt{(3+1{)}^2+(5-3{)}^2+(-2-2{)}^2}$
$=\sqrt{16+4+16}=6$
Now, $A{B}^2+C{A}^2=(3\sqrt{2}{)}^2+(6{)}^2$
$=18+36=54$
$=B{C}^2$
$\therefore \Delta ABC$ is right angled triangle, right angled at $A$ .
$\therefore \angle A=90^{\circ }$