Question

$A(3,2,6),B(1,4,5)$ and $C(3,5,3)$ are the vertices of triangle ABC. Find m$\angle$ABC.

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (A)

$\frac{\pi }{2}$

We know that the direction ratios of the line segment joining $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are $x_2-x_1,y_2-y_1,z_2-z_1$
Therefore direction ratios of BA are $3-1,2-4,6-5$ i.e., $2,-2,1$
a
and direction ratios of BC are $3-1,5-4,3-5$ i.e., $2,1,-2$
Let $m\angle$ ABC $=\theta$, then $\cos \theta =\frac{2\times 2+-2\times 1+1\times -2}{\sqrt{2^2+(-2{)}^2+1^2}\cdot \sqrt{2^2+1^2+(-2{)}^2}}$
$\cos \theta =\frac{4-2-2}{\sqrt{9}-\sqrt{9}}=0$
$\therefore \theta ={90}^o$.