Question

If the co-ordinate 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. $30{}^{\circ }$
• E. $135{}^{\circ }$

Answer 1

Answer: (C)

$A(-1,3,2),B(2,3,5)$ and $C(3,5,-2)$ $\therefore$ $AB=\sqrt{3^2+0+3^2}=\sqrt{18}$ $CA=\sqrt{16+4+16}=6$ and $BC=\sqrt{1+4+49}=\sqrt{54}$ $\because$ $A{B}^2+C{A}^2=B{C}^2$ $\therefore$ $\Delta ABC$ is right angled triangle, right angled at A. $\therefore$ $\angle A=90{}^{\circ }$