Question

The vertices of a triangle ABC are A(2, 3, 1), B(-2, 2, 0) and C(0, 1, -1). Find the cosine of angle ABC.

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{2}{\sqrt{6}}$
• D. None of the above

Answer 1

The correct option is A $\frac{1}{\sqrt{3}}$

Given $A(2,3,1),B(-2,2,0)$ and $C(0,1,-1)$

For $△ABC,$

$a=BC=\sqrt{{(-2-0)}^2+{(2-1)}^2+{(0+1)}^2}\therefore a=\sqrt{6}b=AC=\sqrt{{(2-0)}^2+{(3-1)}^2+{(1+1)}^2}\therefore b=\sqrt{12}c=AB=\sqrt{{(2+2)}^2+(3-2{)}^2+{\left(1\right)}^2}\therefore c=\sqrt{18}\cos \angle ABC=\frac{a^2+c^2-b^2}{2ac}=\frac{18+6-12}{2\times 3\sqrt{2}\times \sqrt{6}}=\frac{1}{\sqrt{3}}$