Question

If the position vectors of $A,B$ and $C$ are respectively $2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k},\overrightarrow{i}-3\overrightarrow{j}-5\overrightarrow{k}$ and $3\overrightarrow{i}-4\overrightarrow{j}-4\overrightarrow{k}$, then ${\cos }^{2}A$ is equal to

• A. $0$
• B. $\frac{6}{41}$
• C. $\frac{35}{41}$
• D. $1$

Answer 1

The correct option is C $\frac{35}{41}$

Let $OA=2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k},OB=\overrightarrow{i}-3\overrightarrow{j}-5\overrightarrow{k}$ and $OC=3\overrightarrow{i}-4\overrightarrow{j}-4\overrightarrow{k}$
Thus $a=OA=\sqrt{6},b=OB=\sqrt{35}$ and $c=OC=\sqrt{41}$
Therefore, $\cos A=\frac{b^2+c^2-a^2}{2bc}$
$=\frac{{\sqrt{35}}^2+{\sqrt{41}}^2-{\sqrt{6}}^2}{2\sqrt{35}\sqrt{41}}$
$\Rightarrow \cos A=\frac{70}{2\sqrt{35}\sqrt{41}}=\frac{\sqrt{35}}{41}$
$\Rightarrow {\cos }^{2}A=\frac{35}{41}$