Question

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

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

Answer 1

Answer: (C)

Let $O A=2 i-j +k, O B=i-3 j-5 k$ and $O C=3 i-4 j-4 k$
$\therefore a=O A=\sqrt{6}, b=O B=\sqrt{35}$
and $c=O C=\sqrt{41}$
$\therefore \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$
$=\frac{\sqrt{35}^{2}+\sqrt{41}^{2}-\sqrt{6}^{2}}{\sqrt{235 41}}$
$\Rightarrow \cos A=\frac{70}{23 \sqrt{55} \sqrt{41}}=\frac{\sqrt{35}}{41}$
$\Rightarrow \cos ^{2} A=\frac{35}{41}$