Question

If the position vectors of $A,B$ and $C$ are respectively $2i-3j-5k,i-3j-5k$ 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 $OA=2i-j+k,OB=i-3j-5k$ and $OC=3i-4j-4k$
$\therefore a=OA=\sqrt{6},b=OB=\sqrt{35}$
and $c=OC=\sqrt{41}$
$\therefore \cos A=\frac{b^2+c^2-a^2}{2bc}$
$=\frac{{\sqrt{35}}^2+{\sqrt{41}}^2-{\sqrt{6}}^2}{\sqrt{23541}}$
$\Rightarrow \cos A=\frac{70}{23\sqrt{55}\sqrt{41}}=\frac{\sqrt{35}}{41}$
$\Rightarrow {\cos }^{2}A=\frac{35}{41}$