If the position vectors of $A, B$ and $C$ are respectively $2…

If the position vectors of $A, B$ and $C$ are respectively $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$, then $\cos ^2 A$ is equal to
  1. $0$
  2. $\frac{6}{41}$
  3. $\frac{35}{41}$
  4. $1$

Solution

Let $\overrightarrow{O A}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \overrightarrow{O B}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ and $ \begin{aligned} & \overrightarrow{O C}=3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\ & \therefore \quad a=|\overrightarrow{\mathrm{OA}}|=\sqrt{6}, b=|\overrightarrow{\mathrm{OB}}|=\sqrt{35} \\ & \text { and } \quad c=|\overrightarrow{O C}|=\sqrt{41} \\ & \therefore \quad \cos A=\frac{b^2+c^2-a^2}{2 b c} \\ & =\frac{(\sqrt{35})^2+(\sqrt{41})^2-(\sqrt{6})^2}{2 \sqrt{35} \sqrt{41}} \\ & \Rightarrow \quad \cos A=\frac{70}{2 \sqrt{35} \sqrt{41}}=\sqrt{\frac{35}{41}} \\ & \Rightarrow \cos ^2 A=\frac{35}{41} \\ & \end{aligned} $

Asked in: AP EAMCET 2008

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