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