Q77 / 160 · TS EAMCET 2008 · Mathematics
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
- A$0$
- B$\frac{6}{41}$
- Correct answerC$\frac{35}{41}$
- D$1$
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
$\begin{aligned}<br />& \overrightarrow{O C}=3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\<br />& \therefore \quad a=|\overrightarrow{\mathrm{OA}}|=\sqrt{6}, b=|\overrightarrow{\mathrm{OB}}|=\sqrt{35} \\<br />& \text {and } \quad c=|\overrightarrow{O C}|=\sqrt{41} \\<br />& \therefore \quad \cos A=\frac{b^2+c^2-a^2}{2 b c} \\<br />& =\frac{(\sqrt{35})^2+(\sqrt{41})^2-(\sqrt{6})^2}{2 \sqrt{35} \sqrt{41}} \\<br />& \Rightarrow \quad \cos A=\frac{70}{2 \sqrt{35} \sqrt{41}}=\sqrt{\frac{35}{41}} \\<br />& \Rightarrow \cos ^2 A=\frac{35}{41}<br />\end{aligned}$
$\begin{aligned}<br />& \overrightarrow{O C}=3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\<br />& \therefore \quad a=|\overrightarrow{\mathrm{OA}}|=\sqrt{6}, b=|\overrightarrow{\mathrm{OB}}|=\sqrt{35} \\<br />& \text {and } \quad c=|\overrightarrow{O C}|=\sqrt{41} \\<br />& \therefore \quad \cos A=\frac{b^2+c^2-a^2}{2 b c} \\<br />& =\frac{(\sqrt{35})^2+(\sqrt{41})^2-(\sqrt{6})^2}{2 \sqrt{35} \sqrt{41}} \\<br />& \Rightarrow \quad \cos A=\frac{70}{2 \sqrt{35} \sqrt{41}}=\sqrt{\frac{35}{41}} \\<br />& \Rightarrow \cos ^2 A=\frac{35}{41}<br />\end{aligned}$
Hence, the correct answer is option: 'C'.