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Let $\mathbf{p}=\hat{\mathbf{i}}+2 \mathbf{j}-\mathbf{k}, \mathbf{q}=2 \mathbf{i}-\mathbf{j}+\mathbf{k}$. If $\mathbf{a}$ and $\mathbf{b}$ are two vectors such that $\mathbf{p}=\mathbf{a}-2 \mathbf{b}$, $\mathbf{q}=2 \mathbf{a}+\mathbf{b}$, then the angle between $\mathbf{a}$ and $\mathbf{b}$ is
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Verified Answer
The correct answer is:
$\cos ^{-1}\left(\frac{3}{2 \sqrt{221}}\right)$
$\because \mathbf{p}=\mathbf{a}-2 \mathbf{b}$ and $\mathbf{q}=2 \mathbf{a}+\mathbf{b}$
So,
$\begin{aligned} 5 \mathbf{a} & =\mathbf{p}+2 \mathbf{q}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})+2(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ & =5 \hat{\mathbf{i}}+\hat{\mathbf{k}}\end{aligned}$
$\Rightarrow \quad \mathbf{a}=\hat{\mathbf{i}}+\frac{\hat{\mathbf{k}}}{5}$
Similarly, $\mathbf{b}=\mathbf{q}-2 \mathbf{a}=(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})-2 \hat{\mathbf{i}}-\frac{2 \hat{\mathbf{k}}}{5}$
$=-\hat{\mathbf{j}}+\frac{3 \hat{\mathbf{k}}}{5}$
So, angle between $\mathbf{a}$ and $\mathbf{b}$ is
$\begin{aligned} & \theta=\cos ^{-1}\left|\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \mid \mathbf{b} \|}\right|=\cos ^{-1}\left(\frac{\frac{3}{25}}{\frac{\sqrt{26}}{5} \times \frac{\sqrt{34}}{5}}\right) \\ & =\cos ^{-1}\left(\frac{3}{2 \sqrt{13 \times 17}}\right)=\cos ^{-1}\left(\frac{3}{2 \sqrt{221}}\right)\end{aligned}$
Hence, option (a) is correct.
So,
$\begin{aligned} 5 \mathbf{a} & =\mathbf{p}+2 \mathbf{q}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})+2(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ & =5 \hat{\mathbf{i}}+\hat{\mathbf{k}}\end{aligned}$
$\Rightarrow \quad \mathbf{a}=\hat{\mathbf{i}}+\frac{\hat{\mathbf{k}}}{5}$
Similarly, $\mathbf{b}=\mathbf{q}-2 \mathbf{a}=(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})-2 \hat{\mathbf{i}}-\frac{2 \hat{\mathbf{k}}}{5}$
$=-\hat{\mathbf{j}}+\frac{3 \hat{\mathbf{k}}}{5}$
So, angle between $\mathbf{a}$ and $\mathbf{b}$ is
$\begin{aligned} & \theta=\cos ^{-1}\left|\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \mid \mathbf{b} \|}\right|=\cos ^{-1}\left(\frac{\frac{3}{25}}{\frac{\sqrt{26}}{5} \times \frac{\sqrt{34}}{5}}\right) \\ & =\cos ^{-1}\left(\frac{3}{2 \sqrt{13 \times 17}}\right)=\cos ^{-1}\left(\frac{3}{2 \sqrt{221}}\right)\end{aligned}$
Hence, option (a) is correct.
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