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If $\quad \overrightarrow{\mathbf{a}}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \quad \overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} \quad$ and $\overrightarrow{\mathbf{c}}=-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$, then the angle between $2 \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}}$ and $\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}$ is
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2116 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi}{3}$
Now,
$\begin{aligned}
2 \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}} & =2(-\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})-(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\
& =\hat{\mathbf{j}}+\hat{\mathbf{k}}
\end{aligned}$
and
$\begin{aligned}
\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}} & =-\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}+2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} \\
& =\hat{\mathbf{i}}+\hat{\mathbf{k}}
\end{aligned}$
Let $\theta$ be the angle between $2 \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}}$ and $\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}$.
$\begin{array}{ll}\therefore & \cos \theta=\frac{(\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{k}})}{\sqrt{1^2+1^2} \sqrt{1^2+1^2}} \\ \Rightarrow & \cos \theta=\frac{1}{\sqrt{2} \sqrt{2}}=\frac{1}{2}\end{array}$
$\Rightarrow \quad \theta=\frac{\pi}{3}$
$\begin{aligned}
2 \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}} & =2(-\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})-(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\
& =\hat{\mathbf{j}}+\hat{\mathbf{k}}
\end{aligned}$
and
$\begin{aligned}
\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}} & =-\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}+2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} \\
& =\hat{\mathbf{i}}+\hat{\mathbf{k}}
\end{aligned}$
Let $\theta$ be the angle between $2 \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}}$ and $\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}$.
$\begin{array}{ll}\therefore & \cos \theta=\frac{(\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{k}})}{\sqrt{1^2+1^2} \sqrt{1^2+1^2}} \\ \Rightarrow & \cos \theta=\frac{1}{\sqrt{2} \sqrt{2}}=\frac{1}{2}\end{array}$
$\Rightarrow \quad \theta=\frac{\pi}{3}$
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