If overrightarrow{a}=-hat{i}+hat{j}+2 hat{k}, | vedclass

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(B) First, we calculate the vectors $2 \overrightarrow{a}-\overrightarrow{c}$ and $\overrightarrow{a}+\overrightarrow{b}$.$2 \overrightarrow{a}-\overr

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$\frac{\pi}{3}$

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(B) First, we calculate the vectors $2 \overrightarrow{a}-\overrightarrow{c}$ and $\overrightarrow{a}+\overrightarrow{b}$.$2 \overrightarrow{a}-\overrightarrow{c} = 2(-\hat{i}+\hat{j}+2 \hat{k}) - (-2 \hat{i}+\hat{j}+3 \hat{k}) = (-2\hat{i}+2\hat{j}+4\hat{k}) + (2\hat{i}-\hat{j}-3\hat{k}) = \hat{j}+\hat{k}$.$\overrightarrow{a}+\overrightarrow{b} = (-\hat{i}+\hat{j}+2 \hat{k}) + (2 \hat{i}-\hat{j}-\hat{k}) = \hat{i}+\hat{k}$.Let $	heta$ be the angle between these two vectors. The cosine of the angle is given by $\cos 	heta = \frac{(\hat{j}+\hat{k}) \cdot (\hat{i}+\hat{k})}{|\hat{j}+\hat{k}| |\hat{i}+\hat{k}|}$.Calculating the dot product: $(\hat{j}+\hat{k}) \cdot (\hat{i}+\hat{k}) = (0)(1) + (1)(0) + (1)(1) = 1$.Calculating the magnitudes: $|\hat{j}+\hat{k}| = \sqrt{0^2+1^2+1^2} = \sqrt{2}$ and $|\hat{i}+\hat{k}| = \sqrt{1^2+0^2+1^2} = \sqrt{2}$.Thus, $\cos 	heta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.Therefore, $	heta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

If $\overrightarrow{a}=-\hat{i}+\hat{j}+2 \hat{k}$, $\overrightarrow{b}=2 \hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{c}=-2 \hat{i}+\hat{j}+3 \hat{k}$, then the angle between $2 \overrightarrow{a}-\overrightarrow{c}$ and $\overrightarrow{a}+\overrightarrow{b}$ is

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