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If $\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ find a unit vector parallel to the vector $2 \vec{a}-\vec{b}+3 \vec{c}$.
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Verified Answer
We have,
$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}, \vec{c}=\hat{i}-2 \hat{j}+\hat{k}$
$2 \vec{a}-\vec{b}+3 \vec{c}=2(\hat{i}+\hat{j}+\hat{k})-(2 \hat{i}-\hat{j}+3 \hat{k})+3(\hat{i}-2 \hat{j}+\hat{k})$
$=3 \hat{i}-3 \hat{j}+2 \hat{k}$
$\therefore \quad|\overrightarrow{\mathrm{a}}|=\sqrt{3^2+(-3)^2+2^2}=\sqrt{9+9+4}=\sqrt{22}$
$\therefore \quad$ unit vector parallel to $2 \vec{a}-\vec{b}+3 \vec{c}$
$=\frac{3}{\sqrt{22}} \hat{\mathrm{i}}-\frac{3}{\sqrt{22}} \hat{\mathrm{j}}+\frac{2}{\sqrt{22}} \hat{\mathrm{k}}$
$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}, \vec{c}=\hat{i}-2 \hat{j}+\hat{k}$
$2 \vec{a}-\vec{b}+3 \vec{c}=2(\hat{i}+\hat{j}+\hat{k})-(2 \hat{i}-\hat{j}+3 \hat{k})+3(\hat{i}-2 \hat{j}+\hat{k})$
$=3 \hat{i}-3 \hat{j}+2 \hat{k}$
$\therefore \quad|\overrightarrow{\mathrm{a}}|=\sqrt{3^2+(-3)^2+2^2}=\sqrt{9+9+4}=\sqrt{22}$
$\therefore \quad$ unit vector parallel to $2 \vec{a}-\vec{b}+3 \vec{c}$
$=\frac{3}{\sqrt{22}} \hat{\mathrm{i}}-\frac{3}{\sqrt{22}} \hat{\mathrm{j}}+\frac{2}{\sqrt{22}} \hat{\mathrm{k}}$
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