Question:

If \(\vec{a}=\hat{i}+\hat{j}+\hat{k},\vec{b}=2\hat{i}-\hat{j}+3\hat{k}\) and \(\vec{c}=\hat{i}-2\hat{j}+\hat{k}\),find a unit vector parallel to the vector \(2\vec{a}-\vec{b}+3\vec{c}.\)

Updated On: Sep 20, 2023
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Solution and Explanation

We have,
\(\vec{a}=\hat{i}+\hat{j}+\hat{k},\vec{b}=2\hat{i}-\hat{j}+3\hat{k}\),and \(\vec{c}=\hat{i}-2\hat{j}+\hat{k}\)
\(2\vec{a}-\vec{b}+3\vec{c}\)\(=2(\hat{i}+\hat{j}+\hat{k})-(2\hat{i}-\hat{j}+3\hat{k})+3\hat{(i}-2\hat{j}+\hat{k})\)
\(=2/hat{i}+2\hat{j}+2\hat{k}-2\hat{i}+\hat{j}-3\hat{k}+3\hat{i}-6\hat{j}+3\hat{k}\)
\(=3\hat{i}-3\hat{j}+2\hat{k}\)
\(|2\vec{a}-\vec{b}+3\vec{c}|=\sqrt{3^{2}+(-3)^{2}+2^{2}}=\sqrt{9+9+4}=\sqrt{22}\)
Hence,the unit vector along \(2\vec{a}-\vec{b}+3\vec{c}\) is
\(\frac{2\vec{a}-\vec{b}+3\vec{c}}{|2\vec{a}-\vec{b}+3\vec{c}|}\)\(=\frac{3i^-3\hat{j}+2\hat{k}}{\sqrt{22}}=\frac{3}{\sqrt{22}}\hat{i}-\frac{3}{\sqrt{22}}\hat{j}+\frac{2}{\sqrt{22}}\hat{k}\).

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