Question:
Let \(\vec{\alpha}=3\hat{i}+\hat{j}\) and \(\vec{\beta}=2\hat{i}-\hat{j}+3\hat{k}\). If \(\vec{\beta}=\vec{\beta}_1-\vec{\beta}_2\), where \(\vec{\beta}_1\) is parallel to \(\vec{\alpha}\) and \(\vec{\beta}_2\) is perpendicular to \(\vec{\alpha}\), then \(\vec{\beta}_1\times\vec{\beta}_2\) is equal to
Let \(\vec{\alpha}=3\hat{i}+\hat{j}\) and \(\vec{\beta}=2\hat{i}-\hat{j}+3\hat{k}\). If \(\vec{\beta}=\vec{\beta}_1-\vec{\beta}_2\), where \(\vec{\beta}_1\) is parallel to \(\vec{\alpha}\) and \(\vec{\beta}_2\) is perpendicular to \(\vec{\alpha}\), then \(\vec{\beta}_1\times\vec{\beta}_2\) is equal to
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\(\vec{\beta}_1\) is the projection of \(\vec{\beta}\) along \(\vec{\alpha}\); then \(\vec{\beta}_2=\vec{\beta}_1-\vec{\beta}\).
Updated On: Nov 5, 2026
- \(3\hat{i}-9\hat{j}-5\hat{k}\)
- \(\frac{1}{2}\left(-3\hat{i}+9\hat{j}+5\hat{k}\right)\)
- \(-3\hat{i}+9\hat{j}+5\hat{k}\)
- \(\frac{1}{2}\left(3\hat{i}-9\hat{j}+5\hat{k}\right)\)
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The Correct Option is B
Solution and Explanation
\(\vec{\beta}_1\parallel\vec{\alpha}\) and \(\vec{\beta}_2\perp\vec{\alpha}\), so \(\vec{\beta}_1\) is the component of \(\vec{\beta}\) along \(\vec{\alpha}\) and \(-\vec{\beta}_2\) is the component perpendicular to \(\vec{\alpha}\).
\(\vec{\beta}\cdot\vec{\alpha}=6-1+0=5\), \(|\vec{\alpha}|^2=10\), so \(\vec{\beta}_1=\frac{5}{10}\vec{\alpha}=\frac12(3\hat{i}+\hat{j})=\frac32\hat{i}+\frac12\hat{j}\).
\(\vec{\beta}_2=\vec{\beta}_1-\vec{\beta}=-\frac12\hat{i}+\frac32\hat{j}-3\hat{k}\). Check: \(\vec{\beta}_2\cdot\vec{\alpha}=-\frac32+\frac32=0\).
\(\vec{\beta}_1\times\vec{\beta}_2=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ \frac32&\frac12&0\\ -\frac12&\frac32&-3\end{vmatrix}=\hat{i}\left(-\frac32\right)-\hat{j}\left(-\frac92\right)+\hat{k}\left(\frac94+\frac14\right)\)
\(=-\frac32\hat{i}+\frac92\hat{j}+\frac52\hat{k}=\frac12\left(-3\hat{i}+9\hat{j}+5\hat{k}\right)\).
Answer: (2) \(\frac{1}{2}\left(-3\hat{i}+9\hat{j}+5\hat{k}\right)\)
\(\vec{\beta}\cdot\vec{\alpha}=6-1+0=5\), \(|\vec{\alpha}|^2=10\), so \(\vec{\beta}_1=\frac{5}{10}\vec{\alpha}=\frac12(3\hat{i}+\hat{j})=\frac32\hat{i}+\frac12\hat{j}\).
\(\vec{\beta}_2=\vec{\beta}_1-\vec{\beta}=-\frac12\hat{i}+\frac32\hat{j}-3\hat{k}\). Check: \(\vec{\beta}_2\cdot\vec{\alpha}=-\frac32+\frac32=0\).
\(\vec{\beta}_1\times\vec{\beta}_2=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ \frac32&\frac12&0\\ -\frac12&\frac32&-3\end{vmatrix}=\hat{i}\left(-\frac32\right)-\hat{j}\left(-\frac92\right)+\hat{k}\left(\frac94+\frac14\right)\)
\(=-\frac32\hat{i}+\frac92\hat{j}+\frac52\hat{k}=\frac12\left(-3\hat{i}+9\hat{j}+5\hat{k}\right)\).
Answer: (2) \(\frac{1}{2}\left(-3\hat{i}+9\hat{j}+5\hat{k}\right)\)
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