If \(\vec{a}=2\hat{\imath}+3\hat{\jmath}+\hat{k}\) and \(\vec{b}=2\hat{\imath}-\hat{\jmath}+2\hat{k}\), then the vector component of \(\vec{a}\) along \(\vec{b}\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle \frac{1}{3}(2\hat{\imath}-\hat{\jmath}+2\hat{k}) \)
The vector component of \(\vec{a}\) along \(\vec{b}\) is \(\displaystyle\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\,\vec{b}\).
\(\vec{a}\cdot\vec{b} = 4-3+2 = 3\)
\(|\vec{b}|^2 = 4+1+4 = 9\)
So the vector component is
\( \displaystyle \frac{3}{9}(2\hat{\imath}-\hat{\jmath}+2\hat{k}) = \frac{1}{3}(2\hat{\imath}-\hat{\jmath}+2\hat{k}) \)
Why the other options are incorrect
Option (2) \(2\hat{\imath}-\hat{\jmath}+2\hat{k}\) — Incorrect. It is \(\vec{b}\) itself; it leaves out the factor \(\frac{3}{9} = \frac{1}{3}\).
Option (3) \( \displaystyle \frac{1}{9}(2\hat{\imath}-\hat{\jmath}+2\hat{k}) \) — Incorrect. It divides by \(|\vec{b}|^2 = 9\) but leaves out \(\vec{a}\cdot\vec{b} = 3\) in the numerator.
Option (4) \( \displaystyle \frac{3}{14}(2\hat{\imath}+3\hat{\jmath}+\hat{k}) \) — Incorrect. It is \(\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|^2}\,\vec{a}\), the component of \(\vec{b}\) along \(\vec{a}\).