Question

\(\vec{a},\vec{b}\) are two vectors such that \(|\vec{a}|=\sqrt{3}\), \(|\vec{b}|=\sqrt{2}\). If \(\vec{x}\) is a unit vector satisfying \(\vec{x}\times \vec{a}=\vec{b}\), then \(\vec{x}=\)

• A. \(\dfrac{1}{2}\left[(\vec{x}\cdot\vec{a})\vec{a}-\vec{b}\times\vec{a}\right]\)
• B. \(\dfrac{1}{2}\left[\pm(\vec{x}\cdot\vec{a})\vec{a}+(\vec{b}\times\vec{a})\right]\)
• C. \(\dfrac{1}{3}\left[(\vec{x}\cdot\vec{a})\vec{a}+\vec{b}\times\vec{a}\right]\)
• D. \(\dfrac{1}{3}\left[\vec{a}\times\vec{b}\pm\vec{a}\right]\)

Hint:

For vector equations involving cross products, use the identity \[ (\vec{p}\times\vec{q})\times\vec{r} = (\vec{p}\cdot\vec{r})\vec{q} - (\vec{q}\cdot\vec{r})\vec{p} \] and the formula \[ |\vec{p}\times\vec{q}|^2 = |\vec{p}|^2|\vec{q}|^2-(\vec{p}\cdot\vec{q})^2. \]

Answer 1

The Correct Option is D

Step 1: Take cross product of both sides with \(\vec{a}\).
\((\vec{x}\times\vec{a})\times\vec{a} = \vec{b}\times\vec{a}\). Using the identity \((\vec{x}\times\vec{a})\times\vec{a} = (\vec{x}\cdot\vec{a})\vec{a} - |\vec{a}|^2\vec{x}\): \[(\vec{x}\cdot\vec{a})\vec{a} - 3\vec{x} = \vec{b}\times\vec{a}.\]

Step 2: Solve for \(\vec{x}\).
\[\vec{x} = \frac{(\vec{x}\cdot\vec{a})\vec{a} - \vec{b}\times\vec{a}}{3} = \frac{1}{3}\left[(\vec{x}\cdot\vec{a})\vec{a} + \vec{a}\times\vec{b}\right].\] Since \(\vec{x}\cdot\vec{a}\) can be \(\pm|\vec{x}||\vec{a}|\cos\phi=\pm\sqrt{3}\cos\phi\), the general form is \(\tfrac{1}{3}[\vec{a}\times\vec{b}\pm\vec{a}]\) (after computing the scalar).
\[\boxed{\frac{1}{3}[\vec{a}\times\vec{b} \pm \vec{a}]}\]