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: Start with the given relation.
We are given \[ \vec{x}\times \vec{a}=\vec{b} \] Also, \[ |\vec{x}|=1,\qquad |\vec{a}|=\sqrt{3},\qquad |\vec{b}|=\sqrt{2} \]

Step 2: Take cross product with \(\vec{a}\).
Taking cross product with \(\vec{a}\) on the right side, \[ (\vec{x}\times \vec{a})\times \vec{a} = \vec{b}\times \vec{a} \] Using the vector identity, \[ (\vec{x}\times \vec{a})\times \vec{a} = (\vec{x}\cdot\vec{a})\vec{a} - (\vec{a}\cdot\vec{a})\vec{x} \] Since \[ \vec{a}\cdot\vec{a}=|\vec{a}|^2=3, \] we get \[ (\vec{x}\cdot\vec{a})\vec{a}-3\vec{x} = \vec{b}\times\vec{a} \] Now, \[ \vec{b}\times\vec{a}=-(\vec{a}\times\vec{b}) \] Therefore, \[ (\vec{x}\cdot\vec{a})\vec{a}-3\vec{x} = -\vec{a}\times\vec{b} \] So, \[ 3\vec{x} = (\vec{x}\cdot\vec{a})\vec{a} + \vec{a}\times\vec{b} \] Hence, \[ \vec{x} = \frac{1}{3} \left[ (\vec{x}\cdot\vec{a})\vec{a} + \vec{a}\times\vec{b} \right] \]

Step 3: Find \(\vec{x}\cdot\vec{a}\).
From \[ \vec{x}\times\vec{a}=\vec{b}, \] we have \[ |\vec{x}\times\vec{a}|=|\vec{b}| \] So, \[ |\vec{x}\times\vec{a}|^2=|\vec{b}|^2 \] Using \[ |\vec{x}\times\vec{a}|^2 = |\vec{x}|^2|\vec{a}|^2-(\vec{x}\cdot\vec{a})^2 \] we get \[ 2=1^2\cdot(\sqrt{3})^2-(\vec{x}\cdot\vec{a})^2 \] \[ 2=3-(\vec{x}\cdot\vec{a})^2 \] \[ (\vec{x}\cdot\vec{a})^2=1 \] Thus, \[ \vec{x}\cdot\vec{a}=\pm 1 \]

Step 4: Substitute the value.
Now, \[ \vec{x} = \frac{1}{3} \left[ (\pm 1)\vec{a} + \vec{a}\times\vec{b} \right] \] Therefore, \[ \vec{x} = \frac{1}{3} \left[ \vec{a}\times\vec{b} \pm\vec{a} \right] \]

Step 5: Final conclusion.
Hence, \[ \boxed{\vec{x}=\frac{1}{3}\left[\vec{a}\times\vec{b}\pm\vec{a}\right]} \] Therefore, the correct option is \[ \boxed{(4)\ \frac{1}{3}\left[\vec{a}\times\vec{b}\pm\vec{a}\right]} \]