Question

If \[ |\vec a|=2k, \qquad |\vec b|=k \] and \[ |\vec a-\vec b|^2=20k^2-|2\vec a+\vec b|^2, \] then \[ |\vec a\times\vec b| = \ ? \]

• A. \(\sqrt3\,k^2\)
• B. \(k^2\)
• C. \(2\sqrt3\,k^2\)
• D. \(2k^2\)

Hint:

Whenever expressions such as \[ |\vec a+\vec b|^2 \quad\text{or}\quad |\vec a-\vec b|^2 \] appear, expand them using dot products first. The angle between vectors can then be obtained easily.

Answer 1

The Correct Option is A

Step 1: Note what we need.
We want $|\vec a\times\vec b|=|\vec a||\vec b|\sin\theta$. We know $|\vec a|=2k$ and $|\vec b|=k$, so we just need $\sin\theta$, which means we first need $\vec a\cdot\vec b$.

Step 2: Expand the first squared term.
$|\vec a-\vec b|^2=|\vec a|^2+|\vec b|^2-2\vec a\cdot\vec b=4k^2+k^2-2\vec a\cdot\vec b=5k^2-2\vec a\cdot\vec b$.

Step 3: Expand the second squared term.
$|2\vec a+\vec b|^2=4|\vec a|^2+|\vec b|^2+4\vec a\cdot\vec b=16k^2+k^2+4\vec a\cdot\vec b=17k^2+4\vec a\cdot\vec b$.

Step 4: Use the given equation.
The condition is $|\vec a-\vec b|^2=20k^2-|2\vec a+\vec b|^2$. So \[ 5k^2-2\vec a\cdot\vec b=20k^2-(17k^2+4\vec a\cdot\vec b). \]

Step 5: Solve for $\vec a\cdot\vec b$.
The right side is $3k^2-4\vec a\cdot\vec b$. So $5k^2-2\vec a\cdot\vec b=3k^2-4\vec a\cdot\vec b$, giving $2\vec a\cdot\vec b=-2k^2$, hence $\vec a\cdot\vec b=-k^2$.

Step 6: Find $\cos\theta$ and $\sin\theta$.
$\cos\theta=\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}=\dfrac{-k^2}{2k\cdot k}=-\dfrac12$. So $\theta=120^\circ$ and $\sin\theta=\dfrac{\sqrt3}{2}$.

Step 7: Compute the cross product magnitude.
\[ |\vec a\times\vec b|=2k\cdot k\cdot\frac{\sqrt3}{2}=\sqrt3\,k^2. \] \[ \boxed{\sqrt3\,k^2} \]