Question:easy

If \[ |\vec{a}|=13,\qquad |\vec{b}|=5 \] and \[ \vec{a}\cdot\vec{b}=60, \] then \[ |\vec{a}\times\vec{b}|= \]

Show Hint

Remember the identity \[ |\vec{a}\times\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2 \] which is very useful for finding cross products when magnitudes and dot products are known.
Updated On: Jun 25, 2026
  • \(15\)
  • \(20\)
  • \(30\)
  • \(25\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Recall the fundamental vector identity.
For any two vectors $ \vec{a} $ and $ \vec{b} $, we have the identity: \[ |\vec{a}|^2 |\vec{b}|^2 = (\vec{a}\cdot\vec{b})^2 + |\vec{a}\times\vec{b}|^2 \] This follows from $ \vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta $ and $ |\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta $.
Step 2: Substitute the known values.
Given $ |\vec{a}| = 13 $, $ |\vec{b}| = 5 $, $ \vec{a}\cdot\vec{b} = 60 $. So: \[ 13^2 \times 5^2 = 60^2 + |\vec{a}\times\vec{b}|^2 \] \[ 169 \times 25 = 3600 + |\vec{a}\times\vec{b}|^2 \]
Step 3: Compute the left-hand side.
$ 169 \times 25 = 4225 $. So the equation becomes: \[ 4225 = 3600 + |\vec{a}\times\vec{b}|^2 \]
Step 4: Isolate $ |\vec{a}\times\vec{b}|^2 $.
\[ |\vec{a}\times\vec{b}|^2 = 4225 - 3600 = 625 \]
Step 5: Take the square root.
Since magnitude is non-negative, \[ |\vec{a}\times\vec{b}| = \sqrt{625} = 25 \]
Step 6: State the answer.
\[ \boxed{25} \]
Was this answer helpful?
0