Question

If \(\vec a=2\hat i+\hat j+3\hat k\) and \(\vec b=3\hat i+5\hat j-2\hat k\), find \(|\vec a\times\vec b|\).

Hint:

Use the determinant method to find a cross b, then take its magnitude with the square root of the sum of squares.

Answer 1

Step 1: A Different Route Using the Lagrange Identity:
Instead of computing the cross product vector directly, use the identity linking cross product magnitude to dot products.
This identity states $|\vec a\times\vec b|^2=|\vec a|^2|\vec b|^2-(\vec a\cdot\vec b)^2$.

Step 2: Compute the Needed Dot Products:
Find $|\vec a|^2$, $|\vec b|^2$ and $\vec a\cdot\vec b$ separately.
$|\vec a|^2=2^2+1^2+3^2=14$
$|\vec b|^2=3^2+5^2+(-2)^2=38$
$\vec a\cdot\vec b=(2)(3)+(1)(5)+(3)(-2)=6+5-6=5$

Step 3: Apply the Identity:
Substitute these values into the identity.
\[ |\vec a\times\vec b|^2=14\times38-5^2=532-25=507 \]
Taking the square root gives the required magnitude.
\[ |\vec a\times\vec b|=\sqrt{507}=13\sqrt3 \]

Final Answer:
Both methods confirm the same magnitude, 13 root 3. \[ \boxed{|\vec a\times\vec b|=13\sqrt3} \]