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: Understanding the Concept:
The cross product of two vectors gives a new vector that is perpendicular to both of them.
Its magnitude can be found by first computing the vector \(\vec a\times\vec b\) using the determinant method, then finding the length of that resulting vector.

Step 2: Key Formula:
For \(\vec a=a_1\hat i+a_2\hat j+a_3\hat k\) and \(\vec b=b_1\hat i+b_2\hat j+b_3\hat k\),
\[ \vec a\times\vec b=\begin{vmatrix}\hat i & \hat j & \hat k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\end{vmatrix} \]

Step 3: Detailed Explanation:
Here \(a_1=2,a_2=1,a_3=3\) and \(b_1=3,b_2=5,b_3=-2\).
\[ \vec a\times\vec b=\begin{vmatrix}\hat i & \hat j & \hat k\\ 2 & 1 & 3\\ 3 & 5 & -2\end{vmatrix} \]
Expanding along the first row gives the vector \(-17\hat i+13\hat j+7\hat k\).
The magnitude is found using \(|\vec v|=\sqrt{x^2+y^2+z^2}\).
\[ |\vec a\times\vec b|=\sqrt{(-17)^2+13^2+7^2}=\sqrt{289+169+49}=\sqrt{507}=13\sqrt3 \]

Final Answer:
The magnitude of the cross product is 13 root 3. \[ \boxed{|\vec a\times\vec b|=13\sqrt3} \]