Concept:

• Cross Product: For two vectors \(\rm \vec A\) and \(\rm \vec B\) at an angle θ to each other, the cross product is defined as:

  \(\rm \vec A\times \vec B=\vec n|\vec A||\vec B|\sin \theta\), where \(\rm \vec n\) is the unit vector perpendicular to the plane containing the vectors \(\rm \vec A\) and \(\rm \vec B\).

• If \(\rm \vec A = a_1\hat i +a_2\hat j+ a_3\hat k\) and \(\rm \vec B = b_1\hat i +b_2\hat j+b_3 \hat k\), then their cross product is:

  \(\rm \vec A\times\vec B=\begin{vmatrix} \rm \hat i & \rm \hat j & \rm \hat k \\ \rm a_1 & \rm a_2 & \rm a_3 \\ \rm b_1 & \rm b_2 & \rm b_3\end{vmatrix}\).

• The unit vector \(\rm \vec u\) in the direction of a vector \(\rm \vec A\), can be calculated as:

  \(\rm \vec u = \frac{\vec A}{\left| \vec A\right|}\), where \(\rm \left| \vec A\right|\) is the magnitude (length) of the vector \(\rm \vec A\).

Calculation:

We have \(\rm \vec A = 4\hat i +3\hat j+ \hat k\) and \(\rm \vec B = 2\hat i -\hat j+2 \hat k\). Therefore, their cross product will be:

\(\rm \vec A\times\vec B=\begin{vmatrix} \rm \hat i & \ \ \ \rm \hat j & \rm \hat k \\ 4& \ \ \ 3&1\\2&-1&2\end{vmatrix}\)

Expanding along R₁, we get:

= (6 + 1)î + (2 - 8)ĵ + (-4 - 6)k̂

= 7î - 6ĵ - 10k̂

The magnitude of \(\rm \vec A\times \vec B\) is:

\(\rm \left| \vec A\times\vec B\right|=\sqrt{7^2+(-6)^2+(-10)^2}\)

= \(\rm \sqrt{49+36+100}\)

= \(\rm \sqrt{185}\)

The unit vector N̂ along \(\rm \vec A\times\vec B\) will be:

N̂ = \(\rm \frac{\vec A\times\vec B}{\left| \vec A\times\vec B\right|}\)

⇒ N̂ = \(\rm \frac{1}{\sqrt{185}}\left(7\hat{i}-6\hat{j}-10\hat{k}\right)\).