Problem on proving that the cross product gives a perpendicular vector
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Show that $\bfa \times \bfb$ is perpendicular to $\bfa$ by computing a dot product.
Solution
We recall the relationship of dot products and perpendicular vectors:
Dot product of perpendicular vectors is zero
If two vectors have a dot product of zero, they are perpendicular.
Hence, we need to show that $\bfa \cdot (\bfa \times \bfb) = 0$ for all vectors $\bfa$ and $\bfb$.
In order to directly compute the dot product, we need to find the components of all vectors involved.
Let $\bfa = \langle a_1, a_2, a_3\rangle$ and $\bfb = \langle b_1, b_2, b_3 \rangle.$
Recall that
Cross product definition
Determinant of a 3x3 matrix
\begin{align}\mathbf{a} \times \mathbf{b} &= \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}\\ \ \\ &= \begin{vmatrix}a_2 & a_3 \\ b_2 & b_3 \end{vmatrix} \ \mathbf{i} - \begin{vmatrix}a_1 & a_3 \\ b_1 & b_3 \end{vmatrix} \ \mathbf{j} + \begin{vmatrix}a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} \ \mathbf{k} \end{align}
Determinant of a 2x2 matrix
Computing the $2 \times 2$ determinants, we have $$\bfa \times \bfb = (a_2 b_3 - a_3 b_2) \bfi - (a_1 b_3 - a_3 b_1) \bfj + (a_1 b_2 - a_2 b_1) \bfk.$$
To compute the dot product with $\bfa$, recall that
Direct computation of dot product
If $\mathbf{x} = \langle x_1, x_2, x_3 \rangle$ and $\mathbf{y} = \langle y_1, y_2, y_3 \rangle$, the dot product is defined as $$ \mathbf{x} \cdot \mathbf{y} = x_1 y_1 + x_2 y_2 + x_3 y_3.$$
Hence, \begin{align} \bfa \cdot (\bfa \times \bfb) = &a_1(a_2 b_3 - a_3 b_2) - a_2(a_1 b_3 - a_3 b_1) \\&+ a_3(a_1 b_2 - a_2 b_1).\end{align}
Expanding, we get that \begin{align}\bfa\cdot(\bfa \times \bfb) = &a_1 a_2 b_3 - a_1 a_3 b_2 - a_1 a_2 b_3 + a_2 a_3 b_1 \\&+ a_1 a_3 b_2 - a_2 a_3 b_1.\end{align}
Which simplifies to $$\bfa \cdot (\bfa \times \bfb) = 0.$$
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