Find the scalar and vector products of two vectors $\vec{a} = (3 \hat{i} - 4 \hat{j} + 5 \hat{k})$ and $\vec{b} = (-2 \hat{i} + \hat{j} - 3 \hat{k})$.
The scalar product (dot product) is calculated as:
$\vec{a} \cdot \vec{b} = (3 \hat{i} - 4 \hat{j} + 5 \hat{k}) \cdot (-2 \hat{i} + \hat{j} - 3 \hat{k})$
$= (3)(-2) + (-4)(1) + (5)(-3)$
$= -6 - 4 - 15 = -25$
The vector product (cross product) is calculated using the determinant:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -4 & 5 \\ -2 & 1 & -3 \end{vmatrix}$
$= \hat{i}((-4)(-3) - (5)(1)) - \hat{j}((3)(-3) - (5)(-2)) + \hat{k}((3)(1) - (-4)(-2))$
$= \hat{i}(12 - 5) - \hat{j}(-9 + 10) + \hat{k}(3 - 8)$
$= 7 \hat{i} - \hat{j} - 5 \hat{k}$
$\vec{a} \cdot \vec{b} = (3 \hat{i} - 4 \hat{j} + 5 \hat{k}) \cdot (-2 \hat{i} + \hat{j} - 3 \hat{k})$
$= (3)(-2) + (-4)(1) + (5)(-3)$
$= -6 - 4 - 15 = -25$
The vector product (cross product) is calculated using the determinant:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -4 & 5 \\ -2 & 1 & -3 \end{vmatrix}$
$= \hat{i}((-4)(-3) - (5)(1)) - \hat{j}((3)(-3) - (5)(-2)) + \hat{k}((3)(1) - (-4)(-2))$
$= \hat{i}(12 - 5) - \hat{j}(-9 + 10) + \hat{k}(3 - 8)$
$= 7 \hat{i} - \hat{j} - 5 \hat{k}$
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