Geometry and Trigonometry - Vectors: Algebraic representation (HL)

Given vectors $\mathbf{a} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 1 \\ 0 \\ 4 \end{pmatrix}$, find the scalar product $\mathbf{a} \cdot \mathbf{b}$ and the angle $\theta$ between them.

$\mathbf{a} \cdot \mathbf{b} = (2)(1) + (-1)(0) + (3)(4) = 2 + 0 + 12 = 14$ $|\mathbf{a}| = \sqrt{2^2 + (-1)^2 + 3^2} = \sqrt{14}$ $|\mathbf{b}| = \sqrt{1^2 + 0^2 + 4^2} = \sqrt{17}$ $\cos \theta = \frac{14}{\sqrt{14}\sqrt{17}} = \frac{\sqrt{14}}{\sqrt{17}}$ $\theta = \arccos\left(\sqrt{\frac{14}{17}}\right) \approx 25.1^\circ$

We use the algebraic definition of the dot product to find its value, calculate the magnitudes of both vectors, and then use the geometric definition $\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ to find the angle.

Find a vector perpendicular to both $\mathbf{u} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}$ and $\mathbf{v} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}$ using the cross product.

$\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 2 & -1 & 3 \end{vmatrix}$ $= \mathbf{i}(2(3) - (-1)(-1)) - \mathbf{j}(1(3) - (-1)(2)) + \mathbf{k}(1(-1) - 2(2))$ $= \mathbf{i}(6 - 1) - \mathbf{j}(3 + 2) + \mathbf{k}(-1 - 4)$ $= 5\mathbf{i} - 5\mathbf{j} - 5\mathbf{k}$

The cross product of two vectors yields a third vector that is orthogonal (perpendicular) to the plane containing the original two. We expand the determinant along the first row.

Calculate the area of a triangle with vertices $A(1, 0, 2)$, $B(3, -1, 5)$, and $C(1, 4, 0)$.

$\vec{AB} =$\begin{pmatrix} 3-1 \ -1-0 \ 5-2 \end{pmatrix}$=$\begin{pmatrix} 2 \ -1 \ 3 \end{pmatrix}$$\vec{AC} = \begin{pmatrix} 1-1 \ 4-0 \ 0-2 \end{pmatrix}$=$\begin{pmatrix} 0 \ 4 \ -2 \end{pmatrix}$$\vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -1 & 3 \ 0 & 4 & -2 \end{vmatrix} = \begin{pmatrix} (-1)(-2) - (3)(4) \ -(2(-2) - 0) \ 2(4) - 0 \end{pmatrix}$= \begin{pmatrix} -10 \\ 4 \\ 8 \end{pmatrix}$ $\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \sqrt{(-10)^2 + 4^2 + 8^2} = \frac{1}{2} \sqrt{100 + 16 + 64} = \frac{1}{2} \sqrt{180} = 3\sqrt{5}$

First, find two displacement vectors originating from the same vertex. Then calculate their cross product. The area of the triangle is half the magnitude of this cross product vector.

1. Find the magnitude: $|\mathbf{w}| = \sqrt{3^2 + (-4)^2 + 12^2}$ $|\mathbf{w}| = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$
2. Find the unit vector: $\hat{\mathbf{w}} = \frac{1}{|\mathbf{w}|} \mathbf{w} = \frac{1}{13} \begin{pmatrix} 3 \\ -4 \\ 12 \end{pmatrix}$ $\hat{\mathbf{w}} = \begin{pmatrix} \frac{3}{13} \\ -\frac{4}{13} \\ \frac{12}{13} \end{pmatrix}$

The magnitude is the geometric length of the vector. Dividing the vector by its magnitude scales it to a length of 1 (a unit vector) while maintaining its direction.

1. Calculate the cross product $\mathbf{p} \times \mathbf{q}$: $\mathbf{p} \times \mathbf{q} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 0 & 0 \\ 0 & 3 & 0 \end{vmatrix}$ $= \mathbf{i}(0-0) - \mathbf{j}(0-0) + \mathbf{k}(12-0) = \begin{pmatrix} 0 \\ 0 \\ 12 \end{pmatrix}$
2. The area is the magnitude of the cross product: $\text{Area} = |\mathbf{p} \times \mathbf{q}| = \sqrt{0^2 + 0^2 + 12^2} = 12$

Since $\mathbf{p}$ lies on the x-axis and $\mathbf{q}$ lies on the y-axis, they form a rectangle in the xy-plane. The cross product points in the z-direction (perpendicular to the plane), and its magnitude represents the area ($4 \times 3 = 12$).