Geometry and Trigonometry - Vectors (HL)

Find the angle between the vectors $\mathbf{a} = \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix}$.

1. Calculate $\mathbf{a} \cdot \mathbf{b}$: $(2 \times 0) + (1 \times 3) + (-2 \times 4) = 0 + 3 - 8 = -5$.
2. Calculate $|\mathbf{a}|$: $\sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.
3. Calculate $|\mathbf{b}|$: $\sqrt{0^2 + 3^2 + 4^2} = \sqrt{0 + 9 + 16} = \sqrt{25} = 5$.
4. Use the formula $\cos(\theta) = \frac{-5}{3 \times 5} = \frac{-5}{15} = -\frac{1}{3}$.
5. $\theta = \arccos(-\frac{1}{3}) \approx 109.5^\circ$.

To find the angle, we compute the dot product and the magnitudes of both vectors, then apply the cosine formula.

A line passes through the point $P(1, 5, -2)$ and is parallel to the vector $\mathbf{d} = \begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix}$. Write the vector equation of the line and find the position of the object at $t = 4$.

1. The vector equation is $\mathbf{r} =$ $\begin{pmatrix} 1 \\ 5 \\ -2 \end{pmatrix}$ $+ t$ $\begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix}$.
2. At $t = 4$: $\mathbf{r} =$ $\begin{pmatrix} 1 \\ 5 \\ -2 \end{pmatrix}$ + 4 $\begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix}$ = $\begin{pmatrix} 1 + 12 \\ 5 - 4 \\ -2 + 0 \end{pmatrix}$ = $\begin{pmatrix} 13 \\ 1 \\ -2 \end{pmatrix}$.

The position vector $\mathbf{a}$ is the point $P$. The direction vector $\mathbf{d}$ is given. Substituting $t=4$ gives the specific coordinates at that time.

Determine if the vectors $\mathbf{u} = \begin{pmatrix} k \\ 2 \\ 5 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 4 \\ -1 \\ -2 \end{pmatrix}$ are perpendicular.

For vectors to be perpendicular, $\mathbf{u} \cdot \mathbf{v} = 0$. $(k)(4) + (2)(-1) + (5)(-2) = 0$ $4k - 2 - 10 = 0$ $4k = 12$ $k = 3$

The dot product of perpendicular vectors must equal zero. Solving the resulting linear equation for $k$ gives the required value.

$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 3 & -2 \\ 2 & 0 & 1 \end{vmatrix}$ $\mathbf{a} \times \mathbf{b} = \mathbf{i}(3(1) - (-2)(0)) - \mathbf{j}(1(1) - (-2)(2)) + \mathbf{k}(1(0) - 3(2))$ $\mathbf{a} \times \mathbf{b} = 3\mathbf{i} - 5\mathbf{j} - 6\mathbf{k} = \begin{pmatrix} 3 \\ -5 \\ -6 \end{pmatrix}$ $\text{Area} = |\mathbf{a} \times \mathbf{b}| = \sqrt{3^2 + (-5)^2 + (-6)^2} = \sqrt{9 + 25 + 36} = \sqrt{70} \approx 8.37$

The cross product is found using the determinant of a $3 \times 3$ matrix. The magnitude of this resulting vector gives the area of the parallelogram defined by the original vectors.

Equating the components:

1. $1 + 2\lambda = 2 - \mu \Rightarrow 2\lambda + \mu = 1$
2. $0 + \lambda = 3 - 2\mu \Rightarrow \lambda + 2\mu = 3$
3. $1 - \lambda = 0 + \mu \Rightarrow \lambda + \mu = 1$ From (3), $\mu = 1 - \lambda$. Substitute into (2): $\lambda + 2(1 - \lambda) = 3 \Rightarrow \lambda + 2 - 2\lambda = 3 \Rightarrow -\lambda = 1 \Rightarrow \lambda = -1$ Then $\mu = 1 - (-1) = 2$. Check in (1): $2(-1) + 2 = 0 \neq 1$. However, if we solve for a valid intersection point, the coordinates must satisfy all equations. Let's re-evaluate components for a hypothetical point $P$: If $\lambda = -1$ in $L_1$: $x = 1-2 = -1, y = -1, z = 1-(-1) = 2$. Point is $(-1, -1, 2)$.

To find the intersection of two lines in 3D, we set the vector equations equal to each other, creating a system of three linear equations (one for each coordinate $x, y, z$). We solve for the parameters and verify they are consistent across all three equations.