1. **Problem:** Represent the vector $\mathbf{v}$ using the column approach and the $\mathbf{i}, \mathbf{j}, \mathbf{k}$ unit vector approach. 2. **Formula:** A vector $\mathbf{v} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ can also be written as $x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ where $\mathbf{i} = \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$, $\mathbf{j} = \begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}$, $\mathbf{k} = \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$. 3. **Explanation:** The column approach lists components vertically, while the $\mathbf{i}, \mathbf{j}, \mathbf{k}$ approach expresses the vector as a linear combination of unit vectors along coordinate axes. 4. **Example:** For $\mathbf{v} = \begin{bmatrix}3 \\ -2 \\ 5\end{bmatrix}$, the $\mathbf{i}, \mathbf{j}, \mathbf{k}$ form is $3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k}$. --- 1. **Problem:** Find the dot product and cross product of vectors $\mathbf{a} = \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}$ and $\mathbf{b} = \begin{bmatrix}4 \\ -5 \\ 6\end{bmatrix}$. 2. **Formula:** - Dot product: $\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3$ - Cross product: $\mathbf{a} \times \mathbf{b} = \begin{bmatrix}a_2b_3 - a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_1b_2 - a_2b_1\end{bmatrix}$ 3. **Calculation:** - Dot product: $1\times4 + 2\times(-5) + 3\times6 = 4 - 10 + 18 = 12$ - Cross product: $\begin{bmatrix}2\times6 - 3\times(-5) \\ 3\times4 - 1\times6 \\ 1\times(-5) - 2\times4\end{bmatrix} = \begin{bmatrix}12 + 15 \\ 12 - 6 \\ -5 - 8\end{bmatrix} = \begin{bmatrix}27 \\ 6 \\ -13\end{bmatrix}$ --- 1. **Problem:** Determine the condition for two vectors to be orthogonal using the dot product. 2. **Formula:** Two vectors $\mathbf{a}$ and $\mathbf{b}$ are orthogonal if and only if $\mathbf{a} \cdot \mathbf{b} = 0$. 3. **Explanation:** The dot product measures the cosine of the angle between vectors. If it is zero, the angle is $90^\circ$, meaning vectors are perpendicular. --- 1. **Problem:** Show that the cross product of two vectors is orthogonal to both vectors. 2. **Formula:** For vectors $\mathbf{a}$ and $\mathbf{b}$, $\mathbf{a} \times \mathbf{b}$ is orthogonal to both $\mathbf{a}$ and $\mathbf{b}$. 3. **Verification:** Check $\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 0$ and $\mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) = 0$. 4. **Example:** Using previous $\mathbf{a}$ and $\mathbf{b}$, compute dot products with $\mathbf{a} \times \mathbf{b} = \begin{bmatrix}27 \\ 6 \\ -13\end{bmatrix}$: - $\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 1\times27 + 2\times6 + 3\times(-13) = 27 + 12 - 39 = 0$ - $\mathbf{b} \cdot (\mathbf{a} \times \mathbf{b}) = 4\times27 + (-5)\times6 + 6\times(-13) = 108 - 30 - 78 = 0$ --- 1. **Problem:** Find the condition for two vectors to be parallel. 2. **Formula:** Two vectors $\mathbf{a}$ and $\mathbf{b}$ are parallel if $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ or if one is a scalar multiple of the other. 3. **Explanation:** The cross product measures the area of the parallelogram spanned by vectors. Zero area means vectors lie along the same line. --- 1. **Problem:** Find the vector and Cartesian equations of a line passing through point $\mathbf{P}_0 = (1,2,3)$ with direction vector $\mathbf{d} = \begin{bmatrix}4 \\ -1 \\ 2\end{bmatrix}$. 2. **Formula:** - Vector equation: $\mathbf{r} = \mathbf{r}_0 + t\mathbf{d}$ where $\mathbf{r}_0$ is position vector of $\mathbf{P}_0$. - Cartesian equations: $\frac{x - x_0}{d_1} = \frac{y - y_0}{d_2} = \frac{z - z_0}{d_3}$ 3. **Calculation:** - Vector: $\mathbf{r} = \begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix} + t \begin{bmatrix}4 \\ -1 \\ 2\end{bmatrix}$ - Cartesian: $\frac{x - 1}{4} = \frac{y - 2}{-1} = \frac{z - 3}{2}$ --- 1. **Problem:** Find the point of intersection of lines: $\mathbf{r}_1 = \begin{bmatrix}1 \\ 0 \\ 2\end{bmatrix} + s \begin{bmatrix}2 \\ 3 \\ -1\end{bmatrix}$ and $\mathbf{r}_2 = \begin{bmatrix}3 \\ 1 \\ 0\end{bmatrix} + t \begin{bmatrix}1 \\ -1 \\ 4\end{bmatrix}$. 2. **Method:** Set components equal and solve for $s$ and $t$: $x: 1 + 2s = 3 + t$ $y: 0 + 3s = 1 - t$ $z: 2 - s = 0 + 4t$ 3. **Solution:** From $x$: $t = 1 + 2s$ From $y$: $3s = 1 - t \Rightarrow t = 1 - 3s$ Equate $t$: $1 + 2s = 1 - 3s \Rightarrow 5s = 0 \Rightarrow s=0$ Then $t=1 + 0 = 1$ Check $z$: $2 - 0 = 4 \times 1 \Rightarrow 2 = 4$ (False) 4. **Conclusion:** No solution; lines do not intersect. --- 1. **Problem:** Find the shortest distance between skew lines given by $\mathbf{r}_1 = \mathbf{a} + s\mathbf{d}_1$ and $\mathbf{r}_2 = \mathbf{b} + t\mathbf{d}_2$. 2. **Formula:** Distance $D = \frac{|(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{d}_1 \times \mathbf{d}_2)|}{|\mathbf{d}_1 \times \mathbf{d}_2|}$ 3. **Explanation:** The numerator projects the vector between points on lines onto the normal vector to both directions. --- 1. **Problem:** Find the equation of a plane passing through point $\mathbf{P}_0 = (1, -2, 3)$ with normal vector $\mathbf{n} = \begin{bmatrix}4 \\ 5 \\ -6\end{bmatrix}$. 2. **Formula:** Plane equation: $\mathbf{n} \cdot (\mathbf{r} - \mathbf{r}_0) = 0$ or $4(x-1) + 5(y+2) - 6(z-3) = 0$. 3. **Simplify:** $4x - 4 + 5y + 10 - 6z + 18 = 0 \Rightarrow 4x + 5y - 6z + 24 = 0$. --- 1. **Problem:** Find the line of intersection of planes $4x + 5y - 6z + 24 = 0$ and $x - 2y + 3z - 5 = 0$. 2. **Method:** Solve system for parametric form. 3. **Steps:** - Find direction vector as cross product of normals: $\mathbf{n}_1 = \begin{bmatrix}4 \\ 5 \\ -6\end{bmatrix}$, $\mathbf{n}_2 = \begin{bmatrix}1 \\ -2 \\ 3\end{bmatrix}$ - $\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{bmatrix}5\times3 - (-6)(-2) \\ - (4\times3 - (-6)(1)) \\ 4\times(-2) - 5\times1\end{bmatrix} = \begin{bmatrix}15 - 12 \\ -(12 + 6) \\ -8 - 5\end{bmatrix} = \begin{bmatrix}3 \\ -18 \\ -13\end{bmatrix}$ - Set $z = t$, solve for $x,y$ from plane equations: From first: $4x + 5y - 6t + 24 = 0$ From second: $x - 2y + 3t - 5 = 0$ Solve system: From second: $x = 2y - 3t + 5$ Substitute into first: $4(2y - 3t + 5) + 5y - 6t + 24 = 0$ $8y - 12t + 20 + 5y - 6t + 24 = 0$ $13y - 18t + 44 = 0$ $13y = 18t - 44$ $y = \frac{18t - 44}{13}$ Then $x = 2\times \frac{18t - 44}{13} - 3t + 5 = \frac{36t - 88}{13} - 3t + 5 = \frac{36t - 88 - 39t + 65}{13} = \frac{-3t - 23}{13}$ 4. **Parametric line:** $x = \frac{-3t - 23}{13}$, $y = \frac{18t - 44}{13}$, $z = t$. --- 1. **Problem:** Find the distance from point $\mathbf{P} = (2, -1, 3)$ to plane $4x + 5y - 6z + 24 = 0$. 2. **Formula:** Distance $D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$ where plane is $Ax + By + Cz + D = 0$. 3. **Calculation:** $D = \frac{|4(2) + 5(-1) - 6(3) + 24|}{\sqrt{4^2 + 5^2 + (-6)^2}} = \frac{|8 - 5 - 18 + 24|}{\sqrt{16 + 25 + 36}} = \frac{|9|}{\sqrt{77}} = \frac{9}{\sqrt{77}}$. --- 1. **Problem:** Find the angle between two planes with normals $\mathbf{n}_1$ and $\mathbf{n}_2$. 2. **Formula:** Angle $\theta$ between planes is angle between normals: $\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$. --- 1. **Problem:** Find the angle between a line with direction vector $\mathbf{d}$ and a plane with normal vector $\mathbf{n}$. 2. **Formula:** Angle $\phi$ between line and plane satisfies $\sin \phi = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$. --- **Final note:** These problems cover vector representation, operations, line and plane equations, intersections, distances, and angles.