Answers: Vector addition and scalar multiplication
These are the answers to Questions: Addition and scalar multiplication.
Please attempt the questions before reading these answers!
Q1
1.1. For the \(\mathbf{i}\) component, \(4 + 8 = 12\).For the \(\mathbf{j}\) component, \(5 + 2 = 7\).For the \(\mathbf{k}\) component, \(7 + 4 = 11\). So the answer is \(\mathbf{a} + \mathbf{b} = 12\mathbf{i} + 7\mathbf{j} + 11\mathbf{k}\).
1.2. \(\mathbf{a} + \mathbf{b} = 2\mathbf{i} + 3\mathbf{j} + 9\mathbf{k}\).
1.3. \(\mathbf{a} - \mathbf{b} = 2\mathbf{i} - 11\mathbf{j} + 14\mathbf{k}\).
1.4. You can solve this by doing addition componentwise. \(\mathbf{i}\) component: \(4- (3+11) = -10\), \(\mathbf{j}\) component: \(12- (-3-4) = 19\), \(\mathbf{k}\) component: \(-7-(-2+9) = -14\). So the answer is \(-10\mathbf{i} + 19\mathbf{j} -14\mathbf{k}\).
Q2
2.1. \(\mathbf{a} + \mathbf{b} = \begin{bmatrix}4x \\7y\\0\end{bmatrix}\)
2.2. \(\mathbf{a} - \mathbf{b} = \begin{bmatrix}7 \\ 3y - 2x \\ -z\end{bmatrix}\)
2.3. \(\mathbf{a} + \mathbf{b} - \mathbf{c} = \mathbf{0}\) or \(\begin{bmatrix}0\\0\\0\end{bmatrix}\).
2.4. \(\mathbf{a}\).
Q3
3.1. \(3\mathbf{u} = (3)5\mathbf{j} + (3)6\mathbf{k} = 15\mathbf{j} + 18\mathbf{k}\).
3.2. \(-6\mathbf{v} = \begin{bmatrix}0\\18\\-42\end{bmatrix}\).
3.3. \(4\mathbf{v} - 3\mathbf{u} = \begin{bmatrix}0\\-27\\10\end{bmatrix}\)
3.4. \(-2\mathbf{w} - (4\mathbf{u} -2\mathbf{v}) = \begin{bmatrix}-4\\-32\\-2\end{bmatrix}\)
Q4
4.1. By the laws of vector addition, \(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\overrightarrow{OA} + \overrightarrow{OB}\), where \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\) are the respective coordinates of \(A\) and \(B\) written in vector form. You can find \(\overrightarrow{AB}\) by solving the above equation. \(\overrightarrow{AB} = \begin{bmatrix}-2-3\\5-4\\7-5\end{bmatrix} = \begin{bmatrix}-5\\1\\2\end{bmatrix}\)
4.2.\(\overrightarrow{AB} = \begin{bmatrix}4\\6\\0\end{bmatrix}\), \(\overrightarrow{AC} = \begin{bmatrix}-2\\-4\\-5\end{bmatrix}\). \(\overrightarrow{AB} - \overrightarrow{AC} = \begin{bmatrix}6\\10\\5\end{bmatrix}\). You can also calculate this by noticing \(\overrightarrow{AB} - \overrightarrow{AC} = \overrightarrow{CA} + \overrightarrow{AB} = \overrightarrow{CB}\). Then \(\overrightarrow{CB} = \begin{bmatrix}6-0\\11-1\\7-2\end{bmatrix} = \begin{bmatrix}6\\10\\5\end{bmatrix}\) as required.
4.3. \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\). \(\begin{bmatrix}1\\5\\9\end{bmatrix} - \begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix} = \begin{bmatrix}6\\7\\-2\end{bmatrix}\). Solving this gives \(A = (-5,-2,11)\).
4.4. Let \(\lambda\) and \(\mu\) be scalars. \(\lambda \mathbf{a} + \mu \mathbf{b} = 13\mathbf{i} -9\mathbf{j}\). This gives you the simultaneous equations \[\begin{align*} 2\lambda + 3\mu &= 13 \tag{\textsf{$\mathbf{i}$ component}}\\ 3\lambda -5\mu &= -9 \tag{\textsf{$\mathbf{j}$ component}} \end{align*}\] Solving this gives \(\mu = 3\), \(\lambda = 2\), which gives the answer \(2\mathbf{a} + 3\mathbf{b}\).
4.5. \(2\begin{bmatrix}2\\5\\z\end{bmatrix} + 3\begin{bmatrix}-1\\-3\\4\end{bmatrix} = \begin{bmatrix}x\\y\\0\end{bmatrix}\). Solving this gives \(x = 3\), \(y = 1\) and \(z = -6\).
4.6. As they are parallel \(\mathbf{a} = \lambda \mathbf{b}\) for some real scalar \(\lambda\). This gives the simultaneous equations \[\begin{align*} x - 7 &= -2\lambda \tag{\textsf{$\mathbf{i}$ component}}\\ 5x+1 &= 8\lambda \tag{\textsf{$\mathbf{k}$ component}} \end{align*}\] Eliminating \(\lambda\) and solving gives \(x = 3\).
Version history and licensing
v1.0: initial version created 08/23 by Renee Knapp, Kin Wang Pang as part of a University of St Andrews STEP project.
- v1.1: edited 05/24 by tdhc.