Magnitude of a vector and unit vectors Problem set B Worked solutions

Find the exact magnitude of each vector.

1. (a) \(\begin{pmatrix}6 \\ -6 \\ 7\end{pmatrix}\)

   • \(|\mathbf{a}|=\sqrt{6^2+(-6)^2+7^2}=\sqrt{121}=11\)

2. (b) \(\begin{pmatrix}2 \\ -4 \\ -7\end{pmatrix}\)

   • \(|\mathbf{a}|=\sqrt{2^2+(-4)^2+(-7)^2}=\sqrt{69}\)

Find the exact value of \(|\mathbf{a}|\) for \(\mathbf{a}=12\mathbf{i}-4\mathbf{j}+3\mathbf{k}\).

• \(|\mathbf{a}|=\sqrt{12^2+(-4)^2+3^2}=\sqrt{169}=13\)
• Square each coefficient: \(12^2\), \((-4)^2\), \(3^2\).

Write down the two unit vectors parallel to \(\begin{pmatrix}-6 \\ 2 \\ 3\end{pmatrix}\).

• \(|\mathbf{a}|=\sqrt{(-6)^2+2^2+3^2}=\sqrt{49}=7\)
• \(\frac{1}{7}\begin{pmatrix}-6 \\ 2 \\ 3\end{pmatrix}=\begin{pmatrix}- \frac{6}{7} \\ \frac{2}{7} \\ \frac{3}{7}\end{pmatrix}\)
• The other one is the negative: \(\begin{pmatrix}\frac{6}{7} \\ - \frac{2}{7} \\ - \frac{3}{7}\end{pmatrix}\).

Find a unit vector in the same direction as each vector.

1. (a) \(6\mathbf{i}-7\mathbf{j}+6\mathbf{k}\)

   • \(|\mathbf{a}|=\sqrt{6^2+(-7)^2+6^2}=\sqrt{121}=11\)
   • Divide each coefficient by \(11\).

2. (b) \(3\mathbf{i}+\mathbf{j}\)

   • \(|\mathbf{a}|=\sqrt{3^2+1^2+0^2}=\sqrt{10}\)
   • Divide each coefficient by \(\sqrt{10}\).

A submarine dives in the direction of \(\begin{pmatrix}2 \\ 3 \\ 6\end{pmatrix}\) at a speed of 21 m/s. Find its velocity vector.

• \(|\mathbf{d}|=\sqrt{2^2+3^2+6^2}=\sqrt{49}=7\)
• \(21\times\frac{1}{7}\begin{pmatrix}2 \\ 3 \\ 6\end{pmatrix}=\begin{pmatrix}6 \\ 9 \\ 18\end{pmatrix}\)
• Check: the magnitude of the answer is \(21\).

A crane cable pulls a container with a force of magnitude 18 kN in the direction of \(\mathbf{i}-2\mathbf{j}+\mathbf{k}\). Write the force as a vector.

• \(|\mathbf{d}|=\sqrt{1^2+(-2)^2+1^2}=\sqrt{6}\)
• \(\frac{18}{\sqrt{6}}\begin{pmatrix}1 \\ -2 \\ 1\end{pmatrix}\)
• \(=\begin{pmatrix}3 \sqrt{6} \\ - 6 \sqrt{6} \\ 3 \sqrt{6}\end{pmatrix}\)

A hot-air balloon launches from a field. Relative to the launch point its displacement is \(\begin{pmatrix}-7 \\ 15 \\ 9\end{pmatrix}\) metres, where \(\mathbf{k}\) is vertically up. How far is the balloon from the launch point? Give your answer to 3 significant figures.

• \(|\mathbf{d}|=\sqrt{(-7)^2+15^2+9^2}=\sqrt{355}=18.841\ldots\)
• To 3 significant figures: 18.8 m.

The vector \(\mathbf{u}=\begin{pmatrix}2k \\ 3k \\ 6k\end{pmatrix}\) is a unit vector. Find the possible values of \(k\).

• \(|\mathbf{u}|=\sqrt{4k^2+9k^2+36k^2}=7|k|\)
• A unit vector has magnitude 1, so \(7|k|=1\).
• \(k=\pm\frac{1}{7}\)

Find the two vectors of magnitude 8 that are parallel to \(\begin{pmatrix}2 \\ 0 \\ -1\end{pmatrix}\).

• \(|\mathbf{a}|=\sqrt{2^2+0^2+(-1)^2}=\sqrt{5}\)
• \(8\times\frac{1}{\sqrt{5}}\begin{pmatrix}2 \\ 0 \\ -1\end{pmatrix}=\begin{pmatrix}\frac{16 \sqrt{5}}{5} \\ 0 \\ - \frac{8 \sqrt{5}}{5}\end{pmatrix}\)
• The other vector is the negative: \(\begin{pmatrix}- \frac{16 \sqrt{5}}{5} \\ 0 \\ \frac{8 \sqrt{5}}{5}\end{pmatrix}\).