Adding and subtracting vectors Problem set B Worked solutions

Given \(\mathbf{p}=\begin{pmatrix}-5\\8\\1\end{pmatrix}\) and \(\mathbf{q}=\begin{pmatrix}3\\-2\\7\end{pmatrix}\), find:

1. (a) \(\mathbf{p}-\mathbf{q}\)

   • \(\mathbf{p}-\mathbf{q}=\mathbf{p}+(-\mathbf{q})=\begin{pmatrix}-5\\8\\1\end{pmatrix}+\begin{pmatrix}-3\\2\\-7\end{pmatrix}\)
   • \(=\begin{pmatrix}-8\\10\\-6\end{pmatrix}\)

2. (b) \(\mathbf{q}-\mathbf{p}\)

   • \(\mathbf{q}-\mathbf{p}=\begin{pmatrix}3\\-2\\7\end{pmatrix}+\begin{pmatrix}5\\-8\\-1\end{pmatrix}\)
   • \(=\begin{pmatrix}8\\-10\\6\end{pmatrix}\)

Given \(\mathbf{s}=\begin{pmatrix}-8\\5\end{pmatrix}\) and \(\mathbf{t}=\begin{pmatrix}2\\-7\end{pmatrix}\), find the resultant of the two vectors and find the vector \(\mathbf{s}-\mathbf{t}\).

1. (a) The resultant \(\mathbf{s}+\mathbf{t}\)

   • Add the components, row by row
   • \(\mathbf{s}+\mathbf{t}=\begin{pmatrix}-6\\-2\end{pmatrix}\)

2. (b) \(\mathbf{s}-\mathbf{t}\)

   • Subtracting is adding the negative of the second vector
   • \(\mathbf{s}-\mathbf{t}=\begin{pmatrix}-8\\5\end{pmatrix}+\begin{pmatrix}-2\\7\end{pmatrix}\)
   • \(=\begin{pmatrix}-10\\12\end{pmatrix}\)

In the diagram, \(\overrightarrow{PQ}=\mathbf{x}\), \(\overrightarrow{QR}=\mathbf{y}\) and \(\overrightarrow{RS}=\mathbf{z}\). Write each of these in terms of \(\mathbf{x}\), \(\mathbf{y}\) and \(\mathbf{z}\).

1. (a) \(\overrightarrow{PR}\)

   • \(\overrightarrow{PR}=\overrightarrow{PQ}+\overrightarrow{QR}\)
   • \(\overrightarrow{PR}=\mathbf{x}+\mathbf{y}\)

2. (b) \(\overrightarrow{QS}\)

   • \(\overrightarrow{QS}=\overrightarrow{QR}+\overrightarrow{RS}\)
   • \(\overrightarrow{QS}=\mathbf{y}+\mathbf{z}\)

3. (c) \(\overrightarrow{PS}\)

   • \(\overrightarrow{PS}=\overrightarrow{PQ}+\overrightarrow{QR}+\overrightarrow{RS}\)
   • \(\overrightarrow{PS}=\mathbf{x}+\mathbf{y}+\mathbf{z}\)

In the diagram, \(\overrightarrow{PQ}=\mathbf{x}\), \(\overrightarrow{QR}=\mathbf{y}\) and \(\overrightarrow{RS}=\mathbf{z}\). Write each of these in terms of \(\mathbf{x}\), \(\mathbf{y}\) and \(\mathbf{z}\).

1. (a) \(\overrightarrow{SP}\)

   • \(\overrightarrow{PS}=\overrightarrow{PQ}+\overrightarrow{QR}+\overrightarrow{RS}=\mathbf{x}+\mathbf{y}+\mathbf{z}\)
   • \(\overrightarrow{SP}=-\overrightarrow{PS}=-(\mathbf{x}+\mathbf{y}+\mathbf{z})=-\mathbf{x}-\mathbf{y}-\mathbf{z}\)

2. (b) \(\overrightarrow{RP}\)

   • \(\overrightarrow{PR}=\overrightarrow{PQ}+\overrightarrow{QR}=\mathbf{x}+\mathbf{y}\)
   • \(\overrightarrow{RP}=-\overrightarrow{PR}=-(\mathbf{x}+\mathbf{y})=-\mathbf{x}-\mathbf{y}\)

3. (c) \(\overrightarrow{SQ}\)

   • \(\overrightarrow{QS}=\overrightarrow{QR}+\overrightarrow{RS}=\mathbf{y}+\mathbf{z}\)
   • \(\overrightarrow{SQ}=-\overrightarrow{QS}=-(\mathbf{y}+\mathbf{z})=-\mathbf{y}-\mathbf{z}\)

A helicopter flies from a base H to a village J, a displacement of \(\begin{pmatrix}-4\\9\end{pmatrix}\) km, and then from J to a hospital K, a displacement of \(\begin{pmatrix}10\\-6\end{pmatrix}\) km. The first component is east and the second is north.

1. (a) Find the vector \(\overrightarrow{HK}\) from H to K.

   • \(\overrightarrow{HK}=\overrightarrow{HJ}+\overrightarrow{JK}=\begin{pmatrix}-4\\9\end{pmatrix}+\begin{pmatrix}10\\-6\end{pmatrix}\)
   • \(=\begin{pmatrix}6\\3\end{pmatrix}\)

2. (b) Find the vector \(\overrightarrow{KH}\) from K back to H.

   • \(\overrightarrow{KH}=-\overrightarrow{HK}\)
   • \(=\begin{pmatrix}-6\\-3\end{pmatrix}\)

From a lighthouse L, the displacement to a buoy M is \(\begin{pmatrix}-3\\6\end{pmatrix}\) km and the displacement to a wreck N is \(\begin{pmatrix}8\\-2\end{pmatrix}\) km (east, north).

1. (a) Find the vector \(\overrightarrow{MN}\) from M to N.

   • \(\overrightarrow{MN}=\overrightarrow{ML}+\overrightarrow{LN}=-\overrightarrow{LM}+\overrightarrow{LN}\)
   • \(=\begin{pmatrix}3\\-6\end{pmatrix}+\begin{pmatrix}8\\-2\end{pmatrix}\)
   • \(=\begin{pmatrix}11\\-8\end{pmatrix}\)

2. (b) Find the vector \(\overrightarrow{NM}\) from N to M.

   • \(\overrightarrow{NM}=-\overrightarrow{MN}\)
   • \(=\begin{pmatrix}-11\\8\end{pmatrix}\)

In the diagram, \(\overrightarrow{PQ}=\mathbf{x}\), \(\overrightarrow{QR}=\mathbf{y}\) and \(\overrightarrow{SR}=\mathbf{z}\). Write each of these in terms of \(\mathbf{x}\), \(\mathbf{y}\) and \(\mathbf{z}\).

1. (a) \(\overrightarrow{PS}\)

   • \(\overrightarrow{PS}=\overrightarrow{PQ}+\overrightarrow{QR}-\overrightarrow{SR}\)
   • \(\overrightarrow{PS}=\mathbf{x}+\mathbf{y}-\mathbf{z}\)

2. (b) \(\overrightarrow{QS}\)

   • \(\overrightarrow{QS}=\overrightarrow{QR}-\overrightarrow{SR}\)
   • \(\overrightarrow{QS}=\mathbf{y}-\mathbf{z}\)

3. (c) \(\overrightarrow{SP}\)

   • \(\overrightarrow{SP}=\overrightarrow{SR}-\overrightarrow{QR}-\overrightarrow{PQ}\)
   • \(\overrightarrow{SP}=\mathbf{z}-\mathbf{y}-\mathbf{x}\)
   • \(\overrightarrow{SP}=-\mathbf{x}-\mathbf{y}+\mathbf{z}\)

In the quadrilateral WXYZ, \(\overrightarrow{WX}=\begin{pmatrix}-1\\6\\3\end{pmatrix}\), \(\overrightarrow{XY}=\begin{pmatrix}5\\-4\\0\end{pmatrix}\) and \(\overrightarrow{YZ}=\begin{pmatrix}-2\\1\\7\end{pmatrix}\). Find the column vectors \(\overrightarrow{WZ}\) and \(\overrightarrow{ZX}\).

1. (a) \(\overrightarrow{WZ}\)

   • \(\overrightarrow{WZ}=\overrightarrow{WX}+\overrightarrow{XY}+\overrightarrow{YZ}\)
   • \(=\begin{pmatrix}-1\\6\\3\end{pmatrix}+\begin{pmatrix}5\\-4\\0\end{pmatrix}+\begin{pmatrix}-2\\1\\7\end{pmatrix}=\begin{pmatrix}2\\3\\10\end{pmatrix}\)

2. (b) \(\overrightarrow{ZX}\)

   • \(\overrightarrow{XZ}=\overrightarrow{XY}+\overrightarrow{YZ}=\begin{pmatrix}3\\-3\\7\end{pmatrix}\)
   • \(\overrightarrow{ZX}=-\overrightarrow{XZ}=\begin{pmatrix}-3\\3\\-7\end{pmatrix}\)