Exercise Set 6 Answers

These exercises cover the topic of vectors.

Note that the questions are independent of one another, so the definition of a vector \(\mathbf{a}\) in one question does not carry over to another question with a vector called \(\mathbf{a}\).

  1. The points \(A\), \(B\) and \(C\) have position vectors \[\begin{align*} \mathbf{a}&=5\mathbf{i}-6\mathbf{j}+4\mathbf{k}\\ \mathbf{b}&=2\mathbf{i}+5\mathbf{j}-3\mathbf{k}\\ \mathbf{c}&=-\mathbf{i}+4\mathbf{k}. \end{align*}\] Find:

    1. \(5\mathbf{a}-2\mathbf{b}-3\mathbf{c}\)
    2. \(\overrightarrow{AC}+\overrightarrow{AB}\)
    3. \(\mathbf{a}\cdot\mathbf{b}\)
    4. The angle between \(\mathbf{a}\) and \(\mathbf{b}\)
    5. \(\mathbf{a}\times\mathbf{c}\)

    Answers:

    1. \(5\mathbf{a}-2\mathbf{b}-3\mathbf{c}=24\mathbf{i}-40\mathbf{j}+14\mathbf{k}\)

    2. \(\overrightarrow{AC}=\mathbf{c}-\mathbf{a}=-6\mathbf{i}+6\mathbf{j}\)

      \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=-3\mathbf{i}+11\mathbf{j}-7\mathbf{k}\)

      \(\overrightarrow{AC}+\overrightarrow{AB}=-9\mathbf{i}+17\mathbf{j}-7\mathbf{k}\)

    3. \(\mathbf{a}\cdot \mathbf{b}=(5\times 2)+ (-6\times5)+(4\times -3)=-32\)

    4. \(|\mathbf{a}|=\sqrt{77}\)

      \(|\mathbf{b}|=\sqrt{38}\)

      \(\cos(\theta)=\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\frac{-32}{\sqrt{77}\sqrt{38}}=-0.5916\)

      \(\theta=\cos^{-1}(-0.5916)=126.3^\circ\)

    5. \[\begin{align*} \mathbf{a}\times \mathbf{c}&=\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 5&-6&4\\ -1&0&4 \end{vmatrix}\\ &= \mathbf{i}\begin{vmatrix} -6&4\\ 0&4 \end{vmatrix} -\mathbf{j}\begin{vmatrix} 5&4\\ -1&4 \end{vmatrix} +\mathbf{k}\begin{vmatrix} 5&-6\\ -1&0 \end{vmatrix}\\ &=\mathbf{i}(-24-0)-\mathbf{j}(20-(-4))+\mathbf{k}(0-6)\\ &=-24\mathbf{i}-24\mathbf{j}-6\mathbf{k} \end{align*}\]

  2. Let \(\mathbf{p}\) and \(\mathbf{q}\) be vectors.

    1. What is the vector \(\mathbf{r}\) joining the points whose position vectors are \(\mathbf{p}\) and \(2\mathbf{q}\)?
    2. What is the vector \(\mathbf{s}\) joing the points whose position vectors are \(\mathbf{2q}\) and \(-6\mathbf{p}+14\mathbf{q}\)?
    3. What do you notice about \(\mathbf{r}\) and \(\mathbf{s}\)?
    4. What does this tell you about the points whose position vectors are \(\mathbf{p}\), \(2\mathbf{q}\) and \(-6\mathbf{p}+14{q}\)?

    Answers:

    1. \(\mathbf{r}=2\mathbf{q}-\mathbf{p}\)
    2. \(\mathbf{s}=(-6\mathbf{p}+14\mathbf{q})-2\mathbf{q}=12\mathbf{q}-6\mathbf{p}\)
    3. They are parallel: \(\mathbf{s}=6\mathbf{r}\) so they have the same direction.
    4. The points are collinear, i.e. all lie in a straight line.

  3. Let \(\mathbf{u}=5\mathbf{i}+2\mathbf{j}\) and \(\mathbf{v}=3\mathbf{i}-\mathbf{j}-3\mathbf{k}\). Find \((4\mathbf{u}-6\mathbf{v})\cdot(2\mathbf{u}+3\mathbf{v})\).

    Answers:

    \((2\mathbf{i}+14\mathbf{j}+\mathbf{k})\cdot(19\mathbf{i}+\mathbf{j}-9\mathbf{k})=38+14-162=-110.\)

  4. Given that the position vectors of the points \(A\), \(B\), and \(C\) are \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\), write down the position vectors of the midpoints of the lines \(BC\) and \(CA\).

    Answers:

    \(\overline{BC}=\mathbf{c}-\mathbf{b}\)

    \(\overline{OE}=\mathbf{b}+\frac{1}{2}(\mathbf{c}-\mathbf{b})=\frac{1}{2}\mathbf{c}-\frac{1}{2}\mathbf{b}\)

    \(\overline{CA}=\mathbf{a}-\mathbf{c}\)

    \(\overline{OF}=\mathbf{c}+\frac{1}{2}(\mathbf{a}-\mathbf{c})=\frac{1}{2}\mathbf{a}-\frac{1}{2}\mathbf{c}\)

  5. Prove that the vector joining the midpoints of two sides of a triangle is parallel to the third side and half of its length.

    Answers:

    We need to set up some labels for the points of the triangle. Let these have position vectors \(\mathbf{A}, \mathbf{B}, \mathbf{C}\). The vectors defining the sides of the triangle are

    \(\mathbf{a}=\overline{BC}=\mathbf{C}-\mathbf{B}\)

    \(\mathbf{b}=\overline{CA}=\mathbf{A}-\mathbf{C}\)

    \(\mathbf{c}=\overline{AB}=\mathbf{B}-\mathbf{A}\)

    The midpoint of side \(CA\) is

    \(\mathbf{C}+\frac{1}{2}\mathbf{b}=\mathbf{C}+\frac{1}{2}\mathbf{A}-\frac{1}{2}\mathbf{C}=\frac{1}{2}\mathbf{C}+\frac{1}{2}\mathbf{A}\)

    The midpoint of side \(AB\) is

    \(\mathbf{A}+\frac{1}{2}\mathbf{c}=\mathbf{A}+\frac{1}{2}\mathbf{B}-\frac{1}{2}\mathbf{A}=\frac{1}{2}\mathbf{A}+\frac{1}{2}\mathbf{B}\)

    The vector between these midpoints is

    \[(\frac{1}{2}\mathbf{C}+\frac{1}{2}\mathbf{A})-(\frac{1}{2}\mathbf{A}+\frac{1}{2}\mathbf{B})=\frac{1}{2}\mathbf{C}-\frac{1}{2}\mathbf{B}\]

    which is half of the vector \(\mathbf{a}=\overline{BC}\), as required.

  6. Show that the vectors \[\begin{align*} \mathbf{a}&=8\mathbf{i}+2\mathbf{j}-3\mathbf{k}\\ \mathbf{b}&=3\mathbf{i}-6\mathbf{j}+4\mathbf{k} \end{align*}\] are perpendicular to one another (Hint: use the dot product).

    Answers:

    Using the dot product:

    \(\mathbf{a}\cdot\mathbf{b}=(8\times 3)+(2\times-6)+(-3\times 4)=24-12-12=0\)

    and since this is zero and the vectors \(\mathbf{a}\) and \(\mathbf{b}\) are non-zero, these are perpendicular vectors.

  7. A block of mass \(m=1\,\text{kg}\) is sliding down a frictionless slope under gravity. The slope is at an angle of \(45^\circ\) to the horizontal. Recall that the force on the block is given by Newton’s second law: \(\mathbf{F}=m\mathbf{g}\) where \(\mathbf{g}=-9.8\mathbf{k}\,\text{ms}^{-2}\) is the gravitational acceleration. Resolve the force vector in the directions parallel and perpendicular to the slope. (Hint: you need to project the force vector onto unit vectors that are parallel and perpendicular to the slope.)

    Answers:

    We can simplify this to a two dimensional problem by aligning the coordinate axes sensibly. Let’s align the axes so that the \(y\) axis is perpendicular to the slope and the slope is increasing in the \(x\) direction, then we just need the \(\mathbf{i}\) and \(\mathbf{k}\) directions. The unit vector parallel to the (increasing) slope is \(\mathbf{u}=\frac{1}{\sqrt{2}}\mathbf{i}+\frac{1}{\sqrt{2}}\mathbf{k}\) and the unit vector perpendicular to the slope is \(\mathbf{v}=-\frac{1}{\sqrt{2}}\mathbf{i}+\frac{1}{\sqrt{2}}\mathbf{k}\).

    The force vector is \(\mathbf{F}=m\mathbf{g}=-9.8\mathbf{k}\). First projecting onto the parallel vector:

    \(\mathbf{F}\cdot \mathbf{u}=(0\times \frac{1}{\sqrt{2}})+ (-9.8 \times \frac{1}{\sqrt{2}})=-6.9\).

    and then onto the perpendicular vector:

    \(\mathbf{F}\cdot \mathbf{v}=(0\times -\frac{1}{\sqrt{2}})+ (-9.8 \times \frac{1}{\sqrt{2}})=-6.9\).

    Hence we have

    \(\mathbf{F}=-6.9\mathbf{u}-6.9\mathbf{v}\).

  8. Let \[\mathbf{a}=\begin{pmatrix} 2\\ -1\\ 3 \end{pmatrix} ,\qquad \mathbf{b}=\begin{pmatrix} 5\\ 1\\ -4 \end{pmatrix}. \] Show that \((\mathbf{a}-\mathbf{b})\times(\mathbf{a}+\mathbf{b})=2\mathbf{a}\times\mathbf{b}\).

    Answers:

    \(\mathbf{a}-\mathbf{b}=-3\mathbf{i}-2\mathbf{j}+7\mathbf{k}\)

    \(\mathbf{a}+\mathbf{b}=7\mathbf{i}-\mathbf{k}\)

    \[\begin{align*} (\mathbf{a}-\mathbf{b})\times(\mathbf{a}+\mathbf{b})&= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ -3&-2&7\\ 7&0&-1 \end{vmatrix}\\ &= \mathbf{i}\begin{vmatrix} -2&7\\ 0&-1 \end{vmatrix} -\mathbf{j}\begin{vmatrix} -3&7\\ 7&-1 \end{vmatrix} +\mathbf{k}\begin{vmatrix} -3&-2\\ 7&0 \end{vmatrix}\\ &=\mathbf{i}(2-0)-\mathbf{j}(3-49)+\mathbf{k}(0-(-14))\\ &=2\mathbf{i}+46\mathbf{j}+14\mathbf{k} \end{align*}\]

    \[\begin{align*} 2\mathbf{a}\times\mathbf{b}&= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 4&-2&6\\ 5&1&-4 \end{vmatrix}\\ &= \mathbf{i}\begin{vmatrix} -2&6\\ 1&-4 \end{vmatrix} -\mathbf{j}\begin{vmatrix} 4&6\\ 5&-4 \end{vmatrix} +\mathbf{k}\begin{vmatrix} 4&-2\\ 5&1 \end{vmatrix}\\ &=\mathbf{i}(8-6)-\mathbf{j}(-16-30)+\mathbf{k}(4-(-10))\\ &=2\mathbf{i}+46\mathbf{j}+14\mathbf{k} \end{align*}\]

    Hence \((\mathbf{a}-\mathbf{b})\times(\mathbf{a}+\mathbf{b})=2\mathbf{a}\times\mathbf{b}\)

  9. Show that \[(\mathbf{a}+\mathbf{b})\cdot((\mathbf{b}+\mathbf{c})\times (\mathbf{c}+\mathbf{a}))=2\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}).\] The quantity \(\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})\) is known as the scalar triple product: its geometric interpretation is the volume of the parallelepiped defined by the three vectors.

    Answers:

    \[\begin{align*} (\mathbf{a}+\mathbf{b})\cdot((\mathbf{b}+\mathbf{c})\times (\mathbf{c}+\mathbf{a}))&=(\mathbf{a}+\mathbf{b})\cdot(\mathbf{b}\times\mathbf{c}+\mathbf{b}\times\mathbf{a}+\mathbf{c}\times\mathbf{c}+\mathbf{c}\times\mathbf{a})\\ &=(\mathbf{a}+\mathbf{b})\cdot(\mathbf{b}\times\mathbf{c}+\mathbf{b}\times\mathbf{a}+\mathbf{c}\times\mathbf{a})\\ &=\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})+(\mathbf{a}\cdot(\mathbf{b}\times\mathbf{a})+\mathbf{a}\cdot(\mathbf{c}\times\mathbf{a})+\mathbf{b}\cdot(\mathbf{b}\times\mathbf{c})+(\mathbf{b}\cdot(\mathbf{b}\times\mathbf{a})+\mathbf{b}\cdot(\mathbf{c}\times\mathbf{a})\\ &=\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})+\mathbf{b}\cdot(\mathbf{c}\times\mathbf{a})\\ &=\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})+\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})\\ &=2\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c}) \end{align*}\]

  10. Let \[\mathbf{a}=\begin{pmatrix} \alpha\\ 0\\ -1 \end{pmatrix} ,\qquad \mathbf{b}=\begin{pmatrix} 3\\ 2\\ 5 \end{pmatrix} ,\qquad \mathbf{c}=\begin{pmatrix} 2\\ -1\\ 4 \end{pmatrix}. \] Find the values of \(\alpha\) for which \(\mathbf{a}\times\mathbf{b}\) is perpendicular to \(\mathbf{a}\times\mathbf{c}\).

    Answers:

    We find:

    \(\mathbf{a}\times\mathbf{b}=2\mathbf{i}-(5\alpha+3)\mathbf{j}+2\alpha\mathbf{k}\)

    and

    \(\mathbf{a}\times\mathbf{c}=-\mathbf{i}-(4\alpha+2)\mathbf{j}-\alpha\mathbf{k}\).

    Then for perpendicularity we solve \((\mathbf{a}\times\mathbf{b})\cdot(\mathbf{a}\times\mathbf{c})=0\). We have

    \[ (\mathbf{a}\times\mathbf{b})\cdot(\mathbf{a}\times\mathbf{c})=(9\alpha+2)(\alpha+1) \]

    Hence \[\alpha=-\frac{2}{9},\quad\text{or}\quad \alpha={-1}.\]

  11. The torque is the rotational equivalent of linear force. It is given by the product of the magnitude of the force and the perpendicular distance of the line of action of a force from the axis of rotation. We can calculate torque using the cross product: \[\boldsymbol\tau = \mathbf{r}\times \mathbf{F}\] where \(\boldsymbol\tau\) is the torque vector, \(\mathbf{r}\) is the position vector (from the axis of rotation to the point where the force is applied) and \(\mathbf{F}\) is the force vector.

    Find the torque about the point \(\mathbf{i}+2\mathbf{j}-\mathbf{k}\) of a force \(\mathbf{F}=3\mathbf{i}+\mathbf{k}\) acting on a particle at position \(2\mathbf{i}-\mathbf{j}+3\mathbf{k}\).

    Answers:

    \(\mathbf{r}=(2\mathbf{i}-\mathbf{j}+3\mathbf{k})-(\mathbf{i}+2\mathbf{j}-\mathbf{k})=\mathbf{i}-3\mathbf{j}+4\mathbf{k}\)

    Then \[\begin{align*} \mathbf{r}\times \mathbf{F}&= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 1&-3&4\\ 3&0&1 \end{vmatrix}\\ &=-3\mathbf{i}+11\mathbf{j}+9\mathbf{k}. \end{align*}\]