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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.20

Mixed practice: scalar and vector products, lines and distances Problem set B One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

  1. Find the angle between the two vectors.

    1. (a) \(\overrightarrow{OA}=\begin{pmatrix}1\\-2\\4\end{pmatrix}\) and \(\overrightarrow{OB}=\begin{pmatrix}5\\0\\1\end{pmatrix}\). Find the angle between \(\overrightarrow{AB}\) and \(\overrightarrow{OA}\).

      • \(\mathbf{u}=\overrightarrow{AB}=\begin{pmatrix}5\\0\\1\end{pmatrix}-\begin{pmatrix}1\\-2\\4\end{pmatrix}=\begin{pmatrix}4\\2\\-3\end{pmatrix}\), \(\mathbf{w}=\overrightarrow{OA}=\begin{pmatrix}1\\-2\\4\end{pmatrix}\)
      • \(\mathbf{u}\cdot\mathbf{w}=-12\), \(|\mathbf{u}|=\sqrt{29}\), \(|\mathbf{w}|=\sqrt{21}\)
      • \(\cos\theta=\dfrac{-12}{\sqrt{29}\sqrt{21}}\)
      • \(\theta=119^\circ\)

    2. (b) \(P(0, 3, 1)\), \(Q(2, 0, -1)\), \(R(4, 4, 2)\), \(S(1, 1, 5)\). Find the angle between \(\overrightarrow{PR}\) and \(\overrightarrow{QS}\).

      • \(\mathbf{u}=\overrightarrow{PR}=\begin{pmatrix}4\\1\\1\end{pmatrix}\), \(\mathbf{w}=\overrightarrow{QS}=\begin{pmatrix}-1\\1\\6\end{pmatrix}\)
      • \(\mathbf{u}\cdot\mathbf{w}=3\), \(|\mathbf{u}|=\sqrt{18}\), \(|\mathbf{w}|=\sqrt{38}\)
      • \(\cos\theta=\dfrac{3}{\sqrt{18}\sqrt{38}}\)
      • \(\theta=83.4^\circ\)

  2. Use the scalar product and the vector product to find each angle or magnitude.

    1. (a) \(|\mathbf{a}|=9\), \(|\mathbf{b}|=4\) and \(\mathbf{a}\cdot\mathbf{b}=15\). Find the angle between \(\mathbf{a}\) and \(\mathbf{b}\).

      • \(\cos\theta=\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\dfrac{15}{9\times4}\)
      • \(\theta=65.4^\circ\)

    2. (b) \(|\mathbf{c}|=12\), \(|\mathbf{d}|=5\) and \(\mathbf{c}\cdot\mathbf{d}=-42\). Find the angle between \(\mathbf{c}\) and \(\mathbf{d}\).

      • \(\cos\theta=\dfrac{-42}{12\times5}=-0.7000\)
      • \(\theta=134^\circ\) (obtuse, because the scalar product is negative)

    3. (c) \(|\mathbf{a}|=10\), \(\mathbf{a}\cdot\mathbf{b}=24\) and the angle between \(\mathbf{a}\) and \(\mathbf{b}\) is \(25^\circ\). Find \(|\mathbf{b}|\).

      • \(\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta\), so \(24=10|\mathbf{b}|\cos25^\circ\)
      • \(|\mathbf{b}|=\dfrac{24}{10\cos25^\circ}=2.65\)

    4. (d) \(|\mathbf{a}|=8\), \(|\mathbf{b}|=5\) and the angle between them is \(45^\circ\). Find the exact value of \(|\mathbf{a}\times\mathbf{b}|\).

      • \(|\mathbf{a}\times\mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin\theta=8\times5\times\sin45^\circ\)
      • \(\sin45^\circ=\frac{\sqrt{2}}{2}\), so \(|\mathbf{a}\times\mathbf{b}|=20 \sqrt{2}\)

    5. (e) \(|\mathbf{a}|=5\), \(|\mathbf{b}|=6\) and \(|\mathbf{a}\times\mathbf{b}|=24\). Find, in radians, the acute angle between \(\mathbf{a}\) and \(\mathbf{b}\).

      • \(\sin\theta=\dfrac{24}{5\times6}\)
      • Radian mode: \(\theta=\sin^{-1}\left(\dfrac{24}{30}\right)=0.927\) (acute)

    6. (f) \(|\mathbf{a}|=5\), \(|\mathbf{b}|=2\) and \(\mathbf{a}\times\mathbf{b}=\mathbf{i} - 4\mathbf{j} + 8\mathbf{k}\). Find, in radians, the acute angle between \(\mathbf{a}\) and \(\mathbf{b}\).

      • \(|\mathbf{a}\times\mathbf{b}|=\sqrt{81}=9\)
      • \(\sin\theta=\dfrac{9}{5\times2}\), so in radian mode \(\theta=1.12\)

  3. Work with these lines.

    1. (a) Find the acute angle between the lines \(\mathbf{r}=\begin{pmatrix}1\\4\\-2\end{pmatrix}+t\begin{pmatrix}2\\-1\\2\end{pmatrix}\) and \(\mathbf{r}=\begin{pmatrix}3\\0\\5\end{pmatrix}+s\begin{pmatrix}1\\-4\\8\end{pmatrix}\).

      • The directions are \(\mathbf{d}_1=\begin{pmatrix}2\\-1\\2\end{pmatrix}\) and \(\mathbf{d}_2=\begin{pmatrix}1\\-4\\8\end{pmatrix}\)
      • \(\mathbf{d}_1\cdot\mathbf{d}_2=2\times1+(-1)\times(-4)+2\times8=22\)
      • \(|\mathbf{d}_1|=\sqrt{9}\), \(|\mathbf{d}_2|=\sqrt{81}\)
      • \(\cos\theta=\dfrac{22}{\sqrt{9}\sqrt{81}}\) (use the positive value for the acute angle)
      • \(\theta=35.4^\circ\)

    2. (b) Find the acute angle between the lines \(x=3+\lambda , y=-2\lambda , z=1+2\lambda \) and \(x=\mu , y=4-\mu , z=2+3\mu \).

      • The coefficients of the parameters give the directions: \(\mathbf{d}_1=\begin{pmatrix}1\\-2\\2\end{pmatrix}\), \(\mathbf{d}_2=\begin{pmatrix}1\\-1\\3\end{pmatrix}\)
      • \(\mathbf{d}_1\cdot\mathbf{d}_2=1\times1+(-2)\times(-1)+2\times3=9\)
      • \(|\mathbf{d}_1|=\sqrt{9}\), \(|\mathbf{d}_2|=\sqrt{11}\)
      • \(\cos\theta=\dfrac{9}{\sqrt{9}\sqrt{11}}\) (use the positive value for the acute angle)
      • \(\theta=25.2^\circ\)

    3. (c) Show that the lines \(\mathbf{r}=\begin{pmatrix}0\\2\\1\end{pmatrix}+\lambda \begin{pmatrix}3\\-2\\4\end{pmatrix}\) and \(\mathbf{r}=\begin{pmatrix}5\\1\\-3\end{pmatrix}+\mu \begin{pmatrix}2\\1\\-1\end{pmatrix}\) are perpendicular.

      • \(\mathbf{d}_1\cdot\mathbf{d}_2=3\times2+(-2)\times1+4\times(-1)=0\)
      • The scalar product of the directions is 0, so the lines are perpendicular

    4. (d) A line has parametric equations \(x=-1+2\lambda , y=4-3\lambda , z=3+6\lambda \). Write down a vector equation of the line, and find the acute angle it makes with the z-axis.

      • Point \(\begin{pmatrix}-1\\4\\3\end{pmatrix}\) and direction \(\begin{pmatrix}2\\-3\\6\end{pmatrix}\): \(\mathbf{r}=\begin{pmatrix}-1\\4\\3\end{pmatrix}+\lambda \begin{pmatrix}2\\-3\\6\end{pmatrix}\)
      • The z-axis has direction \(\mathbf{k}\). \(\cos\theta=\dfrac{|6|}{\sqrt{49}}\)
      • \(\theta=31.0^\circ\)

  4. Work with perpendicular vectors.

    1. (a) Find a vector that is perpendicular to both \(\mathbf{a}=3\mathbf{i} + \mathbf{j} - 2\mathbf{k}\) and \(\mathbf{b}=-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}\).

      • \(\mathbf{a}\times\mathbf{b}=\begin{pmatrix}3\\1\\-2\end{pmatrix}\times\begin{pmatrix}-1\\2\\2\end{pmatrix}=\begin{pmatrix}1\times2-(-2)\times2\\ (-2)\times(-1)-3\times2\\ 3\times2-1\times(-1)\end{pmatrix}=\begin{pmatrix}6\\-4\\7\end{pmatrix}\)
      • \(\mathbf{a}\times\mathbf{b}=6\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}\) is perpendicular to both
      • Check: \(\mathbf{a}\cdot(\mathbf{a}\times\mathbf{b})=0\) and \(\mathbf{b}\cdot(\mathbf{a}\times\mathbf{b})=0\)

    2. (b) Find a unit vector that is perpendicular to both \(\begin{pmatrix}1\\2\\-1\end{pmatrix}\) and \(\begin{pmatrix}2\\-1\\3\end{pmatrix}\).

      • \(\begin{pmatrix}1\\2\\-1\end{pmatrix}\times\begin{pmatrix}2\\-1\\3\end{pmatrix}=\begin{pmatrix}2\times3-(-1)\times(-1)\\ (-1)\times2-1\times3\\ 1\times(-1)-2\times2\end{pmatrix}=\begin{pmatrix}5\\-5\\-5\end{pmatrix}\)
      • \(|\begin{pmatrix}5\\-5\\-5\end{pmatrix}|=\sqrt{75}\)
      • Divide by the magnitude: \(\dfrac{1}{\sqrt{3}}\begin{pmatrix}1\\-1\\-1\end{pmatrix}\)

    3. (c) Given \(\mathbf{p}=\begin{pmatrix}3\\-1\\2\end{pmatrix}\) and \(\mathbf{q}=\begin{pmatrix}1\\2\\-2\end{pmatrix}\), find the value of \(t\) for which \(\mathbf{p}+t\mathbf{q}\) is perpendicular to \(\begin{pmatrix}2\\1\\3\end{pmatrix}\).

      • \(\mathbf{p}+t\mathbf{q}=\begin{pmatrix}t + 3\\2 t - 1\\2 - 2 t\end{pmatrix}\)
      • Perpendicular: \((t + 3)(2)+(2 t - 1)(1)+(2 - 2 t)(3)=0\)
      • \(11 - 2 t=0\), so \(t=5.5\)

    4. (d) The vectors \(t\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) and \(2 t\mathbf{i} + t\mathbf{j} - 5\mathbf{k}\) are perpendicular. Find the possible values of \(t\).

      • Perpendicular: \(\left(t\right)\left(2 t\right)+\left(-1\right)\left(t\right)+\left(2\right)\left(-5\right)=0\)
      • \(2 t^{2} - t - 10=0\)
      • Polynomial solver: \(t=-2\) or \(t=\frac{5}{2}\)

  5. Properties of the scalar product and the vector product.

    1. (a) Explain why \((\mathbf{a}\times\mathbf{b})\cdot\mathbf{a}=0\).

      • \(\mathbf{a}\times\mathbf{b}\) is perpendicular to \(\mathbf{a}\)
      • Perpendicular vectors have scalar product 0 (\(\cos90^\circ=0\))
      • So \((\mathbf{a}\times\mathbf{b})\cdot\mathbf{a}=0\)

    2. (b) Evaluate \((\mathbf{a}\times\mathbf{b})\cdot(5\mathbf{a}+2\mathbf{b})\).

      • Expand: \(5(\mathbf{a}\times\mathbf{b})\cdot\mathbf{a}+2(\mathbf{a}\times\mathbf{b})\cdot\mathbf{b}\)
      • Each scalar product is 0, because \(\mathbf{a}\times\mathbf{b}\) is perpendicular to both
      • The value is 0

    3. (c) Take \(\mathbf{a}=\begin{pmatrix}3\\-1\\2\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}1\\4\\-2\end{pmatrix}\). Show that \(|\mathbf{a}\times\mathbf{b}|^2+(\mathbf{a}\cdot\mathbf{b})^2=|\mathbf{a}|^2|\mathbf{b}|^2\).

      • \(\mathbf{a}\times\mathbf{b}=\begin{pmatrix}-6\\8\\13\end{pmatrix}\), so \(|\mathbf{a}\times\mathbf{b}|^2=269\)
      • \(\mathbf{a}\cdot\mathbf{b}=-5\), so \((\mathbf{a}\cdot\mathbf{b})^2=25\)
      • Left side: \(269+25=294\)
      • \(|\mathbf{a}|^2|\mathbf{b}|^2=14\times21=294\): the two sides are equal

    4. (d) The points are \(A(0, 1, 2)\), \(B(2, 3, 1)\) and \(C(1, 0, 4)\). Calculate \(\mathbf{p}=\overrightarrow{AB}\times\overrightarrow{AC}\) and \(\mathbf{q}=\overrightarrow{BA}\times\overrightarrow{BC}\). What do you notice about \(\mathbf{p}\) and \(\mathbf{q}\)?

      • \(\overrightarrow{AB}=\begin{pmatrix}2\\2\\-1\end{pmatrix}\), \(\overrightarrow{AC}=\begin{pmatrix}1\\-1\\2\end{pmatrix}\), so \(\mathbf{p}=\begin{pmatrix}3\\-5\\-4\end{pmatrix}\)
      • \(\overrightarrow{BA}=\begin{pmatrix}-2\\-2\\1\end{pmatrix}\), \(\overrightarrow{BC}=\begin{pmatrix}-1\\-3\\3\end{pmatrix}\), so \(\mathbf{q}=\begin{pmatrix}-3\\5\\4\end{pmatrix}\)
      • \(\mathbf{q}=-\mathbf{p}\): same length, opposite direction. Both are perpendicular to the plane of the triangle, and the two orders of the sides turn opposite ways

Shapes in 3D

  1. The triangle ABC has \(A(0, -1, 2)\), \(B(3, 1, 4)\), \(C(-1, 3, 5)\).

    ABC

    1. (a) Find the three angles of the triangle, in degrees.

      • \(\overrightarrow{AB}=\begin{pmatrix}3\\2\\2\end{pmatrix}\), \(\overrightarrow{AC}=\begin{pmatrix}-1\\4\\3\end{pmatrix}\), \(\overrightarrow{BC}=\begin{pmatrix}-4\\2\\1\end{pmatrix}\)
      • At A: \(\cos\theta=\dfrac{11}{\sqrt{17}\sqrt{26}}\), \(\theta=58.5^\circ\)
      • At B: \(\cos\theta=\dfrac{6}{\sqrt{17}\sqrt{21}}\), \(\theta=71.5^\circ\)
      • At C: \(\cos\theta=\dfrac{15}{\sqrt{26}\sqrt{21}}\), \(\theta=50.1^\circ\)
      • Check: \(58.5+71.5+50.1=180.0\), about \(180^\circ\)

    2. (b) Find the area of the triangle.

      • \(\overrightarrow{AB}\times\overrightarrow{AC}=\begin{pmatrix}-2\\-11\\14\end{pmatrix}\)
      • Area \(=\tfrac12\sqrt{321}=8.96\)

  2. The triangle ABC has \(A(0, 1, 2)\), \(B(1, 3, -1)\), \(C(-1, 3, 0)\).

    CAB

    1. (a) Show that the angle at C is a right angle.

      • \(\overrightarrow{CA}=\begin{pmatrix}1\\-2\\2\end{pmatrix}\), \(\overrightarrow{CB}=\begin{pmatrix}2\\0\\-1\end{pmatrix}\)
      • \(\overrightarrow{CA}\cdot\overrightarrow{CB}=0\)
      • The scalar product is 0, so the angle at C is \(90^\circ\)

    2. (b) Find the other two angles, in degrees.

      • At A: \(\cos\theta=\dfrac{9}{\sqrt{9}\sqrt{14}}\), \(\theta=36.7^\circ\)
      • At B: \(\cos\theta=\dfrac{5}{\sqrt{5}\sqrt{14}}\), \(\theta=53.3^\circ\)
      • Check: \(36.7+53.3+90=180.0\)

    3. (c) Find the area of the triangle.

      • \(\overrightarrow{CA}\times\overrightarrow{CB}=\begin{pmatrix}2\\5\\4\end{pmatrix}\)
      • Area \(=\tfrac12\sqrt{45}=3.35\)
      • Check with the right angle: \(\tfrac12|\overrightarrow{CA}||\overrightarrow{CB}|=\tfrac12\sqrt{9}\sqrt{5}=3.35\)

  3. ABCD is a parallelogram. Three of its vertices are \(A(1, -1, 0)\), \(B(4, 0, 2)\), \(C(6, 3, 3)\).

    ABCD

    1. (a) Find the coordinates of D.

      • \(\overrightarrow{DC}=\overrightarrow{AB}\), so \(\mathbf{d}=\mathbf{a}+\mathbf{c}-\mathbf{b}\)
      • \(\mathbf{d}=\begin{pmatrix}1\\-1\\0\end{pmatrix}+\begin{pmatrix}6\\3\\3\end{pmatrix}-\begin{pmatrix}4\\0\\2\end{pmatrix}=\begin{pmatrix}3\\2\\1\end{pmatrix}\)
      • \(D=(3, 2, 1)\)

    2. (b) Find the area of the parallelogram.

      • \(\overrightarrow{AB}=\begin{pmatrix}3\\1\\2\end{pmatrix}\), \(\overrightarrow{AD}=\begin{pmatrix}2\\3\\1\end{pmatrix}\)
      • \(\overrightarrow{AB}\times\overrightarrow{AD}=\begin{pmatrix}-5\\1\\7\end{pmatrix}\)
      • Area \(=|\overrightarrow{AB}\times\overrightarrow{AD}|=\sqrt{75}=8.66\)

  4. ABCD has \(A(0, -1, 3)\), \(B(1, 1, 5)\) and \(D(2, 1, 2)\), and \(\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{AC}\).

    ABCD

    1. (a) Find the coordinates of C.

      • \(\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}=\begin{pmatrix}1\\2\\2\end{pmatrix}+\begin{pmatrix}2\\2\\-1\end{pmatrix}=\begin{pmatrix}3\\4\\1\end{pmatrix}\)
      • \(C=(3, 3, 4)\)

    2. (b) Show that ABCD is a rhombus, by comparing the lengths of \(\overrightarrow{AB}\) and \(\overrightarrow{AD}\).

      • \(|\overrightarrow{AB}|=\sqrt{9}=3\), \(|\overrightarrow{AD}|=\sqrt{9}=3\)
      • A parallelogram with two equal adjacent sides is a rhombus

    3. (c) Show that the diagonals AC and BD are perpendicular.

      • \(\overrightarrow{AC}=\begin{pmatrix}3\\4\\1\end{pmatrix}\), \(\overrightarrow{BD}=\begin{pmatrix}1\\0\\-3\end{pmatrix}\)
      • \(\overrightarrow{AC}\cdot\overrightarrow{BD}=3\times1+4\times0+1\times(-3)=0\)
      • The scalar product is 0, so the diagonals are perpendicular

  5. The cuboid ABCDEFGH has \(A(-1, 2, 1)\), \(B(1, 2, 1)\), \(D(-1, 7, 1)\), \(E(-1, 2, 4)\). E is directly above A, F above B, G above C and H above D.

    253ABCDEFGH

    1. (a) Find the coordinates of C, F, G and H.

      • AB is along the x-axis with length 2, AD along the y-axis with length 5, AE along the z-axis with length 3
      • \(C=(1, 7, 1)\)
      • \(F=(1, 2, 4)\)
      • \(G=(1, 7, 4)\)
      • \(H=(-1, 7, 4)\)

    2. (b) Find the area of the triangle CFH formed by three face diagonals.

      • \(\overrightarrow{CF}=\begin{pmatrix}0\\-5\\3\end{pmatrix}\), \(\overrightarrow{CH}=\begin{pmatrix}-2\\0\\3\end{pmatrix}\)
      • \(\overrightarrow{CF}\times\overrightarrow{CH}=\begin{pmatrix}-15\\-6\\-10\end{pmatrix}\)
      • Area \(=\tfrac12\sqrt{361}=9.50\)

In context

  1. A cyclist rides along a straight road in the direction \(\begin{pmatrix}3\\2\end{pmatrix}\). The wind blows at 6.5 m/s in the direction \(\begin{pmatrix}5\\-12\end{pmatrix}\).

    1. (a) Write the velocity of the wind as a column vector.

      • Unit vector in the direction \(\begin{pmatrix}5\\-12\end{pmatrix}\): \(\dfrac{1}{\sqrt{169}}\begin{pmatrix}5\\-12\end{pmatrix}\)
      • Velocity \(=\frac{13}{2}\times\dfrac{1}{13}\begin{pmatrix}5\\-12\end{pmatrix}=\begin{pmatrix}\frac{5}{2}\\-6\end{pmatrix}\)

    2. (b) Find the component of the wind's velocity in the direction the cyclist is riding.

      • Component \(=\dfrac{\mathbf{v}\cdot\mathbf{d}}{|\mathbf{d}|}\), with \(\mathbf{d}=\begin{pmatrix}3\\2\end{pmatrix}\)
      • \(\mathbf{v}\cdot\mathbf{d}=- \frac{9}{2}\), \(|\mathbf{d}|=\sqrt{13}\)
      • Component \(=-1.25\) m/s Negative: the wind has a part against the direction of travel.

Stretch

  1. Points A and B have position vectors \(\begin{pmatrix}1\\2\\2\end{pmatrix}\) and \(\begin{pmatrix}\lambda\\2\lambda\\2\lambda\end{pmatrix}\) respectively, and \(C(4, 3, 4)\).

    1. (a) Show that B lies on the line OA for every value of \(\lambda\).

      • \(\overrightarrow{OB}=\begin{pmatrix}\lambda\\2\lambda\\2\lambda\end{pmatrix}=\lambda\begin{pmatrix}1\\2\\2\end{pmatrix}=\lambda\overrightarrow{OA}\)
      • OB is a multiple of OA, so B is on the line through O and A

    2. (b) Find the value of \(\lambda\) for which CB is perpendicular to OA.

      • \(\overrightarrow{CB}=\begin{pmatrix}\lambda\\2\lambda\\2\lambda\end{pmatrix}-\begin{pmatrix}4\\3\\4\end{pmatrix}\)
      • Perpendicular to \(\overrightarrow{OA}\): \(\overrightarrow{CB}\cdot\overrightarrow{OA}=0\)
      • \(9\lambda-18=0\), so \(\lambda=2\)

    3. (c) Hence find the exact distance from C to the line OA.

      • With \(\lambda=2\): \(B=(2, 4, 4)\)
      • \(\overrightarrow{BC}=\begin{pmatrix}4\\3\\4\end{pmatrix}-\begin{pmatrix}2\\4\\4\end{pmatrix}=\begin{pmatrix}2\\-1\\0\end{pmatrix}\)
      • Distance \(=|\overrightarrow{BC}|=\sqrt{5}\)

  2. The line \(l\) has equation \(\mathbf{r}=\begin{pmatrix}-2\\4\\1\end{pmatrix}+\lambda \begin{pmatrix}1\\-2\\2\end{pmatrix}\). The point P has coordinates \((3, -1, 7)\).

    1. (a) Find the coordinates of the point M on \(l\) such that PM is perpendicular to \(l\).

      • A point on \(l\) is \((-2+1\lambda, 4-2\lambda, 1+2\lambda)\)
      • \(\overrightarrow{PM}\) is perpendicular to \(\begin{pmatrix}1\\-2\\2\end{pmatrix}\): \(\overrightarrow{PM}\cdot\begin{pmatrix}1\\-2\\2\end{pmatrix}=0\)
      • This gives \(9\lambda=27\), so \(\lambda=3\)
      • \(M=(1, -2, 7)\)

    2. (b) Show that the point \(Q(-3, 6, -1)\) lies on \(l\).

      • Solve \(\begin{pmatrix}-2\\4\\1\end{pmatrix}+\lambda\begin{pmatrix}1\\-2\\2\end{pmatrix}=\begin{pmatrix}-3\\6\\-1\end{pmatrix}\) coordinate by coordinate
      • \(\lambda=-1\) in every coordinate
      • The same value of \(\lambda\) works for all three, so Q is on \(l\)

    3. (c) The point R, different from Q, lies on \(l\) and \(|PR|=|PQ|\). Find the coordinates of R.

      • \(|PR|=|PQ|\) and M is the foot of the perpendicular from P, so M is the midpoint of QR
      • \(\lambda\) for R: \(2\times3-(-1)=7\)
      • \(R=(5, -10, 15)\)

jmaths.xyz · AI HL · AHL 4.20 Mixed practice: scalar and vector products, lines and distances