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AHL 4.20

Mixed practice: scalar and vector products, lines and distances

AHL 4.20

Mixed practice: scalar and vector products, lines and distances

Target: I can find the angle between two vectors from points, from position vectors, or from magnitudes and the scalar product, acute or obtuse. (+5 more)

By the end of this section

Key points

Angles and the scalar product

  • \(\cos\theta=\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\). A negative scalar product means an obtuse angle; 0 means perpendicular.
  • For two lines use the direction vectors, not the position vectors. For the acute angle take the positive value of the cosine.
  • A vector such as \(\overrightarrow{AB}\) is end point minus start point. Use the vectors from the same vertex when you want the angle at that vertex.
  • Perpendicular condition: set the scalar product equal to 0. If \(t\) appears twice you get a quadratic with two values of \(t\).

The vector product

  • \(\mathbf{a}\times\mathbf{b}\) (the vector product, or cross product) is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\). Divide by its magnitude for a unit vector perpendicular to both. Either direction is correct.
  • \(|\mathbf{a}\times\mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin\theta\). For the angle from \(|\mathbf{a}\times\mathbf{b}|\) you get an acute value from \(\sin^{-1}\); say whether the angle is acute or obtuse.
  • \((\mathbf{a}\times\mathbf{b})\cdot\mathbf{a}=0\) and \((\mathbf{a}\times\mathbf{b})\cdot\mathbf{b}=0\), because \(\mathbf{a}\times\mathbf{b}\) is perpendicular to both.
  • \(|\mathbf{a}\times\mathbf{b}|^2+(\mathbf{a}\cdot\mathbf{b})^2=|\mathbf{a}|^2|\mathbf{b}|^2\). It joins the two products: check it with numbers.
  • Swapping the order reverses the vector: \(\overrightarrow{BA}\times\overrightarrow{BC}=-\overrightarrow{AB}\times\overrightarrow{AC}\).

Shapes and areas in 3D

  • Triangle area: \(\tfrac12|\mathbf{a}\times\mathbf{b}|\). Parallelogram area: \(|\mathbf{a}\times\mathbf{b}|\). Use two sides from one vertex.
  • To show a triangle is right-angled, show a scalar product of two sides is 0. Then the area is also half the product of those two lengths.
  • A parallelogram ABCD has \(\overrightarrow{AB}=\overrightarrow{DC}\), so \(\mathbf{d}=\mathbf{a}+\mathbf{c}-\mathbf{b}\). A rhombus also has two equal adjacent sides, and its diagonals are perpendicular.
  • In a cuboid with a vertex at A, the edges from A run along the three axes, so the other vertices come from adding the edge lengths.

Components and distances

  • Component of \(\mathbf{v}\) along \(\mathbf{d}\): \(\dfrac{\mathbf{v}\cdot\mathbf{d}}{|\mathbf{d}|}\). Change a speed and direction into a velocity vector first: scale the direction vector to the speed.
  • Foot of the perpendicular from P to a line: write a general point on the line, then set the scalar product of (that point minus P) with the direction vector equal to 0 and solve for \(\lambda\).
  • The distance from P to the line is the distance from P to the foot. If R is the reflection of Q in the foot M, M is the midpoint of QR, so \(\lambda_R=2\lambda_M-\lambda_Q\).
  • An exact distance is left as a simplified surd.

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Angle between two vectors from four points

Find the angle between \(\overrightarrow{AC}\) and \(\overrightarrow{BD}\), where \(A(1, -1, 2)\), \(B(3, 0, -1)\), \(C(4, 2, 0)\), \(D(0, 3, 3)\).

Working
  1. The two vectors

    \(\overrightarrow{AC}=\begin{pmatrix}4\\2\\0\end{pmatrix}-\begin{pmatrix}1\\-1\\2\end{pmatrix}=\begin{pmatrix}3\\3\\-2\end{pmatrix}\), \(\overrightarrow{BD}=\begin{pmatrix}0\\3\\3\end{pmatrix}-\begin{pmatrix}3\\0\\-1\end{pmatrix}=\begin{pmatrix}-3\\3\\4\end{pmatrix}\)

    End point minus start point.

  2. Scalar product and magnitudes

    \(\mathbf{u}\cdot\mathbf{w}=-8\), \(|\mathbf{u}|=\sqrt{22}\), \(|\mathbf{w}|=\sqrt{34}\)

  3. The cosine

    \(\cos\theta=\dfrac{-8}{\sqrt{22}\sqrt{34}}\)

    A negative cosine means an obtuse angle. That is fine: the question asks for the angle between the vectors.

  4. Inverse cosine

    \(\theta=107^\circ\)

    Degree mode. 3 significant figures.

Video walkthrough coming
An obtuse angle from the scalar product

\(|\mathbf{a}|=6\), \(|\mathbf{b}|=7\) and \(\mathbf{a}\cdot\mathbf{b}=-14\). Find the angle between \(\mathbf{a}\) and \(\mathbf{b}\).

Working
  1. The formula

    \(\cos\theta=\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}\)

    Rearranged from the definition of the scalar product.

  2. Substitute

    \(\cos\theta=\dfrac{-14}{6\times7}=- \frac{1}{3}\)

    A negative cosine.

  3. Inverse cosine

    \(\theta=109^\circ\)

    More than \(90^\circ\): an obtuse angle.

Video walkthrough coming
p + tq perpendicular to a vector

Given \(\mathbf{p}=\begin{pmatrix}3\\-2\\1\end{pmatrix}\) and \(\mathbf{q}=\begin{pmatrix}1\\1\\-2\end{pmatrix}\), find the value of \(t\) for which \(\mathbf{p}+t\mathbf{q}\) is perpendicular to \(\begin{pmatrix}2\\-1\\4\end{pmatrix}\).

Working
  1. Write the vector

    \(\mathbf{p}+t\mathbf{q}=\begin{pmatrix}t + 3\\t - 2\\1 - 2 t\end{pmatrix}\)

  2. Perpendicular means scalar product 0

    \((t + 3)(2)+(t - 2)(-1)+(1 - 2 t)(4)=0\)

  3. Solve

    \(12 - 7 t=0\), so \(t=\frac{12}{7}=1.71\)

Video walkthrough coming
A unit vector perpendicular to two vectors

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

Working
  1. Perpendicular to both: the vector product

    \(\mathbf{a}\times\mathbf{b}=\begin{pmatrix}1\times2-(-2)\times(-1)\\ (-2)\times1-3\times2\\ 3\times(-1)-1\times1\end{pmatrix}=\begin{pmatrix}0\\-8\\-4\end{pmatrix}\)

    Check: its scalar product with each of a and b is 0.

  2. Its length

    \(|\mathbf{a}\times\mathbf{b}|=\sqrt{80}=4 \sqrt{5}=8.94\)

    Dividing by the length gives a vector of length 1.

  3. Divide

    \(\hat{\mathbf{n}}=\dfrac{1}{\sqrt{5}}\begin{pmatrix}0\\-2\\-1\end{pmatrix}\)

    Its negative is also a unit vector perpendicular to both.

Video walkthrough coming
A right-angled triangle in 3D

The triangle PQR has \(P(2, -1, 0)\), \(Q(4, 0, 2)\), \(R(5, -2, 2)\). Show that the angle at Q is \(90^\circ\), then find the area of the triangle.

QPR
Working
  1. Two sides at Q

    \(\overrightarrow{QP}=\begin{pmatrix}-2\\-1\\-2\end{pmatrix}\), \(\overrightarrow{QR}=\begin{pmatrix}1\\-2\\0\end{pmatrix}\)

    A right angle at Q needs the two sides that meet at Q.

  2. Scalar product

    \(\overrightarrow{QP}\cdot\overrightarrow{QR}=(-2)\times1+(-1)\times(-2)+(-2)\times0=0\)

    It is 0, so the angle at Q is \(90^\circ\).

  3. Area from the right angle

    \(\tfrac12|\overrightarrow{QP}||\overrightarrow{QR}|=\tfrac12\times3\times\sqrt{5}=3.35\)

    Half base times height, as the sides are perpendicular.

  4. Check with the vector product

    \(\overrightarrow{QP}\times\overrightarrow{QR}=\begin{pmatrix}-4\\-2\\5\end{pmatrix}\), \(\tfrac12\sqrt{45}=3.35\)

    The two methods agree.

  5. The other angles

    At P: \(36.7^\circ\). At R: \(53.3^\circ\)

    Same method as for any angle: scalar product and magnitudes. They add up to \(90^\circ\).

Video walkthrough coming
The part of a velocity along a direction

A rowing boat travels in the direction \(\begin{pmatrix}5\\12\end{pmatrix}\). The river flows at 3 km/h in the direction \(\begin{pmatrix}3\\-4\end{pmatrix}\). Find the component of the river's velocity in the direction the boat is travelling.

Working
  1. Velocity of the river

    \(\mathbf{v}=3\times\dfrac{1}{5}\begin{pmatrix}3\\-4\end{pmatrix}=\begin{pmatrix}\frac{9}{5}\\- \frac{12}{5}\end{pmatrix}\)

    The direction vector has length 5, not 3. Scale it to the speed.

  2. Component in the boat's direction

    \(\dfrac{\mathbf{v}\cdot\mathbf{d}}{|\mathbf{d}|}=\dfrac{- \frac{99}{5}}{13}=-1.52\)

    Direction: scalar product.

  3. Meaning

    Negative: the river pushes against the boat's direction.

    The component is \(-1.52\) km/h, so the current slows the boat.

Video walkthrough coming
The distance from a point to a line through O

The point B has position vector \(\begin{pmatrix}2\lambda\\-\lambda\\2\lambda\end{pmatrix}\), so B lies on the line through O and \(A(2, -1, 2)\). The point is \(C(6, -2, 2)\). Find \(\lambda\) when CB is perpendicular to OA, and the exact distance from C to the line OA.

Working
  1. CB

    \(\overrightarrow{CB}=\begin{pmatrix}2\lambda\\-\lambda\\2\lambda\end{pmatrix}-\begin{pmatrix}6\\-2\\2\end{pmatrix}\)

    B moves along the line as \(\lambda\) changes.

  2. Perpendicular to OA

    \(\overrightarrow{CB}\cdot\overrightarrow{OA}=9\lambda-18=0\)

    Scalar product 0.

  3. Solve

    \(\lambda=2\), so \(B=(4, -2, 4)\)

    B is the foot of the perpendicular from C.

  4. The distance

    \(\overrightarrow{BC}=\begin{pmatrix}2\\0\\-2\end{pmatrix}\), \(|\overrightarrow{BC}|=\sqrt{8}=2 \sqrt{2}\)

    Exact: leave it as a surd, simplified.

Video walkthrough coming