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AI HL · AHL 4

Vectors

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AHL 4.6

Mixed practice: parallel vectors, unit vectors, displacement, distance and parameters

AHL 4.6

Mixed practice: parallel vectors, unit vectors, displacement, distance and parameters

Target: I can find unknown components that make two vectors parallel. (+4 more)

By the end of this section

Key points

Parallel vectors and parameters

  • Vectors are parallel when one is a multiple of the other: \(\mathbf{b}=k\mathbf{a}\). Scalars are often written \(\lambda\), \(\mu\) or \(k\).
  • To find unknown components, find the multiplier from a pair of known components, then use it on the others.
  • A vector parallel to \(\mathbf{j}\) has zero \(\mathbf{i}\) and \(\mathbf{k}\) components. Equate those components to zero.
  • When a vector is parallel to a given vector, write it as \(\mu\) times that vector and equate components.

Magnitude, unit vectors and distance

  • A unit vector in the direction of \(\mathbf{a}\) is \(\frac{1}{|\mathbf{a}|}\mathbf{a}\). A vector of magnitude \(m\) in that direction is \(m\times\frac{1}{|\mathbf{a}|}\mathbf{a}\).
  • \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\), and the distance \(AB=|\mathbf{b}-\mathbf{a}|\).
  • To find a parameter that gives a magnitude, write \(|\mathbf{v}|\) as a square root of the sum of squares and set it equal to the value. It is a quadratic in the parameter, so expect two answers.
  • Use nSolve with two different starting guesses to find both values.

Geometry with \(\mathbf{a}\) and \(\mathbf{b}\)

  • Take one corner as the origin. Write every other point as a position vector in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
  • A midpoint is half way: add half the side vector. A ratio point \(AP:PB=m:n\) has \(\overrightarrow{AP}=\frac{m}{m+n}\overrightarrow{AB}\).
  • Then \(\overrightarrow{XY}=\overrightarrow{OY}-\overrightarrow{OX}\) gives any vector between two of the points.

On the calculator

Solving for a parameter with nSolve

  • Write the magnitude as a square root of the sum of squares, then set it equal to the given value.
  • nSolve finds one solution near your starting guess. A magnitude equation usually has two, so run it from a guess on each side.
  • Check each value by putting it back into the vector.

Screenshot and video to come

Finding the smallest magnitude from a graph

  • Graph the magnitude as a function of the parameter.
  • Use the minimum tool on the graph. The y-value is the smallest magnitude; the x-value is the parameter at that point.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Finding p and q so that vectors are parallel

The vectors \(\mathbf{a}=\begin{pmatrix}6 \\ -4 \\ p\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}9 \\ q \\ 12\end{pmatrix}\) are parallel. Find \(p\) and \(q\).

Working
  1. Parallel means a multiple

    \(\mathbf{b}=k\mathbf{a}\)

    Every component of \(\mathbf{b}\) is the same multiple of the matching component of \(\mathbf{a}\).

  2. Find k from known components

    \(9=k\times6\ \Rightarrow\ k=\frac{3}{2}\)

    Use the pair where both numbers are known.

  3. Find p

    \(12=\frac{3}{2}p\ \Rightarrow\ p=8\)

    The third components give p.

  4. Find q

    \(q=\frac{3}{2}\times(-4)=-6\)

    The second components give q.

Video walkthrough coming
A combination parallel to a given vector

Given \(\mathbf{a}=\begin{pmatrix}2 \\ -1 \\ 4\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}1 \\ 2 \\ -2\end{pmatrix}\), find the value of \(p\) such that \(\mathbf{a}+p\mathbf{b}\) is parallel to \(\begin{pmatrix}10 \\ 10 \\ -4\end{pmatrix}\).

Working
  1. Write it as a multiple

    \(\mathbf{a}+p\mathbf{b}=\begin{pmatrix}2+p \\ -1+2p \\ 4-2p\end{pmatrix}=\mu\begin{pmatrix}10 \\ 10 \\ -4\end{pmatrix}\)

    A second unknown, μ, is the multiple.

  2. Equate components

    \(2+p=10\mu,\quad -1+2p=10\mu,\quad 4-2p=-4\mu\)

    Three equations, two unknowns.

  3. Use two to find p

    \(2+p=-1+2p\ \Rightarrow\ p=3,\quad \mu=\frac{1}{2}\)

    The first two have the same right-hand side.

  4. Check the third

    \(4-2(3)=-2=-4\times\frac{1}{2}\)

    It works, so the three components agree.

Video walkthrough coming
A combination parallel to a coordinate axis

Given \(\mathbf{x}=\begin{pmatrix}3 \\ -2 \\ 5\end{pmatrix}\) and \(\mathbf{y}=\begin{pmatrix}4 \\ 6 \\ -15\end{pmatrix}\), find \(\lambda\) and \(k\) such that \(\lambda\mathbf{x}+\mathbf{y}=k\mathbf{i}\).

Working
  1. What the axis means

    \(k\mathbf{i}=\begin{pmatrix}k \\ 0 \\ 0\end{pmatrix}\)

    A vector parallel to \(\mathbf{i}\) has zero \(\mathbf{j}\) and \(\mathbf{k}\) components.

  2. Equate components

    \(\lambda\begin{pmatrix}3 \\ -2 \\ 5\end{pmatrix}+\begin{pmatrix}4 \\ 6 \\ -15\end{pmatrix}=\begin{pmatrix}3\lambda+4 \\ -2\lambda+6 \\ 5\lambda-15\end{pmatrix}\)

    Only the \(\mathbf{i}\) component is allowed to be non-zero.

  3. The zero components

    \(-2\lambda+6=0\ \Rightarrow\ \lambda=3,\quad 5\lambda-15=0\ \Rightarrow\ \lambda=3\)

    Both give the same λ. That is the check.

  4. Find k

    \(k=3\lambda+4=13\)

    The \(\mathbf{i}\) component is \(k\).

Video walkthrough coming
A parameter from a magnitude

Given \(\mathbf{a}=\begin{pmatrix}3 \\ -1 \\ 2\end{pmatrix}\) and \(\mathbf{b}=\begin{pmatrix}1 \\ 2 \\ -2\end{pmatrix}\), find the values of \(\lambda\) such that \(|\mathbf{a}+\lambda\mathbf{b}|=7\).

Working
  1. Write the vector

    \(\mathbf{a}+\lambda\mathbf{b}=\begin{pmatrix}3+\lambda \\ -1+2\lambda \\ 2-2\lambda\end{pmatrix}\)

    Each component is a number plus a multiple of λ.

  2. Write the magnitude

    \(|\mathbf{a}+\lambda\mathbf{b}|=\sqrt{(3+\lambda)^2+(-1+2\lambda)^2+(2-2\lambda)^2}=7\)

    Square root of the sum of the squares of the components.

  3. Solve

    From the calculator: nSolve(√((3+λ)²+(−1+2λ)²+(2−2λ)²) = 7, λ) with guesses near 2 and near −2: λ = 2.33 and λ = −1.67

    nSolve finds one solution near a starting guess. Try one guess above 0 and one below.

  4. Both values

    \(\lambda=-1.67\ \text{or}\ \lambda=2.33\)

    A magnitude is a square root, so there are two values of λ.

Video walkthrough coming
Vectors in a triangle with a ratio point

In triangle \(OAB\), \(\overrightarrow{OA}=\mathbf{a}\) and \(\overrightarrow{OB}=\mathbf{b}\). The point \(M\) is the midpoint of \(OB\). The point \(P\) lies on \(AB\) so that \(AP:PB=1:3\). Find \(\overrightarrow{AB}\), \(\overrightarrow{OP}\) and \(\overrightarrow{MP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).

aOABMP
Working
  1. AB

    \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\)

    End minus start.

  2. AP is one quarter of AB

    \(AP:PB=1:3\ \Rightarrow\ \overrightarrow{AP}=\frac{1}{4}(\mathbf{b}-\mathbf{a})\)

    AP is 1 part out of 1 + 3 = 4.

  3. OP

    \(\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}=\mathbf{a}+\frac{1}{4}(\mathbf{b}-\mathbf{a})=\frac{3}{4}\mathbf{a}+\frac{1}{4}\mathbf{b}\)

    Go to A, then a quarter of the way along AB.

  4. MP

    \(\overrightarrow{MP}=\overrightarrow{OP}-\overrightarrow{OM}=\frac{3}{4}\mathbf{a}+\frac{1}{4}\mathbf{b}-\frac{1}{2}\mathbf{b}=\frac{3}{4}\mathbf{a}-\frac{1}{4}\mathbf{b}\)

    M is half way along OB.

Video walkthrough coming
The smallest magnitude of a vector

Find the smallest possible magnitude of the vector \((-2+3t)\mathbf{i}+(1+t)\mathbf{j}+(-4+2t)\mathbf{k}\), where \(t\) is a real number. Give your answer to 3 significant figures.

Working
  1. The magnitude as a function of t

    \(|\mathbf{v}|=\sqrt{(-2+3t)^2+(1+t)^2+(-4+2t)^2}\)

    The magnitude changes with t. We want its lowest point.

  2. Graph it

    From the calculator: graph y = √((−2+3t)²+(1+t)²+(−4+2t)²) and find the minimum.

    The graph is a curve with one lowest point.

  3. Read the minimum

    Minimum at t = 0.929, y = 2.99

    The y-value is the smallest magnitude. The t-value is where it happens.

  4. State

    \(|\mathbf{v}|_{\min}=2.99\)

    Give the magnitude, not the t-value.

Video walkthrough coming