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9709 · 3.7

Vectors — practice questions

Practice and worked examples for 9709 Vectors. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

The points A, B and C have position vectors a=2i+j−k\mathbf{a} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}, b=3i−2j+k\mathbf{b} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k} and c=i+3j+2k\mathbf{c} = \mathbf{i} + 3\mathbf{j} + 2\mathbf{k} respectively. Find the angle ABC.

Show solution outline

To find angle ABC, we need the vectors that form the angle at B, which are BA⃗\vec{BA} and BC⃗\vec{BC}.

  1. Find the vectors BA⃗\vec{BA} and BC⃗\vec{BC}. BA⃗=a−b=(2i+j−k)−(3i−2j+k)=−i+3j−2k\vec{BA} = \mathbf{a} - \mathbf{b} = (2\mathbf{i} + \mathbf{j} - \mathbf{k}) - (3\mathbf{i} - 2\mathbf{j} + \mathbf{k}) = -\mathbf{i} + 3\mathbf{j} - 2\mathbf{k} BC⃗=c−b=(i+3j+2k)−(3i−2j+k)=−2i+5j+k\vec{BC} = \mathbf{c} - \mathbf{b} = (\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) - (3\mathbf{i} - 2\mathbf{j} + \mathbf{k}) = -2\mathbf{i} + 5\mathbf{j} + \mathbf{k}
  2. Calculate the scalar product BA⃗⋅BC⃗\vec{BA} \cdot \vec{BC}. BA⃗⋅BC⃗=(−1)(−2)+(3)(5)+(−2)(1)=2+15−2=15\vec{BA} \cdot \vec{BC} = (-1)(-2) + (3)(5) + (-2)(1) = 2 + 15 - 2 = 15
  3. Calculate the magnitudes of the vectors. | ∣BA⃗∣=(−1)2+32+(−2)2=1+9+4=14 | \vec{BA} | = \sqrt{(-1)^2 + 3^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14} | | --- | --- | --- | | ∣BC⃗∣=(−2)2+52+12=4+25+1=30 | \vec{BC} | = \sqrt{(-2)^2 + 5^2 + 1^2} = \sqrt{4 + 25 + 1} = \sqrt{30} |
  4. Use the scalar product formula to find the angle. cos⁡(∠ABC)=BA⃗⋅BC⃗∣BA⃗∣∣BC⃗∣=151430\cos(\angle ABC) = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}||\vec{BC}|} = \frac{15}{\sqrt{14}\sqrt{30}} cos⁡(∠ABC)=15420≈0.732\cos(\angle ABC) = \frac{15}{\sqrt{420}} \approx 0.732 ∠ABC=cos⁡−1(0.732)≈42.9∘\angle ABC = \cos^{-1}(0.732) \approx 42.9^\circ (to 1 d.p.)

Worked example 2

A line ll passes through the points A(1, 5, -2) and B(4, -1, 4). (i) Find a vector equation for the line ll. (ii) The point C has coordinates (p, 9, -6). Given that C lies on the line ll, find the value of p.

Show solution outline

(i) To find the vector equation, we need a point on the line and a direction vector.

  1. Choose a point on the line. We can use point A, so a=(15−2)\mathbf{a} = \begin{pmatrix} 1 \\ 5 \\ -2 \end{pmatrix}.
  2. Find the direction vector d\mathbf{d} by finding the vector AB⃗\vec{AB}. d=AB⃗=b−a=(4−14)−(15−2)=(3−66)\mathbf{d} = \vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 4 \\ -1 \\ 4 \end{pmatrix} - \begin{pmatrix} 1 \\ 5 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 \\ -6 \\ 6 \end{pmatrix}. We can simplify this direction vector by dividing by a common factor, 3. So we can use d=(1−22)\mathbf{d} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}.
  3. Write the equation of the line. l:r=(15−2)+λ(1−22)l: \mathbf{r} = \begin{pmatrix} 1 \\ 5 \\ -2 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}.

(ii) If C lies on the line, its position vector must satisfy the line's equation for some value of λ\lambda.

  1. Set the general point r\mathbf{r} equal to the position vector of C. (p9−6)=(15−2)+λ(1−22)=(1+λ5−2λ−2+2λ)\begin{pmatrix} p \\ 9 \\ -6 \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \\ -2 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} = \begin{pmatrix} 1+\lambda \\ 5-2\lambda \\ -2+2\lambda \end{pmatrix}.
  2. Equate the components to find λ\lambda. We can use the y or z component. Using the y-component: 9=5−2λ  ⟹  4=−2λ  ⟹  λ=−29 = 5 - 2\lambda \implies 4 = -2\lambda \implies \lambda = -2. Check with the z-component: −6=−2+2λ  ⟹  −4=2λ  ⟹  λ=−2-6 = -2 + 2\lambda \implies -4 = 2\lambda \implies \lambda = -2. The value is consistent.
  3. Use the value of λ\lambda to find p from the x-component. p=1+λ  ⟹  p=1+(−2)  ⟹  p=−1p = 1 + \lambda \implies p = 1 + (-2) \implies p = -1.
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