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Vectors: Question 2

Syllabus 3.7

Structured A2 7 marks

Relative to an origin OO, OA→=a=(3−12),OB→=b=(14−1).\overrightarrow{OA} = \mathbf{a} = \begin{pmatrix}3\\-1\\2\end{pmatrix}, \qquad \overrightarrow{OB} = \mathbf{b} = \begin{pmatrix}1\\4\\-1\end{pmatrix}.

(a) Find AB→\overrightarrow{AB}. [2]

(b) Find ∣AB→∣|\overrightarrow{AB}|, giving your answer in the form n\sqrt{n} for an integer nn. [2]

(c) Find angle AOBAOB, the angle between OA→\overrightarrow{OA} and OB→\overrightarrow{OB}, correct to 11 decimal place. [3]

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Worked solution

Part (a): Finding AB→\overrightarrow{AB}

Since AB→=AO→+OB→=b−a\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = \mathbf{b} - \mathbf{a}:

AB→=(14−1)−(3−12)=(1−34−(−1)−1−2)=(−25−3)\overrightarrow{AB} = \begin{pmatrix}1\\4\\-1\end{pmatrix} - \begin{pmatrix}3\\-1\\2\end{pmatrix} = \begin{pmatrix}1-3\\4-(-1)\\-1-2\end{pmatrix} = \begin{pmatrix}-2\\5\\-3\end{pmatrix}

Part (b): Finding ∣AB→∣|\overrightarrow{AB}|

∣AB→∣=(−2)2+52+(−3)2=4+25+9=38|\overrightarrow{AB}| = \sqrt{(-2)^2 + 5^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38}

Check by recomputing in a different order: (−3)2+(−2)2+52=9+4+25=38(-3)^2 + (-2)^2 + 5^2 = 9 + 4 + 25 = 38, so ∣AB→∣=38|\overrightarrow{AB}| = \sqrt{38} again, the same result.

Part (c): Finding angle AOBAOB

The angle between OA→=a\overrightarrow{OA} = \mathbf{a} and OB→=b\overrightarrow{OB} = \mathbf{b} satisfies

a⋅b=∣a∣∣b∣cos⁡(∠AOB)\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos(\angle AOB)

Scalar product: a⋅b=(3)(1)+(−1)(4)+(2)(−1)=3−4−2=−3\mathbf{a}\cdot\mathbf{b} = (3)(1) + (-1)(4) + (2)(-1) = 3 - 4 - 2 = -3

Magnitudes: ∣a∣=32+(−1)2+22=9+1+4=14,∣b∣=12+42+(−1)2=1+16+1=18|\mathbf{a}| = \sqrt{3^2+(-1)^2+2^2} = \sqrt{9+1+4} = \sqrt{14}, \qquad |\mathbf{b}| = \sqrt{1^2+4^2+(-1)^2} = \sqrt{1+16+1} = \sqrt{18}

Solve for the angle: cos⁡(∠AOB)=a⋅b∣a∣∣b∣=−314×18=−3252=−367=−127≈−0.1890\cos(\angle AOB) = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \frac{-3}{\sqrt{14}\times\sqrt{18}} = \frac{-3}{\sqrt{252}} = \frac{-3}{6\sqrt7} = -\frac{1}{2\sqrt7} \approx -0.1890

∠AOB=cos⁡−1(−0.1890)≈100.9∘\angle AOB = \cos^{-1}(-0.1890) \approx 100.9^\circ

Check by recomputing independently: re-adding the scalar product term-by-term in reverse order, (−1)(4)+(2)(−1)+(3)(1)=−4−2+3=−3(-1)(4) + (2)(-1) + (3)(1) = -4 - 2 + 3 = -3 (matches. Also ∣a∣2∣b∣2=14×18=252|\mathbf{a}|^2|\mathbf{b}|^2 = 14\times18 = 252, and 252=67≈15.8745\sqrt{252} = 6\sqrt7 \approx 15.8745, so cos⁡(∠AOB)≈−3/15.8745≈−0.1890\cos(\angle AOB) \approx -3/15.8745 \approx -0.1890, giving ∠AOB≈100.9∘\angle AOB \approx 100.9^\circ again. Since the cosine is negative, the angle is correctly obtuse) it must not be reported as the acute reference angle cos⁡−1(0.1890)≈79.1∘\cos^{-1}(0.1890)\approx79.1^\circ.

Final answers

  • (a) AB→=(−25−3)\overrightarrow{AB} = \begin{pmatrix}-2\\5\\-3\end{pmatrix}
  • (b) ∣AB→∣=38|\overrightarrow{AB}| = \sqrt{38}
  • (c) ∠AOB≈100.9∘\angle AOB \approx 100.9^\circ