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Vectors: Question 9

Syllabus 3.7

Structured A2 7 marks

Points AA, BB and CC have position vectors OA→=(1−23),OB→=(321),OC→=(4−27)\overrightarrow{OA} = \begin{pmatrix}1\\-2\\3\end{pmatrix}, \qquad \overrightarrow{OB} = \begin{pmatrix}3\\2\\1\end{pmatrix}, \qquad \overrightarrow{OC} = \begin{pmatrix}4\\-2\\7\end{pmatrix} relative to an origin OO.

(a) Find AB→\overrightarrow{AB} and AC→\overrightarrow{AC}. [2]

(b) Use the scalar product to find angle BACBAC, correct to 11 decimal place. [3]

(c) State, with a reason, whether angle BACBAC is exactly 90∘90^\circ. [2]

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Worked solution

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

AB→=OB→−OA→=(321)−(1−23)=(24−2)\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix}3\\2\\1\end{pmatrix} - \begin{pmatrix}1\\-2\\3\end{pmatrix} = \begin{pmatrix}2\\4\\-2\end{pmatrix}

AC→=OC→−OA→=(4−27)−(1−23)=(304)\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \begin{pmatrix}4\\-2\\7\end{pmatrix} - \begin{pmatrix}1\\-2\\3\end{pmatrix} = \begin{pmatrix}3\\0\\4\end{pmatrix}

Part (b): Finding angle BACBAC

The angle at AA between the two vectors that meet there, AB→\overrightarrow{AB} and AC→\overrightarrow{AC}, satisfies

AB→⋅AC→=∣AB→∣∣AC→∣cos⁡(∠BAC)\overrightarrow{AB}\cdot\overrightarrow{AC} = |\overrightarrow{AB}||\overrightarrow{AC}|\cos(\angle BAC)

Scalar product: AB→⋅AC→=(2)(3)+(4)(0)+(−2)(4)=6+0−8=−2\overrightarrow{AB}\cdot\overrightarrow{AC} = (2)(3) + (4)(0) + (-2)(4) = 6 + 0 - 8 = -2

Magnitudes: ∣AB→∣=22+42+(−2)2=4+16+4=24=26,∣AC→∣=32+02+42=9+0+16=25=5|\overrightarrow{AB}| = \sqrt{2^2+4^2+(-2)^2} = \sqrt{4+16+4} = \sqrt{24} = 2\sqrt6, \qquad |\overrightarrow{AC}| = \sqrt{3^2+0^2+4^2} = \sqrt{9+0+16} = \sqrt{25} = 5

Solve for the angle: cos⁡(∠BAC)=−226×5=−2106=−156≈−0.08165\cos(\angle BAC) = \frac{-2}{2\sqrt6 \times 5} = \frac{-2}{10\sqrt6} = -\frac{1}{5\sqrt6} \approx -0.08165

∠BAC=cos⁡−1(−0.08165)≈94.7∘\angle BAC = \cos^{-1}(-0.08165) \approx 94.7^\circ

Check by recomputing independently: re-adding the scalar product in reverse order, (−2)(4)+(4)(0)+(2)(3)=−8+0+6=−2(-2)(4) + (4)(0) + (2)(3) = -8 + 0 + 6 = -2 (matches. Also ∣AB→∣2∣AC→∣2=24×25=600|\overrightarrow{AB}|^2|\overrightarrow{AC}|^2 = 24 \times 25 = 600, so ∣AB→∣∣AC→∣=600=106≈24.495|\overrightarrow{AB}||\overrightarrow{AC}| = \sqrt{600} = 10\sqrt6 \approx 24.495, giving cos⁡(∠BAC)≈−2/24.495≈−0.08165\cos(\angle BAC) \approx -2/24.495 \approx -0.08165 and ∠BAC≈94.7∘\angle BAC \approx 94.7^\circ again) the same result.

Part (c): Is angle BACBAC exactly 90∘90^\circ?

Two vectors are exactly perpendicular only when their scalar product is exactly zero. Here

AB→⋅AC→=−2≠0\overrightarrow{AB}\cdot\overrightarrow{AC} = -2 \neq 0

so angle BACBAC is not exactly 90∘90^\circ. It is close to a right angle (the negative, near-zero cosine explains why the computed value 94.7∘94.7^\circ is only slightly more than 90∘90^\circ), but the exact right-angle condition AB→⋅AC→=0\overrightarrow{AB}\cdot\overrightarrow{AC}=0 is not satisfied.

Final answers

  • (a) AB→=(24−2)\overrightarrow{AB} = \begin{pmatrix}2\\4\\-2\end{pmatrix}, AC→=(304)\overrightarrow{AC} = \begin{pmatrix}3\\0\\4\end{pmatrix}
  • (b) ∠BAC≈94.7∘\angle BAC \approx 94.7^\circ
  • (c) No, since AB→⋅AC→=−2≠0\overrightarrow{AB}\cdot\overrightarrow{AC} = -2 \neq 0, angle BACBAC is not exactly 90∘90^\circ.