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Figure 2 shows a sketch of a triangle ABC - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 1
Question 9
Figure 2 shows a sketch of a triangle ABC. Given \( \overline{AB} = 2i + 3j + k \) and \( \overline{BC} = i - 9j + 3k \), show that \( \angle BAC = 105.9^\circ \) t... show full transcript
Worked Solution & Example Answer:Figure 2 shows a sketch of a triangle ABC - Edexcel - A-Level Maths Pure - Question 9 - 2017 - Paper 1
Step 1
Find \( \overline{AC} \)
Answer
First, we find ( \overline{AC} ) using the equation: [ \overline{AC} = \overline{AB} + \overline{BC} ].
Substituting the given vectors, we have:
[ \overline{AC} = (2i + 3j + k) + (i - 9j + 3k) ] [ \overline{AC} = (2 + 1)i + (3 - 9)j + (1 + 3)k ] [ \overline{AC} = 3i - 6j + 4k ]
Step 2
Find \( |\overline{AB}| \), \( |\overline{BC}| \), and \( |\overline{AC}| \)
Answer
Now, we calculate the lengths of the sides:
- For ( |\overline{AB}| ): [ |\overline{AB}| = \sqrt{(2^2 + 3^2 + 1^2)} = \sqrt{14} ]
- For ( |\overline{BC}| ): [ |\overline{BC}| = \sqrt{(1^2 + (-9)^2 + 3^2)} = \sqrt{91} ]
- For ( |\overline{AC}| ): [ |\overline{AC}| = \sqrt{(3^2 + (-6)^2 + 4^2)} = \sqrt{61} ]
Step 3
Use the Cosine Rule to find \( \angle BAC \)
Answer
To find ( \angle BAC ), we apply the Cosine Rule: [ \cos(BAC) = \frac{|\overline{AB}|^2 + |\overline{AC}|^2 - |\overline{BC}|^2}{2 |\overline{AB}| |\overline{AC}|} ]
Substituting the lengths we calculated: [ \cos(BAC) = \frac{14 + 61 - 91}{2 \sqrt{14} \sqrt{61}} ] [ \cos(BAC) = \frac{-16}{2 \sqrt{14} \sqrt{61}} ]
Now, compute the angle: [ BAC = \cos^{-1}\left( \frac{-16}{2 \sqrt{14} \sqrt{61}} \right) \approx 105.9^\circ ]