Vector Dot Product and Angle

00:00 6 marks

Consider the vectors \(\mathbf{a} = \mathbf{i} + 2\mathbf{j} – \mathbf{k}\) and \(\mathbf{b} = 3\mathbf{i} – \mathbf{j} + 2\mathbf{k}\).

(a) Find \(\mathbf{a} \cdot \mathbf{b}\). [2]

(b) Find the angle between vectors \(\mathbf{a}\) and \(\mathbf{b}\), giving your answer in degrees. [4]

Part a

4 minutes 2 marks

Question

Find \(\mathbf{a} \cdot \mathbf{b}\). [2]

Hint Hint

Hint

Multiply the corresponding components: \(1\) with \(3\), \(2\) with \(-1\), and \(-1\) with \(2\), then add the products.

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Complete Answer

\(\mathbf{a} \cdot \mathbf{b} = 3 + (-2) + (-2) = -1\)

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Part b

6 minutes 4 marks

Question

Find the angle between vectors \(\mathbf{a}\) and \(\mathbf{b}\), giving your answer in degrees. [4]

Hint Hint

Hint

First, compute the magnitudes of \(\mathbf{a}\) and \(\mathbf{b}\). Then use \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|}\) and take the inverse cosine.

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Complete Answer

\(\theta \approx \arccos\left(\frac{-1}{\sqrt{6}\,\sqrt{14}}\right) \approx 96.3^\circ\)

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