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WAEC Further Mathematics 2015 Paper

Question : 16

Total: 40

x = i - 3j

y = 6i + j

, calculate the angle between x and y.

Solution:

Concept:

The angle between two vectors is found using the dot product formula

\overrightarrow{x}\cdot\overrightarrow{y} = |\overrightarrow{x}||\overrightarrow{y}|\cos\theta

Explanation:

Given

\overrightarrow{x} = i - 3j

and

\overrightarrow{y} = 6i + j

Compute the dot product:

\overrightarrow{x}\cdot\overrightarrow{y} = (1)(6) + (-3)(1) = 6 - 3 = 3

Find the magnitude of

\overrightarrow{x}

|\overrightarrow{x}| = \sqrt{1^2 + (-3)^2} = \sqrt{10}

Find the magnitude of

\overrightarrow{y}

|\overrightarrow{y}| = \sqrt{6^2 + 1^2} = \sqrt{37}

Substitute into the dot product formula:

3 = (\sqrt{10})(\sqrt{37})\cos\theta = \sqrt{370}\cos\theta

Therefore,

\cos\theta = \frac{3}{\sqrt{370}} \approx 0.1559

Taking inverse cosine:

\theta = \cos^{-1}(0.1559) \approx 81^\circ

Answer:

The angle between

\overrightarrow{x}

and

\overrightarrow{y}

81^\circ

, which is Option C.

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