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WAEC Further Mathematics 2016 Paper

Question : 35

Total: 40

Given that

r = 3i + 4j

t = -5i + 12j

, find the acute angle between them.

Solution:

Concept:

Use the dot product formula to relate the scalar product of the two vectors to the cosine of the angle between them.

Explanation:

Find the dot product:

\vec{r}\cdot \vec{t} = (3i + 4j)\cdot(-5i + 12j) = (3)(-5) + (4)(12) = -15 + 48 = 33

Find the magnitudes:

|\vec{r}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5

|\vec{t}| = \sqrt{(-5)^2 + 12^2} = \sqrt{169} = 13

Use

\cos\theta = \frac{\vec{r}\cdot\vec{t}}{|\vec{r}||\vec{t}|}

\cos\theta = \frac{33}{5 \times 13} = \frac{33}{65}

Therefore,

\theta = \cos^{-1}\left(\frac{33}{65}\right) \approx 59.5^{\circ}

Since

59.5^{\circ} < 90^{\circ}

, this is the acute angle between the vectors.

Answer:

The acute angle between them is

59.5^{\circ}

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