Find the direction cosines of the vector 4{i} - 3{j}.

Correct Answer is : 45,− 35\;4/5,-\;3/554,−53

Correct Answer is : 45,− 35\;4/5,-\;3/554,−53

Test Index

WAEC Further Mathematics 2010 Paper

Question : 27

Total: 40

4\hat{i} - 3\hat{j}

Solution:

Concept:

For a two-dimensional vector

x\hat{i} + y\hat{j}

, its direction cosines are

\frac{x}{|\mathbf{v}|}

and

\frac{y}{|\mathbf{v}|}

Explanation:

The given vector is

\mathbf{v} = 4\hat{i} - 3\hat{j}

Comparing it with

x\hat{i} + y\hat{j}

, we get

x=4

and

y=-3

First, find the magnitude of the vector:

|\mathbf{v}| = \sqrt{x^2 + y^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5.

Now divide each component by the magnitude:

\frac{x}{|\mathbf{v}|} = \frac{4}{5}, \qquad \frac{y}{|\mathbf{v}|} = \frac{-3}{5}.

Hence, the required direction cosines are

\frac{4}{5}

and

-\frac{3}{5}

Answer:

\frac{4}{5}, -\frac{3}{5}

(Option C).

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