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WAEC Further Mathematics 2012 Paper

Question : 7

Total: 40

\overrightarrow{QR}= (3i + 2j)

\overrightarrow{SR}= (-5i + 3j)

\overrightarrow{SQ}

Solution:

Concept:

Use the triangle law of vector addition along the path

S \to R \to Q

Explanation:

Given

\overrightarrow{QR} = 3\mathbf{i} + 2\mathbf{j}

, the reverse vector is

\overrightarrow{RQ} = -\overrightarrow{QR} = -3\mathbf{i} - 2\mathbf{j}

Also,

\overrightarrow{SR} = -5\mathbf{i} + 3\mathbf{j}

By the triangle law,

\overrightarrow{SQ} = \overrightarrow{SR} + \overrightarrow{RQ}

Substitute the vectors:

\overrightarrow{SQ} = (-5\mathbf{i} + 3\mathbf{j}) + (-3\mathbf{i} - 2\mathbf{j})

Add the

\mathbf{i}

-components and the

\mathbf{j}

-components separately:

\overrightarrow{SQ} = (-5 - 3)\mathbf{i} + (3 - 2)\mathbf{j} = -8\mathbf{i} + \mathbf{j}

Therefore, the required vector is

\overrightarrow{SQ} = -8\mathbf{i} + \mathbf{j}

This result is not listed among the given options, so the options likely contain a misprint.

Answer:

\overrightarrow{SQ} = -8\mathbf{i} + \mathbf{j}

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