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Grade 10 Mathematics Learners’ Book

INNODEMS

Subsection 2.8.5 Multiplication of Vectors by a Scalar

Activity 2.8.5.

Work in groups
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(a)

On a graph paper, draw the \(x\) axis and \(y\) axis.
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(b)

Draw a directed line passing through point \(A(0,2)\) and \(B(2,2)\text{.}\)
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(c)

From point \(B(2,2)\text{,}\) draw directed line to point \(C(4,2)\text{.}\)
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(d)

Determine the coordinate representations of \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\text{..}\)
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(e)

How does \(\overrightarrow{AB}\) relate to \(\overrightarrow{AC}\text{?}\)
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(f)

Discuss and share your findings with the rest of the class.
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\(\textbf{Key Takeaway}\)
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Positive Scalar
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In Figure 2.8.25, the vector \(\mathbf{PQ}\) is represented as \(\mathbf{a}\text{.}\) When we multiply \(\mathbf{a}\) by a positive scalar, say \(2\text{,}\) the length of the vector doubles, making it \(\mathbf{2a}\) as shown in Figure 2.8.26. The direction of the vector remains unchanged, but its magnitude increases.
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Figure 2.8.25.
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In Figure 2.8.26, the vector \(\mathbf{PN}\) is given by: \(\mathbf{PN} = \mathbf{a} + \mathbf{a} = 2\mathbf{a}\text{.}\) This means \(\mathbf{PN}\) has the same direction as \(\mathbf{PQ}\text{,}\) but its magnitude twice that of \(\mathbf{PQ}.\)
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Figure 2.8.26.
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Negative Scalar
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Consider the vector \(\mathbf{AB}\text{,}\)denoted as \(\mathbf{a}\text{,}\) in Figure 2.8.27 The vector points to the right and has a magnitude of \(\mathbf{a}\text{.}\)
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Figure 2.8.27.
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In Figure 2.8.28, the vector \(\mathbf{AB}\) is obtained by multiplying \(\mathbf{AB}\) by \(-2\text{,}\) giving:
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\begin{equation*} \mathbf{AC} = \mathbf{-2} \times \mathbf{a} = \mathbf{-2a} \end{equation*}
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This means that \(\mathbf{AC}\) has twice the magnitude of \(\mathbf{AB}\text{,}\) but its direction is reversed.
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Figure 2.8.28.
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Multiplying a vector by a negative scalar reverses its direction, making it point in the opposite direction.
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Zero Scalar
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When a vector \(\mathbf{a}\text{,}\) as shown in Figure 2.8.29, is multiplied by \(0\text{,}\) its magnitude becomes \(0\text{,}\) resulting in a zero vector.
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Figure 2.8.29.
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\begin{equation*} \mathbf{a} \cdot 0 = 0 \end{equation*}
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Example 2.8.30.

Given the vectors \(u = 2p + 5q\) and \(v = p - 3q\text{,}\) express \(3u + 2v \) in terms of \(p\) and \(q:\)
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Solution.
Substitute the given expressions for u and v.
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\begin{align*} 3u + 2v \amp = 3(2p + 5q) + 2(p - 3q) \\ \amp = 6p + 15q + 2p - 6q \end{align*}
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Combine like terms (terms with p and terms with q).
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\begin{align*} 3u + 2v \amp = (6p + 2p) + (15q - 6q) \\ \amp = 8p + 9q \end{align*}
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Exercises Exercises

1.

Simplify the following:
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  1. \(\displaystyle 5x + 3y - z + 2(3x - z) + (8x - 6y)\)
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  2. \(\displaystyle (\mathbf{a} - \mathbf{b}) + (c - \mathbf{a}) + (\mathbf{b} - c)\)
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  3. \(\displaystyle 4m - 2n + 5(k - m) + 2(m + n)\)
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2.

Given that \(x = 3m - n\) and \(y = n +4m\text{,}\) express the following vectors in terms of \(m\) and \(n:\)
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  1. \(\displaystyle 3x\)
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  2. \(\displaystyle \frac{2}{3} y\)
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  3. \(\displaystyle 6x - 9y\)
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  4. \(\displaystyle 3y - x\)
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3.

A pentagon \(ABCDE\) with \(\overrightarrow{AB} = m\text{,}\) \(\overrightarrow{BC} = n\text{,}\) and \(\overrightarrow{CD} = k\text{.}\) Express the following vectors in terms of \(m\text{,}\) \(n\text{,}\) and \(k:\)
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  1. \(\displaystyle \overrightarrow{AC}\)
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  2. \(\displaystyle \overrightarrow{AD}\)
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  3. \(\displaystyle \overrightarrow{AE}\)
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Checkpoint 2.8.31.

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