Multiplication of a vector by a scalar

Multiplication of a Vector by a Scalar

If a vector U has a magnitude m units and makes an angle $\theta$ with a positive x axis, then doubling the magnitude of U gives a vector with magnitude 2m.

From the figure above, $\overline{OP} = u$ and $\overline{PQ} = u$. So $\overline{OQ} = \overline{OP} + \overline{PQ} = u + u = 2u$ Also $|u| = m$ which implies that

$|u + u| = m + m = 2m$

Generally if U = $(u_1, u_2)$ and t is any non zero real number while $(u_1, u_2)$ are also real numbers, then

$\underline{tu} = t(u_1, u_2)$ $= (tu_1, tu_2)$

It follows therefore that the vector $(tu_1, tu_2)$ is a scalar multiple of vector $(u_1, u_2)$.

$\text{Similarly if} \quad U = u_{zi} i + u_{zj} j,$ $\text{Then}$ $\underline{tu} = tu_1 i + tu_2 j$

Example 1

If a = 3i + 3j and b = 5i + 4j Find – 5a + 3b

Solution

$$-5a + 3b = -5 (3i + 3j) + 3 (5i + 4j)$$

Since $a = 3i + 3j$ and $b = (5i + 4j)$

$$\begin{aligned} &= -5a + 3b = (-15i + -15j) + (15i + 12j) \\ &= (-15i + 15i) + -15j + 12j) \\ &= 0i + -3j \\ &= -3j \end{aligned}$$

Therefore $-5a + 3b = -3j$

Example 2

Given that p = (8, 6) and q = (7, 9). Find 9p – 8q

$$\begin{aligned} \text{Solution} \ \text{Given p} &= (8, 6) \text{ and q} = (7, 9) \\ 9p - 8q &= 9(8, 6) - 8(7, 9) \\ &= (9 \times 8, 9 \times 6) - (8 \times 7, 8 \times 9) \\ &= (72, 54) - (56, 72) \\ &= (72-56), (54-72) = (16, -18) \\ \text{Therefore 9p - 8q} &= (16, -18), \end{aligned}$$