## Solve Lesson 8 Page 62 SBT Math 10 – Kite>

Topic

Find real numbers a and b such that each of the following pairs of vectors are equal:

a) $$\overrightarrow m = (2a + 3;b – 1)$$ and $$\overrightarrow n = (1; – 2)$$

b) $$\overrightarrow u = (3a – 2;5)$$and $$\overrightarrow v = (5;2b + 1)$$

c) $$\overrightarrow x = (2a + b;2b)$$ and $$\overrightarrow y = (3 + 2b;b – 3a)$$

Solution method – See details

$$\overrightarrow a = ({x_1};{y_1})$$ and $$\overrightarrow b = ({x_2};{y_2})$$ are equal if and only if $$\left\{ \begin{ array}{l}{x_1} = {x_2}\\{y_1} = {y_2}\end{array} \right.$$

Detailed explanation

a) $$\overrightarrow m = (2a + 3;b – 1)$$ and $$\overrightarrow n = (1; – 2)$$

$$\overrightarrow m = \overrightarrow n \Leftrightarrow \left\{ \begin{array}{l}2a + 3 = 1\\b – 1 = – 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = – 1\\b = – 1\end{array} \right.$$

b) $$\overrightarrow u = (3a – 2;5)$$and $$\overrightarrow v = (5;2b + 1)$$

$$\overrightarrow u = \overrightarrow v \Leftrightarrow \left\{ \begin{array}{l}3a – 2 = 5\\5 = 2b + 1\end{array} \right. \Leftrightarrow \left\{ \ begin{array}{l}a = \frac{7}{3}\\b = 2\end{array} \right.$$

c) $$\overrightarrow x = (2a + b;2b)$$ and $$\overrightarrow y = (3 + 2b;b – 3a)$$

$$\overrightarrow x = \overrightarrow y \Leftrightarrow \left\{ \begin{array}{l}2a + b = 3 + 2b\\2b = b – 3a\end{array} \right. \Leftrightarrow \left\ { \begin{array}{l}2a – b = 3\\3a + b = 0\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}a = \frac{3} {5}\\b = – \frac{9}{5}\end{array} \right.$$