8.5 Vectors in Cartesian Plane - SPM Additional Mathematics

(A) Vectors in Cartesian Coordinates

1. A unit vector is a vector whose magnitude is one unit.

2. A unit vector that is parallel to the x-axis is denoted by

\underset{˜}{i}

while a unit vector that is parallel to the y-axis is denoted by

\underset{˜}{j}

3. The unit vector can be expressed in columnar form as below:

\begin{array}{l} \vec{O A} = x \underset{˜}{i} + y \underset{˜}{j} \\ = (\begin{matrix} x \\ y \end{matrix}) \leftarrow (Column Vector) \end{array}

4. The magnitudes of the unit vectors are

| \underset{˜}{i} | = | \underset{˜}{j} | = 1.

5. The magnitude of the vector

\vec{O A}

can be calculated using the Pythagoras’ Theorem.

| \vec{O A} | = \sqrt{x^{2} + y^{2}}

(B) Unit Vector in the Direction of a Vector

Unit vector of \underset{˜}{a}, \underset{˜}{\hat{a}} = \frac{x \underset{˜}{i} + y \underset{˜}{j}}{\sqrt{x^{2} + y^{2}}}

Example 1:

\underset{˜}{r} = k \underset{˜}{i} - 8 \underset{˜}{j}

and

| \underset{˜}{r} | = 10

, find the values of k. Determine the unit vector in the direction of

\underset{˜}{r}

for each value of k.

Solution:

\begin{array}{l} Given | \underset{˜}{r} | = 10 \\ \sqrt{x^{2} + y^{2}} = 10 \\ \sqrt{k^{2} + {(- 8)}^{2}} = 10 \\ k^{2} + 64 = 100 \\ k = \pm 6 \\ Unit vector of \hat{\underset{˜}{r}} = \frac{x \underset{˜}{i} + y \underset{˜}{j}}{\sqrt{x^{2} + y^{2}}} \\ When k = 6, When k = - 6 \\ \hat{\underset{˜}{r}} = \frac{6 \underset{˜}{i} - 8 \underset{˜}{j}}{10} = \frac{3 \underset{˜}{i} - 4 \underset{˜}{j}}{5} \hat{\underset{˜}{r}} = \frac{- 6 \underset{˜}{i} - 8 \underset{˜}{j}}{10} = \frac{- 3 \underset{˜}{i} - 4 \underset{˜}{j}}{5} \\ \hat{\underset{˜}{r}} = \frac{1}{5} (3 \underset{˜}{i} - 4 \underset{˜}{j}) \hat{\underset{˜}{r}} = \frac{1}{5} (- 3 \underset{˜}{i} - 4 \underset{˜}{j}) \end{array}

Example 2:

It is given that

\underset{˜}{a} = (\begin{array}{l} 6 \\ 3 \end{array}) and \underset{˜}{b} = (\begin{array}{l} 3 \\ 7 \end{array}) .

(a) Find

\underset{˜}{b} - \underset{˜}{a} and | \underset{˜}{b} - \underset{˜}{a} | .

(b) Hence, find the unit vector in the direction of

\underset{˜}{b} - \underset{˜}{a} .

Solution:

(a)

\begin{array}{l} \underset{˜}{b} - \underset{˜}{a} = (\begin{array}{l} 3 \\ 7 \end{array}) - (\begin{array}{l} 6 \\ 3 \end{array}) = (\begin{array}{l} 3 - 6 \\ 7 - 3 \end{array}) = (\begin{array}{l} - 3 \\ 4 \end{array}) \\ | \underset{˜}{b} - \underset{˜}{a} | = \sqrt{{(- 3)}^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 \end{array}

(b)

\begin{array}{l} The unit vector in the direction of \underset{˜}{b} - \underset{˜}{a} \\ = \frac{1}{5} (\begin{array}{l} - 3 \\ 4 \end{array}) \\ = (\begin{array}{l} - \frac{3}{5} \\ \frac{4}{5} \end{array}) \end{array}