Question

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8 (a) the point (a) is ((2, -3)) and the point (b) is ((7, 4)).

(i) express (overrightarrow{ab}) as a column vector.

(ii) find (|overrightarrow{ab}|).

(iii) (c) is the point such that (overrightarrow{ac} = 3overrightarrow{ab}).
find the coordinates of (c).

Explanation:

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<reasoning>

Calculate column vector AB

To find the column vector \(\overrightarrow{AB}\) from point \(A(x_1, y_1) = (2, -3)\) to point \(B(x_2, y_2) = (7, 4)\), we subtract the coordinates of the initial point \(A\) from the terminal point \(B\):
\[
\overrightarrow{AB} =

$$\begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}$$

=

$$\begin{pmatrix} 7 - 2 \\ 4 - (-3) \end{pmatrix}$$

=

$$\begin{pmatrix} 5 \\ 7 \end{pmatrix}$$

\]

Find magnitude of AB

The magnitude of a vector \(\overrightarrow{AB} =

$$\begin{pmatrix} x \\ y \end{pmatrix}$$

\), denoted as \(|\overrightarrow{AB}|\), is calculated using the Pythagorean theorem:
\[
|\overrightarrow{AB}| = \sqrt{x^2 + y^2}
\]
Substituting our components:
\[
|\overrightarrow{AB}| = \sqrt{5^2 + 7^2} = \sqrt{25 + 49} = \sqrt{74}
\]
Evaluating this numerically:
\[
\sqrt{74} \approx 8.60
\]

Determine coordinates of C

We are given that \(\overrightarrow{AC} = 3\overrightarrow{AB}\).
Let the coordinates of \(C\) be \((x_C, y_C)\). The vector \(\overrightarrow{AC}\) is:
\[
\overrightarrow{AC} =

$$\begin{pmatrix} x_C - 2 \\ y_C - (-3) \end{pmatrix}$$

=

$$\begin{pmatrix} x_C - 2 \\ y_C + 3 \end{pmatrix}$$

\]
Since \(\overrightarrow{AC} = 3\overrightarrow{AB}\):
\[

$$\begin{pmatrix} x_C - 2 \\ y_C + 3 \end{pmatrix}$$

= 3

$$\begin{pmatrix} 5 \\ 7 \end{pmatrix}$$

=

$$\begin{pmatrix} 15 \\ 21 \end{pmatrix}$$

\]
Now, solve for the coordinates of \(C\):
\[
x_C - 2 = 15 \implies x_C = 17
\]
\[
y_C + 3 = 21 \implies y_C = 18
\]
Thus, the coordinates of \(C\) are \((17, 18)\).
</reasoning>

<answer>

Question (i)

\[

$$\begin{pmatrix} 5 \\ 7 \end{pmatrix}$$

\]

Question (ii)

\[
\sqrt{74} \quad (\approx 8.60)
\]

Question (iii)

\[
(17, 18)
\]
</answer>

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Answer:

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<reasoning>

Calculate column vector AB

To find the column vector \(\overrightarrow{AB}\) from point \(A(x_1, y_1) = (2, -3)\) to point \(B(x_2, y_2) = (7, 4)\), we subtract the coordinates of the initial point \(A\) from the terminal point \(B\):
\[
\overrightarrow{AB} =

$$\begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}$$

=

$$\begin{pmatrix} 7 - 2 \\ 4 - (-3) \end{pmatrix}$$

=

$$\begin{pmatrix} 5 \\ 7 \end{pmatrix}$$

\]

Find magnitude of AB

The magnitude of a vector \(\overrightarrow{AB} =

$$\begin{pmatrix} x \\ y \end{pmatrix}$$

\), denoted as \(|\overrightarrow{AB}|\), is calculated using the Pythagorean theorem:
\[
|\overrightarrow{AB}| = \sqrt{x^2 + y^2}
\]
Substituting our components:
\[
|\overrightarrow{AB}| = \sqrt{5^2 + 7^2} = \sqrt{25 + 49} = \sqrt{74}
\]
Evaluating this numerically:
\[
\sqrt{74} \approx 8.60
\]

Determine coordinates of C

We are given that \(\overrightarrow{AC} = 3\overrightarrow{AB}\).
Let the coordinates of \(C\) be \((x_C, y_C)\). The vector \(\overrightarrow{AC}\) is:
\[
\overrightarrow{AC} =

$$\begin{pmatrix} x_C - 2 \\ y_C - (-3) \end{pmatrix}$$

=

$$\begin{pmatrix} x_C - 2 \\ y_C + 3 \end{pmatrix}$$

\]
Since \(\overrightarrow{AC} = 3\overrightarrow{AB}\):
\[

$$\begin{pmatrix} x_C - 2 \\ y_C + 3 \end{pmatrix}$$

= 3

$$\begin{pmatrix} 5 \\ 7 \end{pmatrix}$$

=

$$\begin{pmatrix} 15 \\ 21 \end{pmatrix}$$

\]
Now, solve for the coordinates of \(C\):
\[
x_C - 2 = 15 \implies x_C = 17
\]
\[
y_C + 3 = 21 \implies y_C = 18
\]
Thus, the coordinates of \(C\) are \((17, 18)\).
</reasoning>

<answer>

Question (i)

\[

$$\begin{pmatrix} 5 \\ 7 \end{pmatrix}$$

\]

Question (ii)

\[
\sqrt{74} \quad (\approx 8.60)
\]

Question (iii)

\[
(17, 18)
\]
</answer>

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