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:

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 \(B\):

$$ \overrightarrow{AB} =
LATEXBLOCK0
=
LATEXBLOCK1
=
LATEXBLOCK2
$$

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:

$$ \overrightarrow{AC} =
LATEXBLOCK4
=
LATEXBLOCK5
$$

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

$$
LATEXBLOCK6
= 3
LATEXBLOCK7
=
LATEXBLOCK8
$$

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 \((17, 18)\).

Answer:

Question (i)

$$
LATEXBLOCK0
$$

Question (ii)

$$ \sqrt{74} \quad (\approx 8.60) $$

Question (iii)

$$ (17, 18) $$