Two points P(3, -7) and Q(-2, 5) are given.
(a) Calculate the components of the vector PQ
(b) Calculate the magnitude and direction of PQ
(c) Is PQ equivalent to the vector between the points R(-2, -4) and S(-7, 8)?
The components of PQ are the differences of the components of the final and initial point.
\[ \begin{align}
PQ &= \langle q_1 - p_1, q_2- p_2 \rangle \\
&= \langle -2 - 3, 5 - (-7) \rangle \\
&= \langle -5, 12 \rangle \\
\end{align} \ \]
The magnitude is the square root of the sum of the squares of the differences of the x and y values. The direction is the gradient of the line containing PQ.
\[ \begin{align}
|| PQ || &= \sqrt{(q_1-p_1)^2 + (q_2-p_2)^2} \\
&= \sqrt{(-2-3)^2 + (5-(-7))^2} \\
&= \sqrt{25 + 144} \\
&= 13
\end{align} \ \]
The gradient is
\[ \begin{align}
m &= \frac{q_2-p_2}{q_1-p_1} \\
&= \frac{5 - (-7)}{-2 - 3} \\
&= \frac{12}{-5} \\
\end{align} \ \]
Two vectors are equivalent if they have the same magnitude and direction.
The components of RS are:
\[ \begin{align}
RS &= \langle -7 - (-2), 8 - (-4) \rangle \\
&= \langle -5, 12 \rangle \\
\end{align} \ \]
The magnitude is
\[ \begin{align}
|| RS || &= \sqrt{(-7-(-2))^2 + (8-(-4))^2} \\
&= \sqrt{25 + 144} \\
&= 13
\end{align} \ \]
The gradient is
\[ \begin{align}
m &= \frac{8 - (-4)}{-7 - (-2)} \\
&= \frac{12}{-5} \\
\end{align} \ \]
RS has the same magnitude and direction as PQ. So RS and PQ are equivalent.
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