Question

1a. Given that P = (5,4), Q = (7,3), R = (8,6), and S = (4,1), find the component form of the vector PQ + 3RS.

Answer 1

1a. A
1b. D
2a. C
2b. B
3a. C
3b. B

Answer 2

Answer:

1a) a) $\overrightarrow {\alpha} = (-10,-16)$, 1b) $\| \overrightarrow {\alpha} \| = 2\sqrt{89}$, 2a) $\overrightarrow {\alpha} = (-14,-21)$, 2b) $\| \overrightarrow {\alpha} \| = 7\sqrt{13}$, 3a) $\overrightarrow {\alpha} = (-18,-26)$, 3b) $\| \overrightarrow {\alpha} \| = 10\sqrt{10}$

Step-by-step explanation:

1a) The vectors $\overrightarrow {PQ}$ and $\overrightarrow {RS}$ are determined:

$\overrightarrow {PQ} = (7-5,3-4)$

$\overrightarrow {PQ} = (2, -1)$

$\overrightarrow {RS} = (4-8,1-6)$

$\overrightarrow {RS} = (-4, -5)$

The component form of the resultant vector is:

$\overrightarrow {\alpha} = (2 -12, -1-15)$

$\overrightarrow {\alpha} = (-10,-16)$

1b) The magnitude of the resultant vector is:

$\| \overrightarrow {\alpha} \| = \sqrt{(-10)^{2}+(-16)^{2}}$

$\| \overrightarrow {\alpha} \| = 2\sqrt{89}$

2a) The vectors $\overrightarrow {PQ}$ and $\overrightarrow {RS}$ are determined:

$\overrightarrow {PQ} = (7-5,3-4)$

$\overrightarrow {PQ} = (2, -1)$

$\overrightarrow {RS} = (4-8,1-6)$

$\overrightarrow {RS} = (-4, -5)$

The component form of the resultant vector is:

$\overrightarrow {\alpha} = (2 -16, -1-20)$

$\overrightarrow {\alpha} = (-14,-21)$

2b) The magnitude of the resultant vector is:

$\| \overrightarrow {\alpha} \| = \sqrt{(-14)^{2}+(-21)^{2}}$

$\| \overrightarrow {\alpha} \| = 7\sqrt{13}$

3a) The vectors $\overrightarrow {PQ}$ and $\overrightarrow {RS}$ are determined:

$\overrightarrow {PQ} = (7-5,3-4)$

$\overrightarrow {PQ} = (2, -1)$

$\overrightarrow {RS} = (4-8,1-6)$

$\overrightarrow {RS} = (-4, -5)$

The component form of the resultant vector is:

$\overrightarrow {\alpha} = (2 -20, -1-25)$

$\overrightarrow {\alpha} = (-18,-26)$

3b) The magnitude of the resultant vector is:

$\| \overrightarrow {\alpha} \| = \sqrt{(-18)^{2}+(-26)^{2}}$

$\| \overrightarrow {\alpha} \| = 10\sqrt{10}$