Question

find the angle between the following two vectors: (vec{a} = langle -5, -2
angle) and (vec{b} = langle 4, 4
angle) (round your answer to the nearest whole degree.) select one: a. (117^circ) b. (157^circ) c. (113^circ) d. (31^circ)

Explanation:

Step1: Recall the dot product formula for angle between vectors

The formula to find the angle \( \theta \) between two vectors \( \vec{a} = \langle a_1, a_2
angle \) and \( \vec{b} = \langle b_1, b_2
angle \) is \( \cos\theta=\frac{\vec{a}\cdot\vec{b}}{\vert\vec{a}\vert\vert\vec{b}\vert} \), where \( \vec{a}\cdot\vec{b}=a_1b_1 + a_2b_2 \), \( \vert\vec{a}\vert=\sqrt{a_1^{2}+a_2^{2}} \) and \( \vert\vec{b}\vert=\sqrt{b_1^{2}+b_2^{2}} \).

Step2: Calculate the dot product \( \vec{a}\cdot\vec{b} \)

Given \( \vec{a}=\langle - 5,-2
angle \) and \( \vec{b}=\langle4,4
angle \), then \( \vec{a}\cdot\vec{b}=(-5)\times4+(-2)\times4=-20 - 8=-28 \).

Step3: Calculate \( \vert\vec{a}\vert \)

\( \vert\vec{a}\vert=\sqrt{(-5)^{2}+(-2)^{2}}=\sqrt{25 + 4}=\sqrt{29}\approx5.385 \).

Step4: Calculate \( \vert\vec{b}\vert \)

\( \vert\vec{b}\vert=\sqrt{4^{2}+4^{2}}=\sqrt{16 + 16}=\sqrt{32}\approx5.657 \).

Step5: Calculate \( \cos\theta \)

Substitute the values into the formula: \( \cos\theta=\frac{-28}{\sqrt{29}\times\sqrt{32}}\approx\frac{-28}{5.385\times5.657}\approx\frac{-28}{30.46}\approx - 0.919 \).

Step6: Find \( \theta \)

\( \theta=\arccos(-0.919)\approx157^\circ \) (rounded to the nearest whole degree).

Answer:

b. \( 157^\circ \)