Question

for a vector v, v_x = v cos θ v_y = v sin θ v_x^2 + v_y^2 = v^2 tan θ = v_y/v_x
question 4 (4 points) saved
vector a can be expressed in (i,j,k) notation as 8i - 15j, and vector b as -1i + 8j. determine the angle of the vector 3a + 2b. the angle must be specified as a value between 0 and 360 degrees counter - clockwise with respect to the positive x - axis. your answer: 270 degrees answer units

Explanation:

Step1: Calculate 3A + 2B

Given $\vec{A}=8\vec{i}-15\vec{j}$ and $\vec{B}=- 1\vec{i}+8\vec{j}$.
$3\vec{A}+2\vec{B}=3(8\vec{i}-15\vec{j})+2(-1\vec{i}+8\vec{j})=(3\times8 + 2\times(-1))\vec{i}+(3\times(-15)+2\times8)\vec{j}=(24 - 2)\vec{i}+(-45 + 16)\vec{j}=22\vec{i}-29\vec{j}$

Step2: Find the angle $\theta$

We know that $\tan\theta=\frac{V_y}{V_x}$, where $V_x = 22$ and $V_y=-29$.
$\tan\theta=\frac{-29}{22}\approx - 1.3182$
$\theta=\arctan(-1.3182)$
The principal - value of $\arctan$ is in the range $(-90^{\circ},90^{\circ})$. Since $V_x>0$ and $V_y < 0$, the angle is in the fourth quadrant.
$\theta=\arctan(-1.3182)+360^{\circ}\approx - 52.8^{\circ}+360^{\circ}=307.2^{\circ}$

Answer:

$307.2$ degrees