Question

for a vector v, $v_x = vcos\theta$, $v_y = vsin\theta$, $v_x^{2}+v_y^{2}=v^{2}$, $\tan\theta = v_y/v_x$. 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: 307.2 degrees answer units

Explanation:

Step1: Calculate 3A + 2B

First, find 3A and 2B:
3A = 3(8i - 15j)=24i - 45j
2B = 2(-1i + 8j)=-2i + 16j
Then 3A + 2B=(24i - 45j)+(-2i + 16j)=(24 - 2)i+(-45 + 16)j = 22i - 29j

Step2: Find the angle using the formula $\tan\theta=\frac{V_y}{V_x}$

Here $V_x = 22$ 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