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)
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:
answer units

Explanation:

Step1: Calculate 3A + 2B

First, multiply vector A = 8i - 15j by 3 and vector B=-1i + 8j by 2. Then add them together.
\[

$$\begin{align*} 3\mathbf{A}+2\mathbf{B}&=3(8\mathbf{i}-15\mathbf{j})+2(-1\mathbf{i}+8\mathbf{j})\\ &=(3\times8\mathbf{i}-3\times15\mathbf{j})+(2\times(-1)\mathbf{i}+2\times8\mathbf{j})\\ &=(24\mathbf{i}-45\mathbf{j})+(-2\mathbf{i}+16\mathbf{j})\\ &=(24 - 2)\mathbf{i}+(-45 + 16)\mathbf{j}\\ &=22\mathbf{i}-29\mathbf{j} \end{align*}$$

\]

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

Here, $V_x = 22$ and $V_y=-29$. So, $\tan\theta=\frac{-29}{22}$.
$\theta=\arctan(\frac{-29}{22})\approx - 52.8^{\circ}$.
Since we want the angle between 0 and 360 degrees counter - clockwise with respect to the positive x - axis, we add 360 degrees to the negative angle.
$\theta\approx360^{\circ}- 52.8^{\circ}=307.2^{\circ}$

Answer:

$307.2^{\circ}$