If vector $\overrightarrow { \mathrm { A } }$ is added to vector $\vec { B }$ ,the result is $- 9 \hat { \mathbf { i } } - 8 \hat { \mathbf { j } }$ .If $\vec { B }$ is subtracted from $\overrightarrow { \mathrm { C } }$ ,the result is $5 \tilde { \mathbf { i } } + 4 \hat { \mathbf { j } }$ .What is the direction of $\vec { B }$ (to the nearest degree)?
A)225$\degree$
B)221$\degree$
C)230$\degree$
D)236$\degree$
E)206$\degree$

Question

Quizplus · Accepted Answer

Let's represent vector $\vec { B }$ as $\begin{pmatrix}a\b\end{pmatrix}$.
From the given information, we have:
$\overrightarrow { \mathrm { A } }+\vec { B }=-9 \hat { \mathbf { i } } - 8 \hat { \mathbf { j } }\Rightarrow \begin{pmatrix}a\b\end{pmatrix}=-9 \hat { \mathbf { i } } - 8 \hat { \mathbf { j } }-\overrightarrow { \mathrm { A } }$
$\overrightarrow { \mathrm { C } }-\vec { B }=5 \tilde { \mathbf { i } } + 4 \hat { \mathbf { j } }\Rightarrow \overrightarrow { \mathrm { C } }- \begin{pmatrix}a\b\end{pmatrix}=5 \tilde { \mathbf { i } } + 4 \hat { \mathbf { j } }$
Substituting the value of $\begin{pmatrix}a\b\end{pmatrix}$ in the above equation, we get:
$$\overrightarrow { \mathrm { C } }- (-9 \hat { \mathbf { i } } - 8 \hat { \mathbf { j } }-\overrightarrow { \mathrm { A } })=5 \tilde { \mathbf { i } } + 4 \hat { \mathbf { j } }$$
$$\Rightarrow (\overrightarrow { \mathrm { C } }+\overrightarrow { \mathrm { A } }+9 \hat { \mathbf { i } } + 8 \hat { \mathbf { j } })-\begin{pmatrix}a\b\end{pmatrix}=5 \tilde { \mathbf { i } } + 4 \hat { \mathbf { j } }$$
$$\Rightarrow \begin{pmatrix}a\b\end{pmatrix}+(\overrightarrow { \mathrm { C } }+\overrightarrow { \mathrm { A } }+9 \hat { \mathbf { i } } + 8 \hat { \mathbf { j } })=5 \tilde { \mathbf { i } } + 4 \hat { \mathbf { j } }$$
$$\Rightarrow -9 \hat { \mathbf { i } } - 8 \hat { \mathbf { j } }+(\overrightarrow { \mathrm { C } }+\overrightarrow { \mathrm { A } }+9 \hat { \mathbf { i } } + 8 \hat { \mathbf { j } })=5 \tilde { \mathbf { i } } + 4 \hat { \mathbf { j } }$$
$$\Rightarrow \overrightarrow { \mathrm { C } }+\overrightarrow { \mathrm { A } }=22 \hat { \mathbf { i } } + 20 \hat { \mathbf { j } }$$
Now we can find the angle between $\overrightarrow {\mathrm { A }}$ and $\vec { B }$ by finding the dot product of the two vectors and then dividing it by the product of their magnitudes.
$$\overrightarrow {\mathrm { A }} \cdot \vec { B } = |\overrightarrow {\mathrm { A }}||\vec { B }|\cos\theta$$
$$\Rightarrow \overrightarrow {\mathrm { A }} \cdot \vec { B } = | 22 \hat { \mathbf { i } } + 20 \hat { \mathbf { j } }||\vec { B }|\cos\theta$$
$$\Rightarrow (-9a-8b) = \sqrt{22^2 + 20^2}\sqrt{a^2+b^2}\cos\theta$$
Now substituting the value of $\begin{pmatrix}a\b\end{pmatrix}$ in the above equation, we get:
$$(-9(-9) - 8(-8))=\sqrt{22^2 + 20^2}\sqrt{(-9)^2+(-8)^2}\cos\theta$$
$$\Rightarrow 15=2\sqrt{145}\cos\theta$$
$$\Rightarrow \cos\theta = \frac{15}{2\sqrt{145}}$$
$$\Rightarrow \theta \approx 221^\circ$$
Therefore, the direction of $\vec{B}$ is 221$^\circ$ to the nearest degree, which is option $\boxed{\mathrm{(B)}}$.

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