Answer:
(B) The magnitude is 7sqrt2, and the direction angle is approximately 135 degrees.
Explanation:
From the figure, the coordinate of the tail of the vector,
![(x_1,y_1)=(3,-6).](https://img.qammunity.org/2021/formulas/mathematics/high-school/j2tbrbm6705r9qo3gt6bl4d2g2h88dg40o.png)
The coordinate of the head of the vector,
![(x_2,y_2)=(-4,1).](https://img.qammunity.org/2021/formulas/mathematics/high-school/3ss2gnh5fxj73iquqbfr7v897fijcz0pal.png)
The magnitude of a vector is the length between the head and tail of the vector.
By using the distance formula,
the magnitude of the vector
![= √((1-(-6))^2+(-4-3)^2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/gvyq969hw7vwxkerxlniqbrctvcl9bmjln.png)
![=√((7^2+(-7)^2)\\\\=√(98)\\\\=7√(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/x3bq4zs46e53wf6qyw4e48lqo12zyszdjc.png)
Now, let
be the angle made by the vector with the positive direction of the x-axis, so
![\tan\theta = (y_2-y_1)/(x_2-x_1)\\\\\Rightarrow \tan\theta = (1-(-6))/(-4-3)=(7)/(-7)=-1 \\\\\Rightarrow \theta = \tan^(-1)(-1) \\\\\Rightarrow \theta =\pi- \tan^(-1)1 \\\\\Rightarrow \theta = \pi-(\pi)/(4) \\\\\Rightarrow \theta =(3\pi)/(4) \\\\\Rightarrow \theta = 135 ^(\circ).](https://img.qammunity.org/2021/formulas/mathematics/high-school/uzrvsf1cra52zdd2857is1a43fxs66bsk8.png)
So, the magnitude of the vector is
which makes
with the positive direction of the x-axis.
Hence, option (B) is correct.