Given the vector $\vec{u}=\langle 4,5\rangle$, find the magnitude and angle in which the vector points (measured counterclockwise from the positive $x$-axis, $0 \leq \theta<2 \pi$ ) \[ \begin{array}{l} \|\vec{u}\|= \\ \theta= \end{array} \] Question Help: $\square$ Video $1 \square$ Video $2 \square$ Video $3 \square$ Video 4 Submit Question
Solution
Step 1 ：Given the vector $\vec{u}=\langle 4,5\rangle$, we are asked to find the magnitude and angle in which the vector points (measured counterclockwise from the positive $x$-axis, $0 \leq \theta<2 \pi$ ).
Step 2 ：The magnitude of a vector $\vec{u}=\langle x,y\rangle$ is given by the formula $\|\vec{u}\|=\sqrt{x^2+y^2}$. Substituting $x=4$ and $y=5$ into the formula, we get $\|\vec{u}\|=\sqrt{4^2+5^2} = \sqrt{16+25} = \sqrt{41} \approx 6.40$.
Step 3 ：The angle $\theta$ which the vector points can be found using the formula $\theta=\arctan(\frac{y}{x})$. Substituting $x=4$ and $y=5$ into the formula, we get $\theta=\arctan(\frac{5}{4}) \approx 0.896$ radians or $51.34$ degrees.
Step 4 ：Since the vector $\vec{u}=\langle 4,5\rangle$ lies in the first quadrant, we don't need to adjust the angle.
Step 5 ：Final Answer: The magnitude of the vector $\vec{u}=\langle 4,5\rangle$ is $\boxed{6.40}$ and the angle which the vector points is $\boxed{51.34}$ degrees measured counterclockwise from the positive $x$-axis.
