Find the vector $\mathbf { v }$ that has a magnitude of 8 and is in the same direction as $\mathbf { u }$, where $\mathbf { u } = \langle - 6 , - 4 \rangle$
A) $\mathbf { v } = \left\langle \frac { - 24 } { \sqrt { 13 } } , \frac { - 16 } { \sqrt { 13 } } \right\rangle$
B) $\mathbf { v } = \left\langle \frac { - 3 } { \sqrt { 13 } } , \frac { - 2 } { \sqrt { 13 } } \right\rangle$
C) $\mathbf { v } = \left\langle \frac { - 8 } { \sqrt { 13 } } , \frac { - 4 } { \sqrt { 13 } } \right\rangle$
D) $\mathbf { v } = \left\langle \frac { - 2 } { \sqrt { 13 } } , \frac { 2 } { \sqrt { 13 } } \right\rangle$
E) $ \mathbf { v } = \left\langle \frac { 4 } { \sqrt { 13 } } , \frac { 6 } { \sqrt { 13 } } \right\rangle$

Question

Quizplus · Accepted Answer

To find a vector with a specific magnitude in the same direction as another vector, you first find the unit vector in the direction of the given vector and then multiply it by the desired magnitude. The magnitude of $\mathbf{u} = \langle -6, -4 \rangle$ is $\sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = \sqrt{4 \cdot 13} = 2\sqrt{13}$. The unit vector in the direction of $\mathbf{u}$ is $\frac{1}{2\sqrt{13}} \langle -6, -4 \rangle = \langle \frac{-6}{2\sqrt{13}}, \frac{-4}{2\sqrt{13}} \rangle = \langle \frac{-3}{\sqrt{13}}, \frac{-2}{\sqrt{13}} \rangle$. To get a vector of magnitude 8 in the same direction, multiply the unit vector by 8: $8 \langle \frac{-3}{\sqrt{13}}, \frac{-2}{\sqrt{13}} \rangle = \langle \frac{-24}{\sqrt{13}}, \frac{-16}{\sqrt{13}} \rangle$.

Find the Vector v\mathbf { v }v That Has a Magnitude of 8 and Is in the Magnitude

Find the Vector $\mathbf { v }$ That Has a Magnitude of 8 and Is in the Magnitude