Questionin Exercises 39 and 40 , find the vector v with the given magnitude and the same direction as $u$ .
39. $|\mathrm{v}|=2, \mathrm{u}=\langle 3,-3\rangle$

Studdy Solution

STEP 1

1. We are given a vector $\mathbf{u} = \langle 3, -3 \rangle$ .
2. We need to find a vector $\mathbf{v}$ with magnitude $|\mathbf{v}| = 2$ .
3. The vector $\mathbf{v}$ must have the same direction as $\mathbf{u}$ .

STEP 2

1. Calculate the magnitude of the vector $\mathbf{u}$ .
2. Find the unit vector in the direction of $\mathbf{u}$ .
3. Scale the unit vector to have the magnitude of 2.

STEP 3

Calculate the magnitude of the vector $\mathbf{u} = \langle 3, -3 \rangle$ .
The magnitude $|\mathbf{u}|$ is given by:
$|\mathbf{u}| = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$

STEP 4

Find the unit vector in the direction of $\mathbf{u}$ .
The unit vector $\mathbf{u}_{\text{unit}}$ is:
$\mathbf{u}_{\text{unit}} = \frac{\mathbf{u}}{|\mathbf{u}|} = \frac{\langle 3, -3 \rangle}{3\sqrt{2}} = \left\langle \frac{3}{3\sqrt{2}}, \frac{-3}{3\sqrt{2}} \right\rangle = \left\langle \frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right\rangle$

STEP 5

Scale the unit vector to have a magnitude of 2.
The vector $\mathbf{v}$ is:
$\mathbf{v} = 2 \times \mathbf{u}_{\text{unit}} = 2 \times \left\langle \frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right\rangle = \left\langle \frac{2}{\sqrt{2}}, \frac{-2}{\sqrt{2}} \right\rangle = \langle \sqrt{2}, -\sqrt{2} \rangle$
The vector $\mathbf{v}$ with the given magnitude and the same direction as $\mathbf{u}$ is:
$\boxed{\langle \sqrt{2}, -\sqrt{2} \rangle}$

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