How do you find the vector v with the given magnitude of 5 and in the same direction as $u = < 3, 3 >$ $u=<3,3>$ ?

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Answer 1 · 2017-05-03T18:30:27

Please see the explanation.

Given: $∣ ∣ \to v ∣ ∣ = 5$ $|vecv|= 5$

We know the formula for the magnitude

$∣ ∣ \to v ∣ ∣ = \sqrt{x^{2} + y^{2}} [1]$ $|vecv|= sqrt(x^2+y^2)" [1]"$

This problem is made simple by the fact that the x component of the vector $\to u$ $vecu$ is the same as the y component.

$y = x$ $y = x$

Substitute x for y into equation [1]:

$∣ ∣ \to v ∣ ∣ = \sqrt{x^{2} + x^{2}} [2]$ $|vecv|= sqrt(x^2+x^2)" [2]"$

Substitute 5 for the magnitude into equation [2]:

$5 = \sqrt{x^{2} + x^{2}}$ $5= sqrt(x^2+x^2)$

$25 = 2 x^{2}$ $25= 2x^2$

$\frac{25}{2} = x^{2}$ $25/2=x^2$

$x = 5 \frac{\sqrt{2}}{2}$ $x = 5sqrt2/2$

We know that $y = x$ $y = x$ :

$y = 5 \frac{\sqrt{2}}{2}$ $y = 5sqrt(2)/2$

The vector is $< 5 \frac{\sqrt{2}}{2}, 5 \frac{\sqrt{2}}{2} >$ $< 5sqrt(2)/2, 5sqrt(2)/2 >$

How do you find the vector v with the given magnitude of 5 and in the same direction as u=<3,3>u=<3,3>u=<3,3>?