Find two perpendicular vectors $\vec u$ and $\vec v$ such that one of these vectors is twice as long as the other, and their sum is $[6,8]$

Question

Find two perpendicular vectors $\vec u$ and $\vec v$ such that one of these vectors is twice as long as the other, and their sum is $[6,8]$

So I have come up with:

$$\vec u + \vec v = [6,8] \\ \vec u \cdot \vec v = 0 \\ |\vec u| = 2|\vec v|$$

I was able to find the magnitudes of $\vec v$ and $\vec u$. However I cannot find the answer to the question.

$$|\vec u + \vec v| = \sqrt{36+64} \\ |\vec u + \vec v|^2= 10^2 \\ |\vec u |^2+2 \vec u \cdot \vec v + |\vec v|^2 = 100 \\ |\vec u |^2 + |\vec v|^2 = 100 \\ (2|\vec u|)^2+|\vec v|^2 = 100 \\ 5 |\vec v|^2 = 100 \\ |\vec v| = 2 \sqrt{5}$$ and $$ \\ |\vec u | = 4 \sqrt{5} $$

How do I find the vector components of $\vec u$ and $\vec v$??

Thank you

Answer 1

HINT: If $\vec u$ and $\vec v$ are perpendicular, then note that $\vec u = (a,b)$ for some numbers $a$ and $b$ and $\vec v$ is some scalar multiple $c(-b,a)$. In this case, since $|\vec u|=2|\vec v|$, what must $c$ be? Now write down the equations that say $\vec u+\vec v = (6,8)$.

Answer 2

If $u=(a,b)$, then, since $v$ is orthogonal to $u$ and twice as long, $v=\pm(-2b,2a)$. Let us suppose that $v=(-2b,2a)$, Then$$u+v=(6,8)\iff\left\{\begin{array}{l}a-2b=6\\b+2a=8.\end{array}\right.$$Can you take it from here?

Answer 3

Let $u=(x,y)$ and $v=(-2 y, 2 x)$

These vectors are perpendicular and $ V$ is twice as long as $u$.

You want the sum to be $(6,8)$, that is $ x-2y =6, y+2x=8$

Solving the system we get $x=22/5$ and y=-4/5$

Thus we have $u=(22/5,-4/5)$ and $v=(8/5, 44/5)$

Answer 4

A complex number $(a+bi)$ is a vector.

$|2 + i| = \sqrt 5\\ (2+i)^2 = 3 + 4i$

$4$ and $2i$ are perpendicular, and one is twice the magnitude of the other.

$4(2+i) + 2i(2+i) = (8+4i) + (-2 + 4i) = 6+8i$

In the more traditional vocabulary

$(8,4) + (-2,4) = (6,8)$

and $\|(8,4)\| = 2\|(-2,4)\|$

And $(8,4)\cdot(-2,4) = 0$

Answer 5

Here’s a geometric approach: Draw the circle that has the segment from the origin to $\vec s = [6,8]$ as its diameter. Every point on this circle other than the endpoints of the diameter corresponds to a vector $\vec v$ such that $\vec w = \vec s-\vec v$ is perpendicular to $\vec v$. (Recall the inscribed angle theorem.)

These three vectors form a right triangle, so the ratio $\|\vec w\|/\|\vec v\|$ is the tangent of the angle $\phi$ between $\vec s$ and $\vec v$. The condition on the lengths of these two vectors means that $\tan\phi = 2$ or $\tan\phi = 1/2$. Working out the four possible solutions is then a straightforward application of trigonometry or the Pythagorean theorem.