Question related to cartesian vectors

**dalbir4444** · May 29th 2009, 07:26 PM

Find two perpendicular vectors such that one of these vectors is twice as long as the other, and their sum is the vector [6,8]. I tried the following approach.

Let's say that u = [u1,u2] and v=[v1,v2].

Then we would have the u1v1 = -u2v2. I also got the following equations.
u1+v1=6
u2+v2=8
2v1^2 + 2v2^2 = u1^2 + u2^2. I'm having trouble combining the equation to get an equation with only one variable.

**red_dog** · May 29th 2009, 10:01 PM

Substitute $u_1=6-v_1, \ u_2=6-v_2$ in the first and the last equation:

$\left\{\begin{array}{ll}v_1^2+v_2^2-6v_1-6v_2=0\\v_1^2+v_2^2+12v_1+12v_2=72\end{array}\righ t.$

Substract the first from the second:

$v_1+v_2=4\Rightarrow v_2=4-v_1$

Substitute v2: $v_1^2-4v_1-4=0$

Now solve the quadratic.

**Grandad** · May 29th 2009, 10:26 PM

Hello dalbir4444

Originally Posted by dalbir4444

Find two perpendicular vectors such that one of these vectors is twice as long as the other, and their sum is the vector [6,8]. I tried the following approach.

Let's say that u = [u1,u2] and v=[v1,v2].

Then we would have the u1v1 = -u2v2. I also got the following equations.
u1+v1=6
u2+v2=8
2v1^2 + 2v2^2 = u1^2 + u2^2. I'm having trouble combining the equation to get an equation with only one variable.

$u_1v_1 +u_2v_2 = 0$ (1)

Also if $|\vec{u}|=2|\vec{v}|$ , when we square both sides we get:

$u_1^2 +u_2^2 = 4(v_1^2+v_2^2)$ (2)

So:

$u_1+v_1=6 \Rightarrow u_1^2 + 2u_1v_1+v_1^2 = 36$

$u_2+v_2=8 \Rightarrow u_2^2 + 2u_2v_2+v_2^2 = 64$

Add and use (1): $u_1^2 + u_2^2+v_1^2+v_2^2 = 100$

$\Rightarrow 5(v_1^2+v_2^2) = 100$ , from (2)

$\Rightarrow v_1^2 + v_2^2 = 20$ (3)

Now substitute for $u_1, u_2$ into (1):

$(6-v_1)v_1 + (8-v_2)v_2=0$

$\Rightarrow 6v_1+8v_2-(v_1^2+v_2^2)=0$

$\Rightarrow 6v_1+8v_2-20=0$ , from (3)

$\Rightarrow v_1=\tfrac13(10-4v_2)$

Substitute into (3):

$\tfrac19(100-80v_2+16v_2^2)+v_2^2=20$

$\Rightarrow 25v_2^2 -80v_2-80=0$

$\Rightarrow 5v_2^2-16v_2-16=0$

$\Rightarrow (5v_2+4)(v_2-4)=0$

So one solution is $v_2=4$ which gives $\vec{v} = -2\vec{i}+4\vec{j}, \vec{u} =8\vec{i}+4\vec{j}$ , and another solution is $v_2=-\tfrac45$ , which gives $\vec{v} = \tfrac{22}{5}\vec{i}-\tfrac45\vec{j}, \vec{u} =\tfrac85\vec{i}+\tfrac{44}{5}\vec{j}$

Grandad

**dalbir4444** · May 30th 2009, 04:54 AM

Shouldn't u1^2 + u2^2 = 2(v1^2 + v2^2), rather than equaling 4(v1^2 +v2^2), since one of the criteria is that one of these vectors is twice as long as the other. You final answer was that

. This makes |u| four times as long as |v|.

**Grandad** · May 30th 2009, 05:09 AM

Hello dalbir4444

Originally Posted by dalbir4444

Shouldn't u1^2 + u2^2 = 2(v1^2 + v2^2), rather than equaling 4(v1^2 +v2^2), since one of the criteria is that one of these vectors is twice as long as the other. You final answer was that

. This makes |u| four times as long as |v|.

As I said in my original posting,

$|\vec{u}| = \sqrt{u_1^2 + u_2^2}=2\sqrt{v_1^2 + v_2^2}$

$\Rightarrow u_1^2 + u_2^2=4(v_1^2 + v_2^2)$

Grandad

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