Thread: Question related to cartesian vectors

1. Question related to cartesian vectors

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.

2. 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: $v_1^2-4v_1-4=0$

3. Vectors

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 $\vec{v} = \tfrac{22}{5}\vec{i}-\tfrac45\vec{j}, \vec{u} =\tfrac85\vec{i}+\tfrac{44}{5}\vec{j}$

4. 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|.

5. Magnitude of a vector

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,

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

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