Dot Products With Orthogonal Vectors

[ME_123]

Homework Statement

The Vector a= -2i -3j and is orthogonal to vector b that has the same length as a. The third vector c has the dot products ca= 8m^2 and cb= 9m^2. What are the components of c? c = ______i + _____ j m

Homework Equations

The Attempt at a Solution

I know that (a⃗ +b⃗ )⋅c⃗ =a⃗ ⋅c⃗ +b⃗ ⋅c. Since I have the dot product of ac and bct and I know (a + b) I get (-5i - j )⋅c⃗ = 17. If I do the dot product of (a+b)c I get -5Ci - Cj = 17. This is how I have done previous problems but I have gotten them wrong. Can someone explain to me if I am going in the right direction or if I am doing it wrong. Thank you.

 

[jedishrfu]

Given A and its components can you easily construct a B which is perpendicular to it?

If you not sure then think about lines and slopes and how you constructed lines perpendicular to a given line.

 

[Ray Vickson]

Homework Statement

The Vector a= -2i -3j and is orthogonal to vector b that has the same length as a. The third vector c has the dot products ca= 8m^2 and cb= 9m^2. What are the components of c? c = ______i + _____ j m

Homework Equations

The Attempt at a Solution

I know that (a⃗ +b⃗ )⋅c⃗ =a⃗ ⋅c⃗ +b⃗ ⋅c. Since I have the dot product of ac and bct and I know (a + b) I get (-5i - j )⋅c⃗ = 17. If I do the dot product of (a+b)c I get -5Ci - Cj = 17. This is how I have done previous problems but I have gotten them wrong. Can someone explain to me if I am going in the right direction or if I am doing it wrong. Thank you.

Get rid of the units---you are using them inconsistently. Your given 'a' has no units, and if 'b' has the same length as 'a' but has units of m, there is, again, an inconsistency.

Anyway, set $\vec{b} = u \vec{i} + v \vec{j}$ and $\vec{c} = x \vec{i} + y \vec{j}$. You know two facts about $u,v$: (i) $\vec{b} \perp \vec{a}$; and (ii) $\text{length}\,(\vec{b} )= \text{length}\, (\vec{a})$. Those suffice to tell you $u,v$, except for an overall sign.

So, now you know $\vec{b}$, up to a sign ambiguity. You are given $\vec{a} \cdot \vec{c}$ and $\vec{b} \cdot \vec{c}$. You can get $x,y$ from these, and so you will know the vector $\vec{c}$. The sign ambiguity in $\vec{b}$ may cause you to arrive at more than one solution---draw a diagram to see why!

 

[RUber]

I think you are too quick to assume you know B.
B must satisfy:
$-2b_x -3b_y = 0$ or $b_x =-\frac32 b_y$
To be the same length, $b_y^2 + \frac94 b_y^2 = 13$ or $b_y = \pm 2$.
So you have two choices for B.

Next, you have a system of equations:
$-2C_i - 3C_j = 9 \\ \pm 3 C_i +\mp 2 C_j = 8$
This will have solutions for each of your directions for B.

 

[ME_123]

Thank you for your help. I actually figured out how to do it. I guess I just over thought the problem and made it more complicated then it should have. Thank you all for taking the time to respond.