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I'm busy studying for my Physics exam this evening and I've come across a vector problem that I cant quite solve. Any help would be appreciated!

We are given the following vectors:

A = 5i - 6.5j
B = -3.5i + 7j

We are also told that a vector C lies in the xy-plane and is perpendicular to A. The scalar product of B and C is equal to 15. With this information we have to calculate the components of the vector C.

I have tried calculating the magnitude of B and then trying to solve the equation:

|B||C|cos{theta} = 15

But I didn't get very far. I have a feeling I'm just not seeing something and making a careless error.

Thanks in advance!

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1  
Let C = $x$i + $y$j. If A and C are perpendicular you know A.C = 0, and you're told B.C = 15. If you multiply out the dot products you get two simultaneous equations for $x$ and $y$. –  John Rennie May 21 '13 at 10:15
    
That makes a lot more sense! Thank you! I tried this on paper now and I got C = -1.69i + 1.29j. Is this right? –  nickcorin May 21 '13 at 10:35
    
@nickcorin: I get another result: $C=\frac{390}{49}i+\frac{300}{49}j$. Which equations do you get when working out A.C=0 and B.C=15? –  fibonatic May 21 '13 at 10:58
    
@fibonatic: we're not supposed to give the answers to homework questions, which is why I specifically avoided doing so. –  John Rennie May 21 '13 at 11:00
    
@nickcorin: have you tried substituting the answers you got into the original equations to check they are correct? –  John Rennie May 21 '13 at 11:01

1 Answer 1

up vote 1 down vote accepted

Here is another approach that is essentially the same as the one suggested by John, but I find it to be less error-prone when doing calculations by hand.

You are given

$$ \mathbf{A} = 5\mathbf{i} - 6.5\mathbf{j}\\ \mathbf{B} = -3.5\mathbf{i} + 7\mathbf{j} $$

You know that $ \mathbf{C} \perp \mathbf{A}$. Therefore,

$$ \mathbf{C} = s \cdot (6.5 \mathbf{i} + 5\mathbf{j}) $$

with $s$ a scale factor. This is because in 2D, for any vector $\alpha = a\mathbf{i} + b\mathbf{j}$, the vector $\beta = -b\mathbf{i} + a\mathbf{j}$ satisfies $\alpha \cdot \beta = 0$ and thus $\alpha \perp \beta$.

Given that $\mathbf{B} \cdot \mathbf{C} = 15$, it should be fairly easy to solve for the scale factor $s$, thus solving for $\mathbf{C}$.

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This makes a lot more sense to me, although I can't see where I should go from here. –  nickcorin May 21 '13 at 12:31
    
@nickcorin: You know that $\mathbf{C}\cdot\mathbf{B}=15$. Just substitute the expression for $\mathbf{C}$ above and solve for $s$. –  Rody Oldenhuis May 21 '13 at 14:14

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