<p>Ok well, I am pretty stumped on this one problem in my AP physics class. We are currently on vectors and dealing with dot products/cross products, etc.. Here is the question, I hope someone can answer it for me and give me a brief explanation on how to approach the answer, it would mean so much to me!</p>
<p>5) Given that (45i + 21j) (ai + bj) = 2050, and that the angle between the two vectors is 50 degrees, determine a and b.</p>
<p>I'm assuming that 2050 is the dot product of the two vectors, yes? I can't answer this one unless I get that information</p>
<p>yes, it is the dot product of the two vectors. sorry for not putting that >.<</p>
<p>I will use * to denote the dot product here.
Alright then, it's pretty easy. The dot product expression is just 45a +21b, and the formula for an angle between two vectors u and v is cosθ=(u*v)/(|u||v|) where |u| is the magnitude of vector u. </p>
<p>I'll put this up just so you have something to read. The rest should be up in a few minutes.</p>
<p>Let's make some substitutions.</p>
<p>u*v=2050 and 45a+21b=2050, so a=(2050-21b)/45
|u| = sqrt(45^2+21^2)
|v| = sqrt(a^2+b^2)</p>
<p>So we now have
cos (50 degrees)=2050/[sqrt(45^2+21^2)sqrt(a^2+b^2)]
Substitute the expression above for a will give us
cos (50 degrees)=2050/[sqrt(45^2+21^2)sqrt([(2050-21b)/45]^2+b^2)]</p>
<p>From here on, it just degenerates into an algebra problem, which I assume you can do. Once you find b, just plug it in to the expression for a to get a. If you still need any help or confirmation, just say so.</p>
<p>Speaking of vectors...is there a program (maybe, mathematica or something) that lets you graph vectors in 3d space if you input the values for i, j, and k?</p>
<p>thanks for setting it up. I gotta eat dinner now, but I'll attempt the problem later. If you have AIM, it would probably be quicker to ask you questions since I don't want to hassle you to go back and forth on this thread.</p>
<p>As for explanations, to find the magnitude of a vector, you can just imagine the coordinates i and j as x and y respectively, and since the x and y-axis are perpendicular to each other, you can form a right triangle by forming a hypotenuse. The length of this hypotenuse is the magnitude of the vector, and can be calculated by Pythagoras' Theorem. </p>
<p>Solving this problem is just like solving a system of equations. Just find an expression for one variable and substitute that expression for any instance of the variable in the other equation.</p>
<p>I don't have AIM (probably should get one soon)</p>
<p>Well, if you're still watching this.. i got 16.6 for a and b= 61.9765. Duno if i'm right.</p>
<p>Looks good to me.</p>
<p>Ah thanks. My teacher really didn't seem to have faith in us since in the actual problem he gave us, he actually wrote "I doubt that many, if any, of you will get the correct answer. Not that it's hard, it's just you haven't been exposed to such a problem. If anyone gets the correct answer, I will give them some money."
- I'm still waiting on the money =)</p>
<p>Ah, so what did he say when you showed him the answer?</p>