physicsvectors
Let vector B = 3m at 60 degree. Let vector C have the same magnitude as vector A. Vector C has a direction angle greater than of vector A by 25 degree. Dot product of A and B is 30. Dot product of B and C is 35. What is the magnitude and angle of vector A.
I know how to get the x and y components of vector B. I also know how the dot product is working but I'm stuck in forming equation so that I could get all their components. Any hint on how to attack this problem?
Well, in a general way the dot product is an operation defined between two objects that satisfies some properties (check out this page on wikipedia for example: http://en.wikipedia.org/wiki/Dot_product), for a 3D vector it turns out that if you define the dot product as: $$ \mathbf{A} \cdot \mathbf{B} = a_xb_x + a_yb_y + a_zb_z $$ these properties are fullfilled. Note that $ a_xb_x + a_yb_y + a_zb_z $ is ascalar and not a vector anymore, because you are summing up products between scalar quantities (i.e. the components of your vectors).
Now consider the dot product between the vector $ \mathbf{A}= a_x \mathbf{i} + a_y \mathbf{j} + a_z \mathbf{k} $ and the vector $ \mathbf{i} $, using the formula above, you get: $$ \mathbf{A} \cdot \mathbf{i} = a_x = \mid \mathbf{A} \mid cos\theta $$ where $\theta$ is the angle between $ \mathbf{A} $ and $ \mathbf{i} $, i.e. $ \mathbf{A} \cdot \mathbf{i}$ gives you the projection of $ \mathbf{A} $ on $\mathbf{i} $ (try to sketch the vectors in 2D). So, in a general way, you have that $ \mathbf{A} \cdot \mathbf{B} $ gives you the projection of $\mathbf{A} $ on $\mathbf{B}$, i.e $ \mid \mathbf{A} \mid \mid \mathbf{B} \mid cos\theta $.
The components axes $x^\prime$ and $y^\prime$ are inclined by angle $\theta$.
Use the sine rule.
Best Answer
Let's assume that vectors A and C have magnitude A, and the angle of vector A is $\alpha$. Then, the angle of vector C is going to be $\alpha+25$. Here are all components: $$ A_x=A\cos(\alpha)\\ A_y=A\sin(\alpha)\\ B_x=3\cos(60)\\ B_y=3\sin(60)\\ C_x=A\cos(\alpha+25)\\ C_y=A\sin(\alpha+25) $$ The dot product between A and B can be written as $$ A\cdot B=A_xB_x+A_yB_y=3A(\cos(\alpha)\cos(60)+\sin(\alpha)\sin(60))=30 $$ or $$A\cos(\alpha-60)=10$$ Similarly, for B and C: $$ B\cdot C=B_xC_x+B_yC_y=3A(\cos(60)\cos(\alpha+25)+\sin(60)\sin(\alpha+25))=35 $$ or $$3A\cos(35-\alpha)=35$$ From these equations, take the ratio, and you will get an equation in only $\alpha$, that you might need to solve numerically. Then just replace it into one of them, and you get A