mill said:Homework Statement
If |v|=4, |w|=3 and the angle between and is pi/3, find |2v −w|
Homework Equations
##cosθ=\frac {v dot w} {|v||w|} ##The Attempt at a Solution
## 6= v dot w ##
This is as far as I got. How would I find the separate values of v and w for the equation?
ehild said:Very good! How is the absolute value of a vector defined with dot product by itself?
You do not need to know the individual vectors. You need the absolute value of 2v-w.
ehild
Start from that definition. What then is ##\sqrt{(2v-w) \cdot (2v-w)}##?mill said:u dot u = ##|u|^2## so |u|= ##\sqrt {u dot u} ## ?
To find the third equation, you will need to use the vector cross product. Take the cross product of the two given vectors to get a third vector. Then, set the components of this third vector equal to a new variable, say "c". This variable, along with the original two vectors, will form the third equation.
The formula for finding the third equation is:
c = a x b
where c is the third vector, a and b are the given vectors, and x represents the cross product operation.
No, the dot product only gives a scalar value, while the third equation requires a vector. Therefore, the cross product must be used to find the third equation.
Yes, the cross product can only be used if the given vectors are in 3-dimensional space. If the vectors are in 2-dimensional space, then the third equation cannot be found using the cross product method.
Yes, the third vector obtained from the cross product can be thought of as the normal vector to the plane formed by the original two vectors. This normal vector is perpendicular to both vectors and can be used to find the third equation of the plane.