Two vectors 2a+b and a=3b are perpendicular. Find the angle between a and b, if |a| = 2|b|.
To find the angle between vectors a and b, we can use the dot product. The dot product of two vectors a and b is defined as the product of their magnitudes (lengths) and the cosine of the angle between them.
Given that |a| = 2|b|, we can express vector a in terms of vector b:
a = 3b
Now, we need to find the dot product of vectors a and b. Let's substitute the values:
(2a + b) · a = 0
Using the distributive property, we expand the equation:
2(a · a) + (b · a) = 0
Simplifying further, we substitute a = 3b:
2(3b · 3b) + (b · 3b) = 0
Expanding and simplifying the equation:
18(b · b) + 3(b · b) = 0
21(b · b) = 0
Since the dot product of vector b with itself (b · b) will be non-zero if and only if vector b is non-zero, we can conclude that b must be the zero vector.
However, this contradicts the given information, |a| = 2|b|. Therefore, there is no solution for the angle between vectors a and b that satisfies all the given conditions.
Is a = 3b supposed to be a - 3b? or a + 3b?
If a = 3b is one of the vectors, then you would be saying that a and b are collinear, with |a| = 3|b|. That would contradict your later statement that |a| = 2|b|