If |a-b| = 1/2|b| and (a-b) is perpendicular to a then what is the angle between a and b
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Answered by
1
Use definition of dot-product.
(A-B).B = 0 because (A-B) and B are perpendicular.
A.B-B.B=0
|A|*|B|*cos(alpha) - |B|^2 = 0
Now we use the fact that |A|=2|B|
2*|B|*|B|*cos(alpha) - |B|^2 = 0
2*|B|^2*cos(alpha) -|B|^2=0
cos(alpha) = 1/2
alpha = pi/3.
Answered by
3
Answer: as they have said that (a-b) is perpendicular to a do a dot product b/w (a-b) and a. after that substitute the result into the first statement. on ongoing solving you will arrive to the solution.
Explanation:
[a-b]=1/2[b]
a²+b²-2abcos(α)=1/4b²_____(1)
(a-b).a=0
a .a - a .b=0
a²-abcos(α)=0
a²=ab cos(α)_____________(2)
substitute 2 into 1
on solving we get α=π/6=30°
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