The magnitude of vectors a b and c are 12 5 and 13
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If the magnitude of two vectors vector A vector B and vector c are respectively 12, 5 and 13 units and vector A + vector b is equal to vector C then the angle between vector a and vector b is
Ask for details Follow Report by Pankajvishwakar 01.06.2018
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qwerty12345m
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Since you have given options, short solution to this is:
12, 5 and 13 are right triangle triplets i.e. 144 + 25 = 169, clearly ABC can form a right angled triangle.
Given in the statement, A + B = C, with respect to vectors, A and B are height - base pair.
This brings us to the conclusion that: Angle between A and B must be 90deg or pi/2.
PS: The question has 2 same options. (typo?).
Note: This is just to come up with a solution with MCQ. This is not the correct/proper solution/explanation.
Proper solution:
Given,
m(A) = 12, m(B) = 5, m(C) = 13, where m(X) stands for magnitude of X.
A + B = C
squaring both sides,
A.A+ B.B + 2 A.B = C.C
note 'dot’ product i.e. A.A is equal to m(A)ˆ2.
=> m(A)ˆ2 + m(B)ˆ2 + 2 m(A)m(B)cos(ab) = m(C)ˆ2 , where ab is the angle between A and B.
Substituting the values in the above equation,
=> 144 + 25 + 60cos(ab) = 169
=> 169 + 60cos(ab) = 169
=> 60cos(ab) = 0
=> cos(ab) = 0
=> cos(ab) = cos(pi/2) or cos(90deg)
ab = pi/2 or 90deg, where ab is the angle between A and B.