Question

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

Answer 1

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.

Answer 2

Answer:

The angle between two vectors vector A vector B is $90^0$

Explanation:

• Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector.
• The magnitude of resultant vector is given by
  $C = \sqrt{a^2+b^2+2ab Cos \alpha }$
  Where $\alpha$ is the angle between the vectors vector A and vector B
• So on putting the values we get
  $13 = \sqrt{12^2 + 5^2+2*12*5*cos \alpha } \\13*13 = 144+25+120 *cos \alpha\\169 - 169 = 120 * cos \alpha\\0 = cos \alpha\\\alpha = 90^0$
• So the angle between two vectors vector A vector B is $90^0$
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