# Can the magnitude of the resultant vector of two given vectors be less than the magnitude of any of the given vector?Explain.

### Asked by | 17th Jun, 2013, 03:07: PM

Expert Answer:

### Yes, if the two vectors are at a sufficiently large obtuse angle.

^{If C = A + B, where A, B, C are vectors, then C is the "resultant."}

^{}

^{The law of cosines says, he magnitudes, A,B,C, are related as follows,}

C^{2}=A^{2}+B^{2}+2AB cosine(theta),

where theta is the angle between the vectors A and B. When theta is zero, then C has the maximum length, equal to the lengths of A and B added. When theta is 180 degrees, then C has the minimum length of the difference of the length of A and of B. Somewhere in between, the length of C will equal the length of the longer component and for larger angles be smaller.

To be specific, suppose that A is the longer of the two, then the resultant, C, has the same length as A at one special angle which we will call theta*.

A^{2}=A^{2}+B^{2}+2AB cosine(theta*)

cosine(theta*)=-B/(2A).

The answer to the question is then, that for angles greater than theta* the resultant is smaller than the larger component. (Greater means, of course, greater than theta* and up to 360-theta*.)

Note that if we ask whether the resultant can be smaller than the smaller of the two component vectors, then the answer is again yes and the above equation holds true when A is the smaller with the condition that it is not smaller than half the length of B. When the smaller vector is less than half the length of the larger component, then the resultant may equal the length of the larger but can never be made equal to the length of the smaller component.

^{If C = A + B, where A, B, C are vectors, then C is the "resultant."}

^{}

^{The law of cosines says, he magnitudes, A,B,C, are related as follows,}

C

^{2}=A

^{2}+B

^{2}+2AB cosine(theta),

where theta is the angle between the vectors A and B. When theta is zero, then C has the maximum length, equal to the lengths of A and B added. When theta is 180 degrees, then C has the minimum length of the difference of the length of A and of B. Somewhere in between, the length of C will equal the length of the longer component and for larger angles be smaller.

To be specific, suppose that A is the longer of the two, then the resultant, C, has the same length as A at one special angle which we will call theta*.

A

^{2}=A

^{2}+B

^{2}+2AB cosine(theta*)

cosine(theta*)=-B/(2A).

The answer to the question is then, that for angles greater than theta* the resultant is smaller than the larger component. (Greater means, of course, greater than theta* and up to 360-theta*.)

Note that if we ask whether the resultant can be smaller than the smaller of the two component vectors, then the answer is again yes and the above equation holds true when A is the smaller with the condition that it is not smaller than half the length of B. When the smaller vector is less than half the length of the larger component, then the resultant may equal the length of the larger but can never be made equal to the length of the smaller component.

### Answered by | 17th Jun, 2013, 04:43: PM

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