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Consider the two vectors $\vec{A}$ and $\vec{B}$ . The sum of their vectors ie: $|\vec{A}+\vec{B}|$ , if $|\vec{A}|>|\vec{B}|$

1) is equal to $|\vec{A}|+|\vec{B}|$

2)must be less than $|\vec{A}|+|\vec{B}|$

3)cannot be greater than $|\vec{A}|+|\vec{B}|$

4)must be equal to $|\vec{A}|-|\vec{B}|$

Ok so i initially got my answer as 3) as sum of vectors is less than or equal to $|\vec{A}|+|\vec{B}|$ but the answer key shows that the answer is 2) and 3)

But I don't understand why it must be less than $|\vec{A}|+|\vec{B}|$

We have many examples where angle is $0°$ where adition of vectors is equal to $|\vec{A}|+|\vec{B}|$ .

Please suggest your explanations .

gucci
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  • My explanation is that whoever wrote that problem has trouble with English. You are correct that it is quite possible for $|\vec A + \vec B| = |\vec A| + |\vec B|$, which the wording of (2) does not allow. (2) is not a correct answer. However your explanation is a bit off: $|\vec A + \vec B| = |\vec A| + |\vec B|$ when the angle between them is $180^\circ$, not when it is $0^\circ$. – Paul Sinclair May 16 '19 at 02:43

1 Answers1

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To answer your question -

Consider the "triangle inequality",

$$||\mathbf{A+B}|| \le ||\mathbf{A}||+||\mathbf{B}||$$

This relation can be proven by using the law of cosines

$$||\mathbf{A+B}||^2=||\mathbf{A}||^2+||\mathbf{B}||^2-2||\mathbf{A}||.||\mathbf{B}||\cos(\mathbf{A,B}),$$

where $$0\le(\mathbf{A,B})\le\pi$$

For proof, see this link

user142523
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