Ncert Exercises & Solutions

It is found that $|\vec{A}+\vec{B}|=|\vec{A}|$. This necessarily implies:

1.	$\vec{B}=0$
2.	$\vec{A},$ $\vec{B}$ are antiparallel
3.	$\vec{A}$ and $\vec{B}$ are perpendicular
4.	$\vec{A}.\vec{B}\leq0$

(4) Hint: Use the parallelogram method to find the magnitude of the resultant.

Step 1: Find the magnitude of the resultant.

$\begin{array}{l} {Given that | A + B| = |A| or |A + B|}^{2} {= |A|}^{2} \\ \Rightarrow | A |^{2} + | B |^{2} + 2 | A | | B | \cos θ = | A |^{2} \end{array}$

where $θ$ is the angle between A and B.

Step 2: Solve it to find the necessary conditions.

$\begin{array}{l} \Rightarrow | B | (| B | + 2 | A | \cos θ) = 0 \\ \Rightarrow | B | = 0 or | B | + 2 | A | \cos θ = 0 \\ \Rightarrow \cos θ = - \frac{| B |}{2 | A |} \end{array}$

If A and B are antiparallel, then $θ$ = 180 $°$
Hence, from Eq. (i)

$- 1 = - \frac{| B |}{2 | A |} \Rightarrow | B | = 2 | A |$

Hence, the correct answer will either |B| = 0 or A and B are antiparallel |Bl = 2|A|

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1.	\(\vec{B}=0\)
2.	\(\vec{A},\) \(\vec{B}\) are antiparallel
3.	\(\vec{A}\) and \(\vec{B}\) are perpendicular
4.	\(\vec{A}.\vec{B}\leq0\)