Ncert Exercises & Solutions

For two vectors $\vec A$ and $\vec B$ , | $\vec A$ + $\vec B$ |=| $\vec A$ - $\vec B$ | is always true when:

(a)	\| $\vec A$ \| = \| $\vec B$ \| ≠ $0$
(b)	$\vec A\perp\vec B$
(c)	\| $\vec A$ \| = \| $\vec B$ \| ≠ $0$ and $\vec A$ and $\vec B$ are parallel or antiparallel.
(d)	when either \| $\vec A$ \| or \| $\vec B$ \| is zero.

Choose the correct option from the given ones:

1.	(a), (d)
2.	(b), (c)
3.	(b), (d)
4.	(a), (b)

Hint: Use the parallelogram method.

Step: Find the magnitude of the resultant on both sides.

Given, |A+ B| = |A - B|

$\begin{array}{l} \Rightarrow \sqrt{| A |^{2} + | B |^{2} + 2 | A | | B | \cos θ} = \sqrt{| A |^{2} + | B |^{2} - 2 | A | | B | \cos θ} \\ \Rightarrow | A |^{2} + | B |^{2} + 2 | A | | B | \cos θ = | A |^{2} + | B |^{2} - 2 | A | | B | \cos θ \end{array}$ $\begin{array}{l} \Rightarrow \sqrt{| A |^{2} + | B |^{2} + 2 | A | | B | \cos θ} = \sqrt{| A |^{2} + | B |^{2} - 2 | A | | B | \cos θ} \\ \Rightarrow | A |^{2} + | B |^{2} + 2 | A | | B | \cos θ = | A |^{2} + | B |^{2} - 2 | A | | B | \cos θ \end{array}$
$\begin{array}{l} \Rightarrow 4 | A | | B | \cos θ = 0 \\ \Rightarrow | A | | B | \cos θ = 0 \\ \Rightarrow | A | = 0 or | B | = 0 or \cos θ = 0 \\ \Rightarrow θ = 90 \end{array}$ $\begin{array}{l} \Rightarrow 4 | A | | B | \cos θ = 0 \\ \Rightarrow | A | | B | \cos θ = 0 \\ \Rightarrow | A | = 0 or | B | = 0 or \cos θ = 0 \\ \Rightarrow θ = 90 \end{array}$

When $θ = 90^{\circ}$ $θ = 90^{\circ}$ , we can say that $A ⊥ B$ $A ⊥ B$ .
Hence, option (3) is the correct answer.

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