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

(3) Hint: Use the parallelogram method.

Step 1: Find the magnitude of the resultant on both sides.

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

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

When $θ = 90^{\circ}$ , we can say that $A ⊥ B$ .

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(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.