Let and be the vectors of the same magnitude such that . Then is : [2025]

Question

Let $\vec{a}$ and $\vec{b}$ be the vectors of the same magnitude such that $\frac{| \vec{a} + \vec{b} | + | \vec{a} - \vec{b} |}{| \vec{a} + \vec{b} | - | \vec{a} - \vec{b} |} = \sqrt{2} + 1$ . Then $\frac{| \vec{a} + \vec{b} |^{2}}{| \vec{a} |^{2}}$ is : [2025]

Smart Achievers · Accepted Answer

(4)

We have, $\frac{| \vec{a} + \vec{b} | + | \vec{a} - \vec{b} |}{| \vec{a} + \vec{b} | - | \vec{a} - \vec{b} |} = \sqrt{2} + 1$

Apply componendo and dividendo, we get

$\Rightarrow \frac{2 | \vec{a} + \vec{b} |}{2 | \vec{a} - \vec{b} |} = \frac{\sqrt{2} + 2}{\sqrt{2}}$

$\Rightarrow | \vec{a} + \vec{b} | = (1 + \sqrt{2}) | \vec{a} - \vec{b} |$

$\Rightarrow | \vec{a} + \vec{b} |^{2} = (3 + 2 \sqrt{2}) | \vec{a} - \vec{b} |^{2}$

$\Rightarrow 2 | \vec{a} |^{2} + 2 \vec{a} \cdot \vec{b} = (3 + 2 \sqrt{2}) (2 | \vec{a} |^{2} - 2 \vec{a} \cdot \vec{b})$ [ $∵$ $| \vec{a} |$ $=$ $| \vec{b} |$ ]

$\Rightarrow 2 | \vec{a} |^{2} (2 + 2 \sqrt{2}) = 2 \vec{a} \cdot \vec{b} (4 + 2 \sqrt{2})$

$\Rightarrow \frac{\vec{a} \cdot \vec{b}}{| \vec{a} |^{2}} = \frac{2 + 2 \sqrt{2}}{4 + 2 \sqrt{2}} = \frac{1}{\sqrt{2}}$

Now, we have

$\frac{| \vec{a} + \vec{b} |^{2}}{| \vec{a} |^{2}} = 1 + \frac{| \vec{b} |^{2}}{| \vec{a} |^{2}} + \frac{2 \vec{a} \cdot \vec{b}}{| \vec{a} |^{2}} = 1 + 1 + 2 (\frac{1}{\sqrt{2}}) = 2 + \sqrt{2}$

Solution

Q.
Let $\vec{a}$ and $\vec{b}$ be the vectors of the same magnitude such that $\frac{| \vec{a} + \vec{b} | + | \vec{a} - \vec{b} |}{| \vec{a} + \vec{b} | - | \vec{a} - \vec{b} |} = \sqrt{2} + 1$ . Then $\frac{| \vec{a} + \vec{b} |^{2}}{| \vec{a} |^{2}}$ is : [2025]

1 $4 + 2 \sqrt{2}$

2 $2 + 4 \sqrt{2}$

3 $1 + \sqrt{2}$

4 $2 + \sqrt{2}$

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Solution

Q. Let a→ and b→ be the vectors of the same magnitude such that |a→+b→|+|a→–b→||a→+b→|–|a→–b→|=2+1. Then |a→+b→|2|a→|2 is : [2025]

1 4+22

2 2+42

3 1+2