If -a-b-1- -a-b-7- -a-4 then -b-

The correct option is C 3 |a⃗ + b⃗|^2 #160; + |a⃗ b⃗|^2 = 2|a⃗|^2 + 2 |b⃗|^2 ⇒ 1 + 49 = 2 × 16 + 2|b⃗|^2 ⇒ 50 = 32 + 2 |b⃗|^2 ⇒ 2|b ...

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Byju's Answer

Standard VIII

Mathematics

Factors of Natural Numbers

If |a⃗+b⃗|=...

Question

If $∣ ∣ \to a + \to b ∣ ∣ = 1, ∣ ∣ \to a - \to b ∣ ∣ = 7, | \to a | = 4$ then $∣ ∣ \to b ∣ ∣ =$

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Solution

The correct option is C $3$
$| \to a + \to b |^{2} + | \to a - \to b |^{2} = 2 | \to a |^{2} + 2 | \to b |^{2}$

$\Rightarrow 1 + 49 = 2 \times 16 + 2 | \to b |^{2}$

$\Rightarrow 50 = 32 + 2 | \to b |^{2}$

$\Rightarrow 2 | \to b |^{2} = 50 - 32 = 18$

$\Rightarrow | \to b |^{2} = \frac{18}{2} = 9 \Rightarrow | \to b | = 3$

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If ∣∣→a+→b∣∣=1,∣∣→a−→b∣∣=7,|→a|=4 then ∣∣→b∣∣=

The correct option is C 3|→a+→b|2+|→a−→b|2=2|→a|2+2|→b|2⇒1+49=2×16+2|→b|2⇒50=32+2|→b|2⇒2|→b|2=50−32=18⇒|→b|2=182=9⇒|→b|=3

If $∣ ∣ \to a + \to b ∣ ∣ = 1, ∣ ∣ \to a - \to b ∣ ∣ = 7, | \to a | = 4$ then $∣ ∣ \to b ∣ ∣ =$