Let a- - i - 2j - 3k if b- is a vector such that a-b- - -b-2 and -a- - b- - -7- then -b- -

The correct option is B √(7) We have |a⃗|=√(1^2+( 2)^2+3^2)=√(14) Now square both sides |a⃗ b⃗|=√(7) ⇒ |a⃗|^2+|b⃗|^2 2a⃗·b⃗=7 ⇒ 14+|b⃗ ...

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Byju's Answer

Standard XII

Mathematics

Applications of Dot Product

Let a⃗ = i ...

Question

Let $\to a = i - 2 j + 3 k$ if $\to b$ is a vector such that $\to a \cdot \to b = | \to b |^{2}$ and $| \to a - \to b | = \sqrt{7}$ , then $| \to b | =$ ____

7

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\sqrt{7}

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Solution

The correct option is B $\sqrt{7}$
We have $| \to a | = \sqrt{1^{2} + (- 2)^{2} + 3^{2}} = \sqrt{14}$
Now square both sides $| \to a - \to b | = \sqrt{7}$
$\Rightarrow | \to a |^{2} + | \to b |^{2} - 2 \to a \cdot \to b = 7$
$\Rightarrow 14 + | \to b |^{2} - 2 | \to b |^{2} = 7$ , [using the given condition]
$\Rightarrow | \to b |^{2} = 14 - 7 = 7$
$\Rightarrow | \to b | = \sqrt{7}$

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Let →a=i−2j+3k if →b is a vector such that →a⋅→b=|→b|2 and |→a−→b|=√7, then |→b|= ____

The correct option is B √7We have |→a|=√12+(−2)2+32=√14Now square both sides |→a−→b|=√7⇒|→a|2+|→b|2−2→a⋅→b=7⇒14+|→b|2−2|→b|2=7, [using the given condition]⇒|→b|2=14−7=7⇒|→b|=√7

Let $\to a = i - 2 j + 3 k$ if $\to b$ is a vector such that $\to a \cdot \to b = | \to b |^{2}$ and $| \to a - \to b | = \sqrt{7}$ , then $| \to b | =$ ____