Let a→=i^-2j^+3k^. If b→is a vector such that a→·b→=b→2nd a→-b→=7,then b→ equals

a→=i^−2j^+3k^Now, |a→|=1+4+9=14Also, |a→−b→|=7|a→−b→|2=7(a→−b→)⋅(a→−b→)=7|a→|2+|b→|2−2a→⋅b→=714+|b→|2−2|b→|2=7 ∵ given a→⋅b→=|b→|2⇒|b→|2=7⇒|b→|=7

Let a→=i^-2j^+3k^. If b→is a vector such that a→·b→=b→2nd a→-b→=7,then b→ equals

a→=i^−2j^+3k^Now, |a→|=1+4+9=14Also, |a→−b→|=7|a→−b→|2=7(a→−b→)⋅(a→−b→)=7|a→|2+|b→|2−2a→⋅b→=714+|b→|2−2|b→|2=7 ∵ given a→⋅b→=|b→|2⇒|b→|2=7⇒|b→|=7

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Let $\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k} .$ If $\vec{b}$ is a vector such that $\vec{a} \cdot \vec{b} = {|\vec{b}|}^{2} nd |\vec{a} - \vec{b}| = \sqrt{7,}$ then $|\vec{b}|$ equals

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$\sqrt{7}$

answer is C.

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Detailed Solution

$\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}$
Now, $| \vec{a} | = \sqrt{1 + 4 + 9} = \sqrt{14}$
Also, $| \vec{a} - \vec{b} | = \sqrt{7}$

$\begin{array}{l} | \vec{a} - \vec{b} |^{2} = 7 \\ (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) = 7 \\ | \vec{a} |^{2} + | \vec{b} |^{2} - 2 \vec{a} \cdot \vec{b} = 7 \\ 14 + | \vec{b} |^{2} - 2 | \vec{b} |^{2} = 7 [∵ given \vec{a} \cdot \vec{b} = | \vec{b} |^{2}] \end{array}$
$\begin{array}{l} \Rightarrow | \vec{b} |^{2} = 7 \\ \Rightarrow | \vec{b} | = \sqrt{7} \end{array}$