Let \( \vec{a}=i-2 j+3 k \) if \( \vec{b} \) is a vector such that \( \vec{a} \cdot \vec{b}=|\vec{b}|^{2} \)…
Let \( \vec{a}=i-2 j+3 k \) if \( \vec{b} \) is a vector such that \( \vec{a} \cdot \vec{b}=|\vec{b}|^{2} \) and \( |\vec{a}-\vec{b}|=\sqrt{7} \), then \( |\vec{b}|= \)
\( 07 \)
\( 14 \)
\( \sqrt{7} \)
\( 21 \)
Solution
Given that, $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k} \rightarrow(1)$
$\vec{a} \cdot \vec{b}=|\vec{b}|^{2} \rightarrow(2)$
and $|\vec{a}-\vec{b}|=\sqrt{7} \rightarrow(3)$
Now, $|\vec{a}|=\sqrt{1+4+9}=\sqrt{14}$
Squaring both sides of Eq. (3), we get
$|\vec{a}|^{2}+|\vec{b}|^{2}-2|\vec{a}| \vec{b} \mid \cdot \cos \theta=7$
$\Rightarrow 14+|\vec{b}|^{2}-2|\vec{b}|^{2}=7$
$\Rightarrow 7=|\vec{b}|^{2} \Rightarrow|\vec{b}|=\sqrt{7}$