KCET · Maths · Sequences and Series
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 \( |\vec{b}|= \)
- A \( 07 \)
- B \( 14 \)
- C \( \sqrt{7} \)
- D \( 21 \)
Answer & Solution
Correct Answer
(C) \( \sqrt{7} \)
Step-by-step Solution
Detailed explanation
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
\(|\vec{a}|^{2}+|\vec{b}|^{2}-2|\vec{a}| \vec{b} \mid \cdot \cos \theta=7\)
\(\Rightarrow 14+|\vec{b}|^{2}-2|\vec{b}|^{2}=7\)
\(\Rightarrow 7=|\vec{b}|^{2} \Rightarrow|\vec{b}|=\sqrt{7}\)
\(\vec{a} \cdot \vec{b}=|\vec{b}|^{2} \rightarrow(2)\)
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
\(|\vec{a}|^{2}+|\vec{b}|^{2}-2|\vec{a}| \vec{b} \mid \cdot \cos \theta=7\)
\(\Rightarrow 14+|\vec{b}|^{2}-2|\vec{b}|^{2}=7\)
\(\Rightarrow 7=|\vec{b}|^{2} \Rightarrow|\vec{b}|=\sqrt{7}\)
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