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a,b,c are three vectors such that \(|\mathbf{a}|=\mathbf{l},|\mathbf{b}|=2,|\mathbf{c}|=3\) and \(\mathbf{b}. \mathbf{c}=\mathbf{0}\). If the projection of \(\mathbf{b}\) along \(\mathbf{a}\) is equal to the projection of \(\mathbf{c}\) along \(\mathbf{a}\), then \(|2 \mathbf{a}+3 \mathbf{b}-3 \mathbf{c}|=\)
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Verified Answer
The correct answer is:
11
Given,
and
\(\begin{gathered}
|\mathbf{a}|=1,|\mathbf{b}|=2,|\mathbf{c}|=3 \\
\mathbf{b} \cdot \mathbf{c}=0
\end{gathered}\)
Now, \(|2 \mathbf{a}+3 \mathbf{b}-3 \mathbf{c}|\)
\(\begin{gathered}
=\sqrt{4|\mathbf{a}|^2+9|\mathbf{b}|^2+9|\mathbf{c}|^2+12 \mathbf{a} \cdot \mathbf{b}-18 \mathbf{b} \cdot \mathbf{c}-12 \mathbf{a} \cdot \mathbf{c}} \\
=\sqrt{4+9(4)+9(9)+12 \mathbf{a} \cdot \mathbf{b}-0-12 \mathbf{a} \cdot \mathbf{c}} \\
=\sqrt{4+36+81+12 \mathbf{a} \cdot \mathbf{b}-12 \mathbf{a} \cdot \mathbf{c}}
\end{gathered}\)
Since, projection of \(\mathbf{b}\) along \(\mathbf{a}\) is equal to projection of \(\mathbf{c}\) along
\(\begin{aligned}
\mathbf{a} & \Rightarrow \mathbf{a} \cdot \mathbf{b}=\mathbf{a} \cdot \mathbf{c} \\
& =\sqrt{121+6 \mathbf{a} \cdot \mathbf{b}-6 \mathbf{a} \cdot \mathbf{b}}=\sqrt{121}=11
\end{aligned}\)
and
\(\begin{gathered}
|\mathbf{a}|=1,|\mathbf{b}|=2,|\mathbf{c}|=3 \\
\mathbf{b} \cdot \mathbf{c}=0
\end{gathered}\)
Now, \(|2 \mathbf{a}+3 \mathbf{b}-3 \mathbf{c}|\)
\(\begin{gathered}
=\sqrt{4|\mathbf{a}|^2+9|\mathbf{b}|^2+9|\mathbf{c}|^2+12 \mathbf{a} \cdot \mathbf{b}-18 \mathbf{b} \cdot \mathbf{c}-12 \mathbf{a} \cdot \mathbf{c}} \\
=\sqrt{4+9(4)+9(9)+12 \mathbf{a} \cdot \mathbf{b}-0-12 \mathbf{a} \cdot \mathbf{c}} \\
=\sqrt{4+36+81+12 \mathbf{a} \cdot \mathbf{b}-12 \mathbf{a} \cdot \mathbf{c}}
\end{gathered}\)
Since, projection of \(\mathbf{b}\) along \(\mathbf{a}\) is equal to projection of \(\mathbf{c}\) along
\(\begin{aligned}
\mathbf{a} & \Rightarrow \mathbf{a} \cdot \mathbf{b}=\mathbf{a} \cdot \mathbf{c} \\
& =\sqrt{121+6 \mathbf{a} \cdot \mathbf{b}-6 \mathbf{a} \cdot \mathbf{b}}=\sqrt{121}=11
\end{aligned}\)
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