a,b,c are three vectors such that \(|\mathbf{a}|=\mathbf{l},|\mathbf{b}|=2,|\mathbf{c}|=3\) and \(\mathbf{b}…

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 \(|2 \mathbf{a}+3 \mathbf{b}-3 \mathbf{c}|=\)
  1. 3
  2. \(\sqrt{22}\)
  3. 9
  4. 11

Solution

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 \(\begin{aligned} \mathbf{a} & \Rightarrow \mathbf{a} \cdot \mathbf{b}=\mathbf{a} \cdot \mathbf{c} \\ & =\sqrt{121+6 \mathbf{a} \cdot \mathbf{b}-6 \mathbf{a} \cdot \mathbf{b}}=\sqrt{121}=11 \end{aligned}\)

Asked in: AP EAMCET 2019 (22 Apr Shift 1)

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