If $\mathbf{a}$ is collinear with $\mathbf{b}=3 \hat{i}+6 \hat{j}+6 \hat{k}$ and $\mathbf{a} \cdot \mathbf{b}=27$, then $|\mathbf{a}|=$

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Aicharya · Accepted Answer

Let $\mathbf{a}=x \hat{i}+y \hat{j}+z \hat{k}, \mathbf{b}=3 \hat{i}+6 \hat{j}+6 \hat{k}$ $\mathbf{a} \cdot \mathbf{b}=27$ According to question, $\mathbf{a}$ is a collinear with $\mathbf{b}$, then $\mathbf{a}=\lambda \mathbf{b}$ ...(i) $\Rightarrow \quad \mathbf{a} \cdot \mathbf{b}=27$ $\Rightarrow \quad \lambda \mathbf{b} \cdot \mathbf{b}=27$ [from Eq. (i)] $\Rightarrow \quad \lambda|\mathbf{b}|^2=27$ $\lambda=\frac{27}{\left(\sqrt{(3)^2+(6)^2+(6)^2}\right)^2}=\frac{27}{(9)^2}$ $\lambda=\frac{1}{3}$ $\mathbf{a}=\frac{1}{3}(3 \hat{i}+6 \hat{j}+6 \hat{k}) \Rightarrow \mathbf{a}=\hat{i}+2 \hat{j}+2 \hat{k}$ So, $|\mathbf{a}|=\sqrt{1+(2)^2+(2)^2}$ $=\sqrt{9}=3$ units

If $\mathbf{a}$ is collinear with $\mathbf{b}=3 \hat{i}+6 \hat{j}+6 \hat{k}$ and $\mathbf{a} \cdot…

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