If overline{a}=2 hat{i}+3 hat{j}+4 hat{k},ove | vedclass

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(B) Let $\overline{r} = x \hat{i} + y \hat{j} + z \hat{k}$.Given $\overline{r} 	imes \overline{a} = \overline{b}$,we have:$\begin{vmatrix} \hat{i} &

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$\sqrt{155}$

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(B) Let $\overline{r} = x \hat{i} + y \hat{j} + z \hat{k}$.Given $\overline{r} 	imes \overline{a} = \overline{b}$,we have:$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ x & y & z \ 2 & 3 & 4 \end{vmatrix} = \hat{i} - 2 \hat{j} + \hat{k}$$(4y - 3z) \hat{i} - (4x - 2z) \hat{j} + (3x - 2y) \hat{k} = \hat{i} - 2 \hat{j} + \hat{k}$Comparing coefficients:$4y - 3z = 1$ $(1)$$4x - 2z = 2 \Rightarrow 2x - z = 1 \Rightarrow z = 2x - 1$ $(2)$$3x - 2y = 1 \Rightarrow 2y = 3x - 1 \Rightarrow y = \frac{3x - 1}{2}$ $(3)$Given $\overline{r} \cdot \overline{c} = 3$,we have $x + y - z = 3$ $(4)$Substituting $(2)$ and $(3)$ into $(4)$:$x + \frac{3x - 1}{2} - (2x - 1) = 3$$2x + 3x - 1 - 4x + 2 = 6$$x + 1 = 6 \Rightarrow x = 5$Using $x=5$ in $(2)$ and $(3)$:$z = 2(5) - 1 = 9$$y = \frac{3(5) - 1}{2} = 7$Thus,$\overline{r} = 5 \hat{i} + 7 \hat{j} + 9 \hat{k}$$|\overline{r}| = \sqrt{5^2 + 7^2 + 9^2} = \sqrt{25 + 49 + 81} = \sqrt{155}$

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