If $\mathbf{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \mathbf{b}=2 \hat{i}+3 \hat{j}+\hat{k}$, $\mathbf{c}=8 \hat{i}+13 \hat{j}+9 \hat{k}$ and $x \mathbf{a}+y \mathbf{b}+z \mathbf{c}=0$, then $\frac{x y}{z^2}=$

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Aicharya · Accepted Answer

$ \begin{aligned} & x \mathbf{a}+y \mathbf{b}+z \mathbf{c}=0 \ & \Rightarrow(x+2 y+8 z) \hat{i}+(2 x+3 y+13 z) \hat{j} \ & \quad+(3 x+y+9 z) \hat{k}=0 \ & \Rightarrow \quad x+2 y+8 z=0 \ & 2 x+3 y+13 z=0 \ & 3 x+y+9 z=0 \end{aligned} $ From Eqs. (i) and (ii), $ \begin{aligned} & -y-3 z=0 \Rightarrow y=-3 z \ & 	herefore \quad x-6 z+8 z=0 \ & \Rightarrow \quad x=-2 z \ & 	herefore \quad \frac{x y}{z^2}=\frac{(-2 z)(-3 z)}{z^2}=6 \ & \end{aligned} $

If $\mathbf{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \mathbf{b}=2 \hat{i}+3 \hat{j}+\hat{k}$, $\mathbf{c}=8…

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