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