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Show that the points $(\hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})$ and $3(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$ are equidistant from the plane $\overrightarrow{\mathrm{r}} \cdot(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+9=0$ and lies on opposite side of it.
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Verified Answer
To show: Given points $(\hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})$ and $3(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$ are equidistant from.
$\overrightarrow{\mathrm{r}} \cdot(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+9=0$, we first find out the mid-point of the points which is $2 \hat{i}+\hat{j}+3 \hat{k}$.
On substituting $\overrightarrow{\mathrm{r}}$ by the mid-point in plane, we get
$$
\begin{aligned}
&\text { LHS }=(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+9 \\
&=10+2-21+9=0=\text { RHS }
\end{aligned}
$$
Hence, the two points lie on opposite sides of the plane are equidistant from the plane.
$\overrightarrow{\mathrm{r}} \cdot(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+9=0$, we first find out the mid-point of the points which is $2 \hat{i}+\hat{j}+3 \hat{k}$.
On substituting $\overrightarrow{\mathrm{r}}$ by the mid-point in plane, we get
$$
\begin{aligned}
&\text { LHS }=(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+9 \\
&=10+2-21+9=0=\text { RHS }
\end{aligned}
$$
Hence, the two points lie on opposite sides of the plane are equidistant from the plane.
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