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# Show that the points $(\hat i-\hat j+3 \hat k)\; and\;3(\hat i+\hat j+\hat k)$ are equidistant from the plane $\overrightarrow{r}.(5 \hat i+2 \hat j-7 \hat k)+9=0$ and lies on opposite side of it.

Toolbox:
• Distance of a point from plane is $\bigg| \large\frac{ax_1+by_1+cz_1+d}{\sqrt{a^2+b^2+c^2}} \bigg|$
• If two points are in opposite sides then substitute the coordinates in the equation. The result should be of opposite sign of the plane.
Step 1
Name the points and convert the plane equation in cartesian form $A(1, -1, 3) B(3,3,3)$
$P \rightarrow 5x=2y-72+9=0$
Step 2
$d_1$ between $AP = \bigg| \large\frac{5-2-21+9}{\sqrt{25+4+49}} \bigg| = \large\frac{9}{\sqrt{78}}$
$d_2$ between $BP \bigg|\large \frac{15+6-21+9}{\sqrt{78}} \bigg| = \large\frac{9}{\sqrt{78}}$
Step 3
$d_1 = d_2$
Step 4
$(5$ x $1)+(2x-1)+(3x-7)+9 \: \: Q \: is -ve.$
$(5$ x $3)+(2$ x $3)+(-7$ x $3)+9=9 \: is \: +ve$

edited Apr 5, 2013