Show that the points i - j +3 k and 3 i +3 j +3 k are equidistant from the plane r (5 i +2 j -7 k )+9=0

We know that, distance of a point P of position vector a from the plane r n =d is given by p= | a n -d | | n | ...(i) Let D_1 be the distance of point ( i - j + k ) from the plane r (5 i +2 j -7 k )+9=0, then D_1= | ( i - j +3 k ) (5 i +2 j -7 k )+9 (5)^2+(2)^2+(-7)^2 | [Using equation (i)] = |(1)(5)+(-1)(2)+(3)(-7)+9 25+4+49 | = |5-2-21+9 78 | = |-9 78 | D_1=9 78 units ...(ii) Again, let D_2 be the distance of point (3 i -3 j +3 k ) from the plane r (5 i +2 j -7 k )+9=0, then using equation (i) we get, D_2= | (3 i -3 j +3 k ) (5 i +2 j -7 k )+9 (5)^2+(2)^2+(-7)^2 | = |(3)(5)+(3)(2)+(3)(-7)+9 25+4+49 | = |15+6-21+9 78 | = |9 78 | D_2=9 78 units ...(iii) From equation (i) and (iii) D_1=D_2 Distance of point ( i - j +3 k ) from plane r (5 i +2 j -7 k )+9=0 = Distance of point (3 i -3 j +3 k ) from plane r (5 i +2 j -7 k )+9=0

Show that the points i - j +3 k and 3 i +3 j +3 k are equidistant…

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Question

Show that the points $\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ and $3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}}$ are equidistant from the plane $\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9=0$

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Answer

We know that, distance of a point P of position vector $\vec{\text{a}}$ from the plane $\vec{\text{r}}\cdot\hat{\text{n}}=\text{d}$ is given by
$\text{p}=\frac{\big|\vec{\text{a}}\cdot\vec{\text{n}}-\text{d}\big|}{|\vec{\text{n}}|}\ ...(\text{i})$
Let $D_1$ be the distance of point $(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})$ from the plane $\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9=0,$ then
$\text{D}_1=\Bigg|\frac{(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9}{\sqrt{(5)^2+(2)^2+(-7)^2}}\Bigg|$ [Using equation (i)]
$=\Bigg|\frac{(1)(5)+(-1)(2)+(3)(-7)+9}{\sqrt{25+4+49}}\Bigg|$
$=\Big|\frac{5-2-21+9}{\sqrt{78}}\Big|$
$=\Big|-\frac{9}{\sqrt{78}}\Big|$
$\text{D}_1=\frac{9}{\sqrt{78}}\text{ units}\ ...(\text{ii})$
Again, let $D_2$ be the distance of point $(3\hat{\text{i}}-3\hat{\text{j}}+3\hat{\text{k}})$ from the plane $\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9=0,$ then using equation (i) we get,
$\text{D}_2=\Bigg|\frac{(3\hat{\text{i}}-3\hat{\text{j}}+3\hat{\text{k}})\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9}{\sqrt{(5)^2+(2)^2+(-7)^2}}\Bigg|$
$=\Bigg|\frac{(3)(5)+(3)(2)+(3)(-7)+9}{\sqrt{25+4+49}}\Bigg|$
$=\Big|\frac{15+6-21+9}{\sqrt{78}}\Big|$
$=\Big|\frac{9}{\sqrt{78}}\Big|$
$\text{D}_2=\frac{9}{\sqrt{78}}\text{ units}\ ...(\text{iii})$
From equation (i) and (iii)
$\text{D}_1=\text{D}_2$
Distance of point $(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})$ from plane $\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9=0$ = Distance of point $(3\hat{\text{i}}-3\hat{\text{j}}+3\hat{\text{k}})$ from plane $\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9=0$