Question
Show that the points $(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})$ and $3(\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}})$ are equidistant from the plane $\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9=0$ and lies on opposite side of it.

Answer

To show that these given points $(\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})$ and $3(\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}})$ are equidistant from the plane
$\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9=0,$ we first find out the mid-point of the points which is $(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}).$
On substituting $\vec{\text{r}}$ by the mid-point in plane, we get
$\text{L.H.S.}=(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}})\cdot(5\hat{\text{i}}+\hat{\text{j}}-7\hat{\text{k}})+9$
$=10+2-21+9=0$
$=\text{R.H.S.}$
Hence, the two points lie on opposite sides of the plane are equidistant from the plane.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free