logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
0 votes

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.

Can you answer this question?
 
 

1 Answer

0 votes
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$

 

answered Mar 7, 2013 by thanvigandhi_1
edited Apr 5, 2013 by thanvigandhi_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...