HSC Arts 12th Board ExamMaharashtra State Board
Share
Notifications

View all notifications
Books Shortlist
Your shortlist is empty

Solution for Show that the points (1, –1, 3) and (3, 4, 3) are equidistant from the plane 5x + 2y – 7z + 8 = 0 - HSC Arts 12th Board Exam - Mathematics and Statistics

Login
Create free account


      Forgot password?

Question

Show that the points (1, –1, 3) and (3, 4, 3) are equidistant from the plane 5x + 2y – 7z + 8 = 0

Solution

Let p1 and p2 be the distances of points `hati-hatj+3hatk and 3hati+4hatj+3hatk` from `bar r.(5hati+2hatj-7hatk)+8=0`

The distance of the point A with position vector a from the plane `barr.barn` = p is given by

`d=|bara.barn-p|/|barn|`

`therefore p_1=|(hati-hatj+3hatk).(5hati+2hatj-7hatk)-(-8)|/sqrt(5^2+2^2+(-7)^2)`

=`|1(5)-1(2)+3(-7)+8|/sqrt(25+4+49)`

=`|5-2-21+8|/sqrt(78)=|-10|/sqrt78=10/sqrt78`

`and p_2=|()()-(-8)|/sqrt(5^2+2^2+(-7)^2)`

=`|3(5)+4(2)+3(-7)+8|/sqrt(25+4+49)`

=`|15+8-21+8|/sqrt78`

=`10/sqrt78`

∴ p1 = p2
Hence, points are equidistant from the plane.

  Is there an error in this question or solution?

APPEARS IN

 2015-2016 (July) (with solutions)
Question 3.1.2 | 3.00 marks
Solution for question: Show that the points (1, –1, 3) and (3, 4, 3) are equidistant from the plane 5x + 2y – 7z + 8 = 0 concept: Distance of a Point from a Plane. For the courses HSC Arts, HSC Science (Electronics), HSC Science (General) , HSC Science (Computer Science)
S
View in app×