Home » Ask & Discuss » Mathematics. » Vectors « Back to Discussion

Vectors

Blazing goIITian

Joined:	14 Aug 2007
Post:	407

19 Mar 2008 19:55:22 IST

0 People liked this

465

tough i say !!!!

None

$From\;a\;point\;P(\alpha,\beta,\gamma)\;a\;plane\;is\; drawn\;at\;right\;angle\;to\\\\\;OP\;(O \;is\;origin)\;to\;meet\;the\;coordinate\;axes\;at\;A,B,C\;.\;\\\\Prove \;that\;the\;area\;of\;triangle\;ABC\;is\;\frac{r^5}{2\alpha\beta\gamma}\;\;\\\\where\;r\;is\;measure\;OP.$

Share this article on:

Comments (4)

Ajay Antony

Hot goIITian

Joined: 27 Dec 2007

Posts: 173

19 Mar 2008 20:07:27 IST

Like 1 people liked this

the eq of the plane can b written as a(x-a)+b(y-b)+c(z-c)=0
this meets the coordinate axis at( (a^2+b^2+c^2)/a,(a^2+b^2+c^2)/b,(a^2+b^2+c^2)/c)
therfore the areaof the

is((a^2+b^2+c^2)²/2abc
but frm the eqn we get r^2=(a^2+b^2+c^2)
hence proved.

where a,b,c denotes

Ajay Antony

Hot goIITian

Joined: 27 Dec 2007

Posts: 173

19 Mar 2008 20:08:53 IST

Like 1 people liked this

rate me if im rite and nudge me f im wrong

Ajay Antony

Hot goIITian

Joined: 27 Dec 2007

Posts: 173

19 Mar 2008 20:32:11 IST

Like 0 people liked this

am i correct??dint u see my last line??

ram kumar

Blazing goIITian

Joined: 14 Aug 2007

Posts: 407

19 Mar 2008 20:49:26 IST

Like 0 people liked this

EDITED :

ya ok, problem solved...... thanks annihilator.......!!!

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team