Home » Ask & Discuss » Mathematics. » Analytical Geometry « Back to Discussion

Analytical Geometry

Blazing goIITian

Joined:	4 Nov 2007
Post:	1524

11 Nov 2007 14:18:08 IST

0 People liked this

261

3D doubts

None

this ques is frm 3d( ncert)plz solve it fully.

find the shortest distance b/w lines

i) (x+1)/ 7 = (y+1)/ -6 = (z+1)/1

and

(x-3)/1 = (y-5)/-2 = (z-7)/1

ii) r = (1-t)i^ + (t-2)j^ + (3-2t) k^

r = (s+1)i^ + (2s-t)j^ +(2s+1)k^

Share this article on:

Comments (1)

joy francis

Blazing goIITian

Joined: 19 Feb 2007

Posts: 1802

11 Nov 2007 15:25:14 IST

Like 1 people liked this

if there are two line a+Ac and b+Bc , where a,b,c,d are vectors and A,B are non zero scalars , the shortest distance b/w the lines is given by

d = (a-b).(cxd) / |cxd|

a = -i-j-k

b = 3i+5j+7k

c = 7i-6j+k

d = i-2j+k

a-b = cxd = -4i-6j-8k

& |cxd| = sqrt(116)

.: d = (-4i-6j-8k).(-4i-6j-8k) / sqrt116

= 116/root(116) = root(116) = 2root29.

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

Free Sign Up!

Preparing for IIT-JEE ?

Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor

@ INR 1,995/-

For Quick Info

Find Posts by Topics

Physics.

Topics

9395

20091

3672

2186

7617

2827

2637

Mathematics.

Chemistry.

Biology

Institutes

Parents

Board

Fun Zone