Guest
Displaying Posts 1 - 3 of 3|
#1 Posted 2:22pm 21-02-10 RT P1 45A non-zero vector [o]a [/o] i s parallel to the line of intersection of plane P1 determined by i+j , i - 2j and plane P2 determined by vector 2i + j ,3i + 2k , then angle between [o]a[/o] and vector i - 2j +2k is (A)Π/4 (B) Π/2 (C) Π/3 (D) Π
But there is suffering in life, and there are defeats. No one can avoid them. But it's better to lose some of the battles in the struggles for your dreams than to be defeated without ever knowing what you're fighting for.
|
|
#2 Posted 1:32pm 22-02-10 Re: RT P1 45See first we need to find the eq[p]n[/p] of the normal to the plane P[ss]1[/ss] by finding the cross product of ( i + j ) and (i - 2j ) which comes out to be -3k then similarly eq[p]n[/p] of the normal to the plane P[ss]2[/ss] = 2i - 4j - 3k then to find the eq[p]n[/p] of line passing through the intersection of the abv plane we need to find the cross product P[ss]1[/ss] P[ss]2[/ss].. which comes out to be12i + 6j.. now find dot product 12i + 6j with i -2j + 2k cosθ = 0 so θ = π/2... Ans B
Edited on 00:02am 24-02-10 |
|
#3 Posted 1:45pm 22-02-10 Re: RT P1 45Thanks [159]
But there is suffering in life, and there are defeats. No one can avoid them. But it's better to lose some of the battles in the struggles for your dreams than to be defeated without ever knowing what you're fighting for.
|
Displaying Posts 1 - 3 of 3
