33since it is matrix match i got the answers... but a full method please... [7]
Match the following
Consider the linear equations.
ax+by+cz=0
bx+cy+az=0
cx+ay+bz=0
(A) a+b+c≠0 and a2+b2+c2=ab+bc+ca
(B) a+b+c=0 and a2+b2+c2≠ab+bc+ca
(C) a+b+c≠0 and a2+b2+c2≠ab+bc+ca
(D) a+b+c=0 and a2+b2+c2=ab+bc+ca
(p) the equations represents planes meeting only at a single point.
(q) the equations represent the line x=y=z
(r) the equation represent identical planes.
(s) the equations represents the whole 3-D space.
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10 Answers
33
1ax+by+cz=0 => plane wid dir ratios a,b,c n passing through origin.
bx+cy+az=0 => ..........................b,c,a .................................
cx+ay+bz=0 => ..........................c,a,b .................................
so the planes pass thrugh the origin..
ans should be a.. [12]
62|a b c| [x y z] =0
|b c a|
|c a b|
So we have 2 cases .. determinant zero and determinant non zero
Determinant is
a3+b3+c3 - 3abc
If a+b+c = 0 then
(a+b)3=-c3
will give determinant zero
a2+b2+c2=ab+bc+ca
means (a+b+c)2=3(a2+b2+c2)
if a+b+c=0
thus, a=b=c=0 (Check this statemtn.. jaldi me galti na hog gaya hai) This will represent the whole space..
now does it solve it :)
1(A) a+b+c≠0 and a2+b2+c2=ab+bc+ca
a^2+b^2+c^2=ab+bc+ca
=> (a-b)2 + (b-c)2 + (c-a)2 = 0
=> a=b=c ≠0 (given)
so all the eqns become: x+y+z =0
so A -> r
(B) a+b+c=0 and a2+b2+c2≠ab+bc+ca
a^2+b^2+c^2 ≠ab+bc+ca
=> a ≠b ≠c
but a+b+c=0 => x,y,z are collinear. (property)
so, B-> q
(C) a+b+c≠0 and a2+b2+c2≠ab+bc+ca
a^2+b^2+c^2 ≠ab+bc+ca
ax+by+cz=0 => plane wid dir ratios a,b,c n passing through origin.
bx+cy+az=0 => ..........................b,c,a .................................
cx+ay+bz=0 => ..........................c,a,b .................................
so the planes pass thrugh the origin..
so C-> p
(D) a+b+c=0 and a2+b2+c2=ab+bc+ca
a=b=c =0
=> x, y, z non-coplanar ... and it represents the whole space.
33[1][3]
okieeee.....
samjh me aa gaya...
i was missing a=b=c=nonzero.. [3] in A
tx :)