the two lines `x=ay+b,z=cy+d and x=a\'y+b,z=c\'y+d\'` will be perpendicular, if and only if:
(A) `aa\'+c c' + 1=0`
(B) `aa\'+b b\'+c c'+1=0`
(C) `aa\'+b b\'+c c'=0`
(D) `(a+a\')+(b+b\')+(c+c\')=0`

The two lines x=ay+b,z=cy+d and x=a'y+b', z=c'y +d' are pendicular to each other if

The condition of the lines x=az+b,y=cz+d and x=a_1z+b_1,y=c_1z+d_1 to be perpendicular is (A) ac_1+a_1c+1=0 (B) aa_1+c c_1+1=0 (C) ac_1+b b\'+c c\'=0 (D) aa_1+c c_1-1=0

If the lines x=a_(1)y + b_(1), z=c_(1)y +d_(1) and x=a_(2)y +b_(2), z=c_(2)y + d_(2) are perpendicular, then

Prove that the line x=a y+b ,\ z=c y+d\ a n d\ x=a^(prime)y+b^(prime),\ z=c^(prime)y+d^(prime) are perpendicular if aa^(prime) + c c^(prime) + 1=0

The straight line a x+b y+c=0 , where a b c!=0, will pass through the first quadrant if (a) a c >0,b c >0 (b) ac >0 or b c 0 or a c >0 (d) a c<0 or b c<0

If the lines ax+by+c=0, bx+cy+a=0 and cx+ay+b=0 (a, b,c being distinct) are concurrent, then (A) a+b+c=0 (B) a+b+c=1 (C) ab+bc+ca=1 (D) ab+bc+ca=0

Show that the line represented by equation x=ay+b,z=cy+d in symmetric form is (x-b)/a=y/1=(z-d)/c

The equation of the plane through the intersection of the plane a x+b y+c z+d=0\ a n d\ l x+m y+n+p=0 and parallel to the line y=0,\ z=0. (A) (b l-a m)y+(c l-a n)z+d l-a p=0 (B) (a m-b l)x+(m c-b n)z+m d-b p=0 (c) (n a-c l)d+(b n-c m)y+n d-c p=0 (D) None of these

If a^x=b ,\ b^y=c\ a n d\ c^z=a , prove that x y z=1

If a,b,c are real, a ne 0, b ne 0, c ne 0 and a+b + c ne 0 and 1/a + 1/b + 1/c = 1/(a+b+c) , then (a+b) (b+c)(c+a) =