Samyak Jain
Last Activity: 6 Years ago
Given : a + xb + yc = 0 ...(1) and a x b + b x c + c x a = 6(b x c) ...(2)
To solve this question, we’ll represent a in terms of b and c since RHS in second equation is in terms of (b x c).
We will use the facts that cross product of a vector with itself is a zero vector and negative of cross product
of a vector with another vector is same as cross product of second vector with the first i.e.
a x a = 0 and – (a x b) = (b x a)
a + x
b + y
c =
0 ,
a = – (x
b + y
c)
Now,
a x
b = – (x
b + y
c) x
b = – (x(
b x
b) + y(
c x
b)) = – y(
c x
b) = y(
b x
c) [
b x
b =
0]
& c x a = c x {– (xb + yc)} = – c x (xb + yc) = – (x (c x b) + y(c x c)) = – x(c x b) = x(c x b)
[
c x
c =
0]
(2) becomes y(b x c) + b x c + x(c x b) = 6(b x c) i.e.
(x + y +1)(b x c) = 6(b x c)
Comparing coefficients of b x c on both sides, we get
(x + y +1) = 6
x + y – 5 = 0.