Question

If ai+aj+ck and i+k and ci+cj+bk are coplanar then show that c^2=ab

Answer 1

I think that the answer is this.

Answer 2

Answer:

we need to show  c²=ab if ai+aj+ck and i+k and ci+cj+bk are co-planar

The three vectors are co-planar if their scalar triple product is zero.

$\begin{vmatrix}a&a&c\\ 1&0&1\\ 1&c&b \end{vmatrix}=0$

a(0-c) - a(b-c) + c(c-0)=0

a(-c) - a(b-c) + c(c) = 0

-ac - ab + ac + c²=0

- ab + c²=0

add both the sides by ab, in above expression

c² = ab

Hence proved