Question

The value of [a-b b-c c-a] is?
A. \[0\]
B. \[1\]
C. \[2\]
D. None of the above

Answer 1

Hint: To solve this question first you need to rewrite the question in the form of dot product and cross product that is \[a-b.(b-c\times c-a)\] and then simplify it by doing cross product and opening the bracket

Complete step by step answer:
 First to solve the question that [a-b b-c c-a] we can write it as
\[a-b.(b-c\times c-a)\]
To be able to solve this we further simplify it starting with the bracket and doing cross product of the two vectors b-c and c-a.
\[a-b.(b\times c-b\times a-c\times c+c\times a)\]
Now as we know that while calculating cross product if we take the cross product of two same vectors that the answer of that will be \[0\], therefore
\[a-b.(b\times c-b\times a+c\times a)\]

We can interchange the position of vectors to get positive sign
\[a-b.(b\times c+a\times b+c\times a)\]
Now to further simplify this we need to open the bracket. To open the bracket we will be doing the dot multiplication of all the terms inside the bracket with the vector a-b which gives us
\[a.b\times c+a.a\times c+a.c\times a-b.b\times c-b.a\times b-b.c\times a\]
Now as we know \[a\times c\] gives us a vector that is perpendicular to the vector a and therefore taking the dot product of the perpendicular vector with vector a we get \[0\].

In this manner we can simplify multiple things in this equation that is \[c\times a\] is perpendicular to a, \[b\times c\] is perpendicular to b and \[a\times b\] is also perpendicular to b. This leaves us with
\[a.b\times c-b.c\times a\]
Now we know that \[b.c\times a\] can also be written as \[a.b\times c\] which means that
\[a.b\times c-a.b\times c=0\]
Therefore [a-b b-c c-a] is equal to \[0\].

Hence, the correct answer is option A.

Note: [a-b b-c c-a] is the scalar triple product of thee three vectors which are a-b, b-c and c-a and the absolute value of [a-b b-c c-a] is equal to the volume of a parallelepiped spanned by the vectors a-b, b-c and c-a.