Question

The value of \[\left[ {a - b{\rm{ }}b - c{\rm{ }}c - a} \right]\], where \[a = 1,b = 5\] and \[c = 3\] is
A. 0
B. 1
C. 2
D. 4

Answer 1

Hint: Coplanar vectors are three or more vectors that lie on the same plane. The scalar triple product of three coplanar vectors is zero. So, in this case, we will determine the scalar triple product of three vectors equal to zero.

Formula used: The scalars x, y, and z must not all be zero in order for the vectors a, b, and c to be coplanar.
\[{\bf{xa}} + {\bf{yb}} + {\bf{zc}} = 0\]

Complete step by step solution: Since, \[\vec a - \vec b + \vec b - \vec c + \vec c - \vec a = 0\], therefore the vectors \[\vec a - \vec b,\vec b - \vec c\] and \[\vec c - \vec a\] are coplanar.
 We have been already known that if three vectors are coplanar, then their box product would be equal to zero.
According to the given data, the equation can be formed as, \[[a - b,b - c,c - a] = \{ (a - b) \times (b - c)\} \cdot (c - a)\]
Now, we have to multiply the terms with the terms inside the parentheses, we get\[ = (a \times b - a \times c + b \times c) \cdot (c - a) = a \times b \cdot c - b \times c \cdot a\]
Now on simplifying the above equation, we obtain
\[ = [abc] - [abc] = 0\]
Therefore, the value of\[\left[ {a - b{\rm{ }}b - c{\rm{ }}c - a} \right]\ ]is \[0\]

Thus, Option (A) is correct.

Note: Remember that if there are three vectors, they are coplanar if (a) their scalar triple product is zero, (b) they are linearly dependent, and (c) in the case of n vectors, no more than two vectors are linearly independent. A non-trivial solution is also an equation system with a zero determinant of the coefficient. To answer these kinds of questions, one needs understand how to multiply terms with opposite and same signs. Always remember that the product of one positive and one negative number is always negative, the product of two negative numbers is positive, and the product of two positive numbers is positive.