If veca, vecb and vecc are unit coplanar vectors, then the scalar triple product [(2 veca- vecb)(2 vecb- vecc)(2 vecc- veca)] is

Question

If veca, vecb and vecc are unit coplanar vectors, then the scalar triple product [(2 veca- vecb)(2 vecb- vecc)(2 vecc- veca)] is (A) 0 (B) 1 (C) -√3 (D) √3

Tardigrade Admin · Accepted Answer

Since veca, vecb and vecc are unit coplanar vectors, 2 veca- vecb, 2 vecb- vecc and 2 vecc- veca are also coplanar vectors, being a linear combination of veca, vecb and vecc. Thus, [(2 veca- vecb)(2 vecb- vecc)(2 vecc- veca)]=0

Q. If $a, b$ and $c$ are unit coplanar vectors, then the scalar triple product $[(2 a - b) (2 b - c) (2 c - a)]$ is

$0$

$1$

$- 3$

$3$

Since $a, b$ and $c$ are unit coplanar vectors, $2 a - b$ , $2 b - c$ and $2 c - a$ are also coplanar vectors, being a linear combination of $a, b$ and $c$ .
Thus, $[(2 a - b) (2 b - c) (2 c - a)] = 0$

Q. If a,b and c are unit coplanar vectors, then the scalar triple product [(2a−b)(2b−c)(2c−a)] is

0

1

−3​

3​

Since a,b and c are unit coplanar vectors, 2a−b, 2b−c and 2c−a are also coplanar vectors, being a linear combination of a,b and c. Thus, [(2a−b)(2b−c)(2c−a)]=0

Q. If $a, b$ and $c$ are unit coplanar vectors, then the scalar triple product $[(2 a - b) (2 b - c) (2 c - a)]$ is

$0$

$1$

$- 3$

$3$

Since $a, b$ and $c$ are unit coplanar vectors, $2 a - b$ , $2 b - c$ and $2 c - a$ are also coplanar vectors, being a linear combination of $a, b$ and $c$ .
Thus, $[(2 a - b) (2 b - c) (2 c - a)] = 0$