Question

If $a,b$and $c$ are unit coplanar vectors, then $[2a-b2b-c2c-a]$is equal to $?$

• A. $1$
• B. $0$
• C. $-\sqrt{3}$
• D. $\sqrt{3}$
• E. $6$

Answer 1

The correct option is B

$0$

Find the value of $\mathbf{[}\mathbf{2}\mathit{a}\mathbf{}\mathbf{-}\mathbf{}\mathit{b}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathit{b}\mathbf{-}\mathit{c}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathit{c}\mathbf{-}\mathit{a}\mathbf{]}$:

Given that If $a,b$and $c$ are unit coplanar vectors.

$\Rightarrow [abc]=0$

Now, $[2a-b,2b-c,2c-a]$

$$\begin{gathered} =(2a–b).(2b–c)\times (2c–a)\mathbf{\left[}\mathbf{\because }\mathbf{}\mathbf{[}\mathbf{a}\mathbf{}\mathbf{b}\mathbf{}\mathbf{c}\mathbf{]}\mathbf{=}\mathbf{(}\mathbf{a}\mathbf{.}\mathbf{b}\mathbf{)}\mathbf{\times }\mathbf{c}\mathbf{\right]} \\ =(2a–b).4(b\times c)–2(b\times a)+(c\times a) \\ =8(a.(b\times c\left)\right)–(b.(c\times a\left)\right) \\ =8[abc]–[abc] \\ =7[abc] \\ =0 \end{gathered}$$

Hence, the correct option is (B).