Question

If $a,b,c$ are non-coplanar vectors and $\lambda$ is a real number, then $\left[\lambda \right(a+b){\lambda }^{2}b\lambda c]=[ab+cb]$ for?

Answer 1

Finding the number of values of $\lambda$ that satisfy the equation $\left[\lambda \right(a+b){\lambda }^{2}b\lambda c]=[ab+cb]$:

Given that $a,b,c$ are non coplanar vectors

Taking the LHS,

LHS $=\left[\lambda \right(a+b){\lambda }^{2}b\lambda c]$

$$\begin{gathered} =\lambda \left[\right(a+b){\lambda }^{2}b\lambda c] \\ ={\lambda }^4\left(\right(a+b)\times b).c \\ ={\lambda }^4[abc] \end{gathered}$$

RHS $=[ab\times cb]$

$$\begin{gathered} =(a\times (b+c\left)\right).b \\ =(a\times b+a\times c]b \\ =a\times b.b+a\times c.b \\ =[acb] \\ =–[abc] \end{gathered}$$

Therefore,

$$\begin{gathered} {\lambda }^4[abc]=–[abc] \\ {\lambda }^4=-1 \end{gathered}$$

Since ${\lambda }^4\ne 1$

So no values of $\lambda$ that satisfy the equation

Hence, number of values of $\lambda$ that satisfy the equation is Zero.