Question

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

1. exactly two values of $\lambda$
2. exactly three values of $\lambda$
3. no real value of $\lambda$
4. exactly one value of $\lambda$

Answer 1

In this type of question we have to use the concept of non-coplanar vectors. We know that a finite number of vectors are said to be non-coplanar if they do not lie on the same plane or on the same parallel planes. We know that the scalar triple product of three vectors $a,b,c$ is denoted as $\left[ abc \right]=\left( a\times b \right)\cdot c$. Also by the property of scalar triple product we know that the product is cyclic in nature that is $\left[ abc \right]=\left[ bca \right]=\left[ cab \right]=-\left[ bac \right]=-\left[ cba \right]=-\left[ acb \right]$.

Complete step-by-step solution:
Now we have to find the value of $\lambda$ if $\left[ \lambda \left( a+b \right){{\lambda }^{2}}b\lambda c \right]=\left[ ab+cb \right]$
Let us consider the L.H.S.
$\Rightarrow L.H.S.=\left[ \lambda \left( a+b \right){{\lambda }^{2}}b\lambda c \right]$
As we know that the scalar triple product of three vectors $a,b,c$ is denoted as $\left[ abc \right]=\left( a\times b \right)\cdot c$
$\Rightarrow L.H.S.=\left( \left( \lambda \left( a+b \right)\times {{\lambda }^{2}}b \right)\cdot \lambda c \right)$
Now we have given that the $\lambda$ is a real number
$\Rightarrow L.H.S.={{\lambda }^{4}}\left( \left( a+b \right)\times b \right)\cdot c$
$\Rightarrow L.H.S.={{\lambda }^{4}}\left[ \left( a+b \right)bc \right]$
We know that $\left[ \left( a+b \right)bc \right]=\left[ abc \right]$
$\Rightarrow L.H.S.={{\lambda }^{4}}\left[ abc \right]\cdots \cdots \cdots \left( i \right)$
Now, we will simplify the R.H.S.
$\Rightarrow R.H.S.=\left[ ab+cb \right]$
$\Rightarrow R.H.S.=\left[ a\left( b+c \right)b \right]$
$\Rightarrow R.H.S.=\left( a\times \left( b+c \right) \right)\cdot b$
$\Rightarrow R.H.S.=\left[ acb \right]$
Now we know that, by the property of scalar triple product the product is cyclic in nature that is $\left[ abc \right]=\left[ bca \right]=\left[ cab \right]=-\left[ bac \right]=-\left[ cba \right]=-\left[ acb \right]$.
$\Rightarrow R.H.S.=-\left[ abc \right]\cdots \cdots \cdots \left( ii \right)$
Now, we have given that,
$\Rightarrow R.H.S.=L.H.S.$
Hence, from equation $\left( i \right)$ and $\left( ii \right)$ we can write,

& \Rightarrow {{\lambda }^{4}}\left[ abc \right]=-\left[ abc \right] \\\ & \Rightarrow {{\lambda }^{4}}=-1 \\\ \end{aligned}$$ Hence, there does not exists any real value of $$\lambda $$ **Thus, option (3) is the correct option.** **Note:** In this type of question students have to be well familiar with the concept as well as properties of the scalar triple product. Also students have to remember the relations such as $$\left[ \left( a+b \right)bc \right]=\left[ abc \right]$$, $$\left[ a\left( b+c \right)b \right]=\left[ acb \right]$$, etc.