Question

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

• (A) No value of $\lambda$
• (B) Exactly one value of $\lambda$
• (C) Exactly two values of $\lambda$
• (D) Exactly three values of $\lambda$

Answer 1

The Correct Option is A

Explanation:
Given:Non- coplanar vectors $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$and $[\lambda (\overrightarrow{a}+\overrightarrow{b}){\lambda }^2\overrightarrow{b}\lambda \overrightarrow{c}]=[\overrightarrow{a}\overrightarrow{b}+\overrightarrow{c}\overrightarrow{b}]$We have to find the value of $\lambda$.since, $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are non-coplanar vectors$\Rightarrow [\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]\ne 0$Now, $[\lambda (\overrightarrow{a}+\overrightarrow{b}){\lambda }^2\overrightarrow{b}\lambda \overrightarrow{c}]=[\overrightarrow{a}\overrightarrow{b}+\overrightarrow{c}\overrightarrow{b}]$Using the definition of scalar triple product, we get$\lambda (\overrightarrow{a}+\overrightarrow{b})\cdot ({\lambda }^2\overrightarrow{b}\times \lambda \overrightarrow{c})$$=\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{c})\times \overrightarrow{b})$$=\overrightarrow{a}\cdot (\overrightarrow{b}\times \overrightarrow{b}+\overrightarrow{c}\times \overrightarrow{b})$$=\lambda (\overrightarrow{a}+\overrightarrow{b})\cdot ({\lambda }^2\overrightarrow{b}\times \lambda \overrightarrow{c})$$=\overrightarrow{a}\cdot (0+\overrightarrow{c}\times \overrightarrow{b})$$\overrightarrow{a}\cdot (\overrightarrow{c}\times \overrightarrow{b})$[Using properties of cross product-2]$\Rightarrow {\lambda }^4(\overrightarrow{a}\cdot (\overrightarrow{b}\times \overrightarrow{c}))+\overrightarrow{b}\cdot (\overrightarrow{b}\times \overrightarrow{c})=\overrightarrow{a}\cdot (\overrightarrow{c}\times \overrightarrow{b})$$\Rightarrow {\lambda }^4([\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]+[\overrightarrow{b}\overrightarrow{b}\overrightarrow{c}])=-[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]$[Using properties of scalar triple product-3]$\Rightarrow {\lambda }^4([\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}])=-[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]$$\Rightarrow {\lambda }^4=-1$Which is not true for any real value of $\lambda$.Hence, the correct option is (A).