Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are noncoplanar 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

Answer: (A)

no value of$\lambda$

Given:$[\lambda (\overrightarrow{a}+\overrightarrow{b}),{\lambda }^2\overrightarrow{b},\lambda \overrightarrow{c}]$
$={\lambda }^4[\overrightarrow{a}+\overrightarrow{b},\overrightarrow{b},\overrightarrow{c}]$
$={\lambda }^4[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]+{\lambda }^4[\overrightarrow{b},\overrightarrow{b},\overrightarrow{c}]$
$={\lambda }^4[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]+{\lambda }^4.0$
$={\lambda }^4[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]$
Right side $=[\overrightarrow{a},\overrightarrow{b}+\overrightarrow{c},\overrightarrow{b}]$
$=[\overrightarrow{a},\overrightarrow{b},\overrightarrow{b}]+[\overrightarrow{a},\overrightarrow{c},\overrightarrow{b}]$
$=0+[\overrightarrow{a},\overrightarrow{c},\overrightarrow{b}]$
$=-[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]$
$\therefore {\lambda }^4[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]=-[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]$
$\Rightarrow ({\lambda }^4+1)[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]=0$
But $[\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}]\ne 0\Rightarrow {\lambda }^4+1=0$,no real $\lambda$