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. exactly two values $\lambda$
• B. exactly three values of $\lambda$
• C. no value of $\lambda$
• D. exactly one value of $\lambda$

Answer 1

Answer: (C)

Let $\vec{a}=\left(a_1,a_2,a_3\right),\vec{b}=\left(b_1,b_2,b_3\right)$,
$\vec{c}=\left(c_1,c_2,c_3\right)$
By the given condition,
$\left|\begin{array}{ccc}\lambda \left(a_1+b_1\right)& \lambda \left(a_2+b_2\right)& \lambda \left(a_3+b_3\right)\\ {\lambda }^2{b}_1& {\lambda }^2{b}_2& {\lambda }^2{b}_3\\ \lambda c_1& \lambda c_2& \lambda c_3\end{array}\right|$
$=\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1+c_1& b_2+c_2& b_3+c_3\\ b_1& b_2& b_3\end{array}\right|$
$\Rightarrow {\lambda }^4\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|=\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ c_1& c_2& c_3\\ b_1& b_2& b_3\end{array}\right|$
$\Rightarrow {\lambda }^4=-1$
$\therefore$ no real value of $\lambda$ exists.