Question

If $\left[2\overrightarrow{a}+4\overrightarrow{b}\overrightarrow{c}\overrightarrow{d}\right]=\mathrm{\lambda }\left[\overrightarrow{a}\overrightarrow{c}\overrightarrow{d}\right]+\mathrm{\mu }\left[\overrightarrow{b}\overrightarrow{c}\overrightarrow{d}\right],$ then λ + μ =
(a) 6
(b) −6
(c) 10
(d) 8

Answer 1

(a) 6

$$\begin{gathered} \mathrm{We}\mathrm{have} \\ \left[2\overrightarrow{a}+4\overrightarrow{b}\mathit{}\overrightarrow{c}\mathit{}\overrightarrow{d}\right]=\mathrm{\lambda }\mathit{}\left[\overrightarrow{a}\mathit{}\overrightarrow{c}\mathit{}\overrightarrow{d}\right]\mathit{}\mathit{+}\mathit{}\mu \left[\overrightarrow{b}\mathit{}\overrightarrow{c}\mathit{}\overrightarrow{d}\right] \\ \Rightarrow \left[\left(\mathit{2}\overrightarrow{a}\mathit{+}\mathit{4}\overrightarrow{b}\right)\mathit{\times }\overrightarrow{c}\right]\mathit{.}\overrightarrow{d}=\mathrm{\lambda }\left[\overrightarrow{a}\mathit{}\overrightarrow{c}\mathit{}\overrightarrow{d}\right]\mathit{}\mathit{+}\mathit{}\mu \left[\overrightarrow{b}\mathit{}\overrightarrow{c}\mathit{}\overrightarrow{d}\right]\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathrm{By}\mathrm{definition}\mathrm{of}\mathrm{scalar}\mathrm{triple}\mathrm{product}\right) \\ \Rightarrow \left[\left(2\overrightarrow{\mathrm{a}}\mathit{\times }\overrightarrow{\mathrm{c}}\right)+\left(4\overrightarrow{\mathrm{b}}\mathit{\times }\overrightarrow{\mathrm{c}}\right)\right]\mathit{}\mathit{.}\overrightarrow{d}=\mathrm{\lambda }\left[\overrightarrow{a}\mathit{}\overrightarrow{c}\mathit{}\overrightarrow{d}\right]\mathit{}\mathit{+}\mathit{}\mu \left[\overrightarrow{b}\mathit{}\overrightarrow{c}\mathit{}\overrightarrow{d}\right] \\ \Rightarrow \left(2\overrightarrow{a}\mathit{\times }\overrightarrow{c}\right)\mathit{.}\overrightarrow{d}\mathit{}+\mathit{}\left(4\overrightarrow{b}\mathit{\times }\overrightarrow{c}\right)\mathit{.}\overrightarrow{d}=\mathrm{\lambda }\mathit{}\left[\overrightarrow{a}\mathit{}\overrightarrow{c}\mathit{}\overrightarrow{d}\right]\mathit{}\mathit{+}\mathit{}\mu \left[\overrightarrow{b}\mathit{}\overrightarrow{c}\mathit{}\overrightarrow{d}\right] \\ \Rightarrow \left[2\overrightarrow{a}\mathit{}\overrightarrow{c}\mathit{}\overrightarrow{d}\right]\mathit{}+\mathit{}\left[4\overrightarrow{b}\mathit{}\overrightarrow{c}\mathit{}\overrightarrow{d}\right]=\mathrm{\lambda }\mathit{}\left[\overrightarrow{a}\mathit{}\overrightarrow{c}\mathit{}\overrightarrow{d}\right]\mathit{}\mathit{+}\mathit{}\mu \left[\overrightarrow{b}\mathit{}\overrightarrow{c}\mathit{}\overrightarrow{d}\right] \\ \Rightarrow 2\mathit{}\left[\overrightarrow{a}\mathit{}\overrightarrow{c}\mathit{}\overrightarrow{d}\right]\mathit{}\mathrm{+}4\mathit{}\left[\overrightarrow{b}\mathit{}\overrightarrow{c}\mathit{}\overrightarrow{d}\right]=\mathrm{\lambda }\mathit{}\left[\overrightarrow{a}\mathit{}\overrightarrow{c}\mathit{}\overrightarrow{d}\right]\mathit{}\mathit{+}\mathit{}\mu \left[\overrightarrow{b}\mathit{}\overrightarrow{c}\mathit{}\overrightarrow{d}\right]\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\because \left[\mathrm{\lambda }\overrightarrow{a}\mathit{}\overrightarrow{b}\mathit{}\overrightarrow{c}\right]\mathit{=}\mathrm{\lambda }\left[\overrightarrow{a}\mathit{}\overrightarrow{b}\mathit{}\overrightarrow{c}\right]\mathrm{for}\mathrm{any}\mathrm{scalar}\mathrm{\lambda }\right) \\ \mathrm{Comparing}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \mathrm{\lambda }\mathit{=}2\mathit{} \\ \mathrm{\mu }\mathit{=}4 \\ \therefore \mathrm{\lambda }+\mathrm{\mu }=2+4=6 \end{gathered}$$