Question

If \[\left[ 2 \vec{a} + 4 \vec{b} \vec{c} \vec{d} \right] = \lambda\left[ \vec{a} \vec{c} \vec{d} \right] + \mu\left[ \vec{b} \vec{c} \vec{d} \right],\]  then λ + μ =

विकल्प

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Answer 1

6

We have

\[\left[ 2 \vec{a} + 4 \vec{b} \vec{c} \vec{d} \right] = \lambda \left[ \vec{a} \vec{c} \vec{d} \right] + \mu\left[ \vec{b} \vec{c} \vec{d} \right]\]

\[ \Rightarrow \left[ \left( 2 \vec{a} + 4 \vec{b} \right) \times \vec{c} \right] . \vec{d} = \lambda \left[ \vec{a} \vec{c} \vec{d} \right] + \mu\left[ \vec{b} \vec{c} \vec{d} \right] \left( \text { By definition of scalar triple product } \right)\]

\[ \Rightarrow \left[ \left( 2 \vec{a} \times \vec{c} \right) + \left( 4 \vec{b} \times \vec{c} \right) \right] . \vec{d} = \lambda \left[ \vec{a} \vec{c} \vec{d} \right] + \mu\left[ \vec{b} \vec{c} \vec{d} \right]\]

\[ \Rightarrow \left( 2 \vec{a} \times \vec{c} \right) . \vec{d} + \left( 4 \vec{b} \times \vec{c} \right) . \vec{d} = \lambda \left[ \vec{a} \vec{c} \vec{d} \right] + \mu\left[ \vec{b} \vec{c} \vec{d} \right]\]

\[ \Rightarrow \left[ 2 \vec{a} \vec{c} \vec{d} \right] + \left[ 4 \vec{b} \vec{c} \vec{d} \right] = \lambda \left[ \vec{a} \vec{c} \vec{d} \right] + \mu\left[ \vec{b} \vec{c} \vec{d} \right]\]

\[ \Rightarrow 2 \left[ \vec{a} \vec{c} \vec{d} \right] +4 \left[ \vec{b} \vec{c} \vec{d} \right] = \lambda \left[ \vec{a} \vec{c} \vec{d} \right] + \mu\left[ \vec{b} \vec{c} \vec{d} \right] \left( \because \left[ \lambda \vec{a} \vec{b} \vec{c} \right] = \lambda\left[ \vec{a} \vec{b} \vec{c} \right] \text { for any scalar } \lambda \right)\]

Comparing both sides, we get

\[\lambda = 2 \]

\[\mu = 4\]

\[ \therefore \lambda + \mu = 2 + 4 = 6\]