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प्रश्न
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
6
-6
10
8
उत्तर
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
\[\lambda = 2 \]
\[\mu = 4\]
\[ \therefore \lambda + \mu = 2 + 4 = 6\]
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