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 =

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

If [ 2 → a + 4 → B → C → D ] = λ [ → a → C → D ] + μ [ → B → C → D ] , Then λ + μ = - Mathematics

Question

If $[2 \vec{a} + 4 \vec{b} \vec{c} \vec{d}] = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}],$ then λ + μ =

Options

6
-6
10
8

MCQ

Sum

Solution

We have

$[2 \vec{a} + 4 \vec{b} \vec{c} \vec{d}] = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}]$

$\Rightarrow [(2 \vec{a} + 4 \vec{b}) \times \vec{c}] . \vec{d} = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}] (By definition of scalar triple product)$

$\Rightarrow [(2 \vec{a} \times \vec{c}) + (4 \vec{b} \times \vec{c})] . \vec{d} = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}]$

$\Rightarrow (2 \vec{a} \times \vec{c}) . \vec{d} + (4 \vec{b} \times \vec{c}) . \vec{d} = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}]$

$\Rightarrow [2 \vec{a} \vec{c} \vec{d}] + [4 \vec{b} \vec{c} \vec{d}] = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}]$

$\Rightarrow 2 [\vec{a} \vec{c} \vec{d}] + 4 [\vec{b} \vec{c} \vec{d}] = λ [\vec{a} \vec{c} \vec{d}] + μ [\vec{b} \vec{c} \vec{d}] (∵ [λ \vec{a} \vec{b} \vec{c}] = λ [\vec{a} \vec{b} \vec{c}] for any scalar λ)$

Comparing both sides, we get

$λ = 2$

$μ = 4$

$∴ λ + μ = 2 + 4 = 6$

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Scalar Triple Product of Vectors

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