The vectors a - - - 2- - k- b - - - - c - - are such that a-b-c - -a - -b- then the value of - - - is

⇒ (a×b)×c = λa + μb ⇒ (a . c)b (b . c)a = λa + μb λ = (b . c), μ= (a . c) Hence, λ = (î + ĵ).( î) = 1 μ = (î +2 ĵ k̂ ).( î)= 1 Therefore λ + μ = 1 1 = ...

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Standard XII

Mathematics

Angle between Two Vectors

The vectors a...

Question

The vectors $\to a =^i + 2^j -^k, \to b =^i +^j, \to c = -^i$ are such that $(\to a \times \to b) \times \to c = λ \to a + μ \to b,$ then the value of $λ + μ$ is

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Solution

The correct option is B $0$
$\Rightarrow (\to a \times \to b) \times \to c = λ \to a + μ \to b$
$\Rightarrow (\to a . \to c) \to b - (\to b . \to c) \to a = λ \to a + μ \to b$
$λ = - (\to b . \to c), μ = (\to a . \to c)$
Hence,
$λ = - (^i +^j) . (-^i) = 1$
$μ = (^i + 2^j -^k) . (-^i) = - 1$
Therefore
$λ + μ = 1 - 1 = 0$

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Similar questions

If \to a =^i +^j +^k, \to b =^i +^j, \to c =^i and (\to a \times \to b) \times \to c = λ \to a + μ \to b,

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Angle between Two Vectors

Standard XII Mathematics

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The vectors →a=^i+2^j−^k,→b=^i+^j,→c=−^i are such that (→a×→b)×→c=λ→a+μ→b, then the value of λ+μ is

The correct option is B 0⇒(→a×→b)×→c=λ→a+μ→b ⇒(→a.→c)→b−(→b.→c)→a=λ→a+μ→b λ=−(→b.→c), μ=(→a.→c) Hence, λ=−(^i+^j).(−^i)=1 μ=(^i+2^j−^k).(−^i)=−1 Therefore λ+μ=1−1=0

The vectors $\to a =^i + 2^j -^k, \to b =^i +^j, \to c = -^i$ are such that $(\to a \times \to b) \times \to c = λ \to a + μ \to b,$ then the value of $λ + μ$ is