If →a=^i+^j+^k,→b=^i+^j,→c=^i and (→a×→b)×→c=λ→a+μ→b, then the value of λ+μ is
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Solution
Given,(→a×→b)×→c=λ→a+μ→b ⇒(→a⋅→c)→b−(→b⋅→c)→a=λ→a+μ→b
By comparing, we get ⇒λ=−(→b⋅→c),μ=(→a⋅→c)
On substituting →a,→b,→c, we have, ⇒λ=−((^i+^j)⋅(^i)),μ=((^i+^j+^k)⋅(^i)) ⇒λ=−1,μ=1⇒λ+μ=−1+1=0