Question

If $straight a with rightwards arrow on top comma straight b with rightwards arrow on top comma space straight c with rightwards arrow on top$ are non -coplanar vector λ is a real number then

$open square brackets straight lambda left parenthesis straight a with rightwards arrow on top space plus straight b with rightwards arrow on top right parenthesis space straight lambda squared space straight b with rightwards arrow on top straight lambda space straight c with rightwards arrow on top close square brackets space equals space open square brackets straight a with rightwards arrow on top space straight b with rightwards arrow on top space plus straight c with rightwards arrow on top straight b with rightwards arrow on top close square brackets space for$

exactly one value of λ

no value of λ

exactly three values of λ

exactly two values of λ

Solution

no value of λ

open square brackets straight lambda space left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis close square brackets space straight lambda squared space straight b with rightwards arrow on top space straight lambda space straight c with rightwards arrow on top right square bracket space equals space open square brackets straight a with rightwards arrow on top straight b with rightwards arrow on top plus straight c with rightwards arrow on top space straight b with rightwards arrow on top close square brackets open vertical bar table row straight lambda straight lambda 0 row 0 cell straight lambda squared end cell 0 row 0 0 straight lambda end table close vertical bar open vertical bar table row 1 0 0 row 0 1 1 row 0 1 0 end table close vertical bar rightwards double arrow space straight lambda to the power of 4 space space equals negative 1 Hence space no space real space value space of space straight lambda

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