If `veca and vecb` are two unit vectors perpenicualar to each other and `vecc= lambda_(1)veca+ lambda_2 vecb+ lambda_(3)(vecaxx vecb),` then which of the following is (are) true ?A. `lambda_(1)= veca.vecc`B. `lambda_(2)= |vecb xx vecc|`C. `lambda_(3)= |(vecaxx vecb|xxvecc|`D. `lambda_(1)veca + lambda_(2) vecb + lambda_(3) (vecaxxvecb)`

If `veca and vecb` are two unit vectors perpenicualar to each other an

1.	If `veca and vecb` are two unit vectors perpenicualar to each other and `vecc= lambda_(1)veca+ lambda_2 vecb+ lambda_(3)(vecaxx vecb),` then which of the following is (are) true ?A. `lambda_(1)= veca.vecc`B. `lambda_(2)= \|vecb xx vecc\|`C. `lambda_(3)= \|(vecaxx vecb\|xxvecc\|`D. `lambda_(1)veca + lambda_(2) vecb + lambda_(3) (vecaxxvecb)`
Answer» Correct Answer - a,d Given `vecc= lambda_(1)veca+lambda_(2)vecb + lambda_(3)(vecaxxvecb)` `and veca. vecc= lambda_(1),vecc.vecb=lambda_(2)` ` and vecc. (veca xx vecb) \|veca xx vecb\|^(2) lambda_(3)` ` (1.1 sin 90^(@))^(2) lambda_(3) = lambda_(3)` Hence, ` lambda_(1), + lambda_(2), + lambda_(3) = (veca + vecb + veca xx vecb) . vecc`

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