Find the derivative of `y=2sin3x` .

Answer

`atop, q,r,s;b to p,q;cto p,r;dto r`

Explanation

a. Given equations are consistent if
`(hati + hatj) + lambda(hati + 2hatj -hatk)`
`= (hati + 2hatj) + mu (-hati + hatj + ahatk)`
` Rightarrow 1+lambda = 1 - mu, 2lambda=2+mu, - lambda= amu`
` Rightarrow lambda = 1//3 and mu = -1//3`
`Rightarrow a=1`
b. `veca = lambdahati - 3hatj - hatk`
`vecb = 2 lambdahatj -hatk`
Angle between `veca and vecb` is acute. therefore,
`veca.vecbgt0`
`Rightarrow 2lambda^(2)-3lambda+1gt0`
`or (2lamda-1) (lambda-1)gt0`
`or lambdain(-oo,1/2)cup ( 1,oo)`
Also`vecb` makes an obtuse angle with the axis . therefore,
`vecb.hati lt0 Rightarrow lambdalt0`
`vecb.hatj lt0 Rightarrow lambdalt0`
Combining these two , we get `lambda = -4,-2`
C. if vectors ` 2hati - hatj + hatk, hati + 2hatj + (1+a)hatk and 3hati + ahatj + 5hatk` are coplanar, then
`|{:(2,-1,1),(1,2,1+a),(3,a,5):}|=0`
`or a^(2)+2a-8=0`
`or (a+4)" (a-2)=0`
`or a=-4,2`
d. `vecA = 2hati+ lambdahatj + 3hatk`
` B = 2hati + lambdahatj + hatk`
` C = 3hati + hatj + 0.hatk`
` vecA+ lambda vecB= 2(1 + lambda)hati + (lambda + lambda^(2))hatj + (3 + lambda)hatk`
Now, `(vecA + lambdavecB) bot vecC`. therefore,
`(vecA + lambdavecB) . vecC =0`
`or 6(1+ lambda) + (lambda + lambda^(2)) +0=0`
`or lambda^(2) + 7lambda + 6=0`
`or (lambda + 6) (lambda +1 ) =0`
`or lambda = -6, =-1`
` Rightarrow |2lambda|= 12,2`