Differentiate `y = cos (3x^2 + 2)`.

Answer

` a to q' b to s ; c to p ; d to r`

Explanation

a. `|veca + vecb + vecc| = sqrt6`
`Rightarrow veca^(2)+vecb^(2)+vecc^(2)+ 2veca.vecb+ 2vecb.vecc + 2vecc.veca=6`
`|veca|=1`
b. `veca` is perpendicular to `vecb + vecc`
`Rightarrow veca. vecb+veca.vecc=0`
`vecb` is perpendicular to `veca +vecb`
`Rightarrow vecb.vecb + vecb.vecc=0`
`vecc` is perpendicular to `veca + vecb`
`Rightarrow vecc. veca + veca .vecc.vecb=0`
Form (i), (ii) and (iii), we get
`veca.vecb = vecb .vecc =vecc.veca=0`
` |veca+vecb +vecc|=7`
=c. `(veca.vecc)(vecb.vecd)-(vecb.vecc)(veca.vecd)=21`
d. we know that `[veca xx vecb vecb xx vecc vecc xx veca] = [veca vecb vecc]^(2)`
`and [veca vecbvecc]^(2)=|{:(veca.veca,veca.vecb,veca.vecc),(vecb.veca,vecb.vecb,vecb.vecc),(vecc.veca,vecc.vecb,vecc.vecc):}|=|{:(4,2,2),(2,4,2),(2,2,4):}|`
32
`[veca vecb vecc] = 4sqrt2`