Find the area of a parallelogram whose adjacent sides are determined by the vectors `vec a = 2 hat i - hat j +3 hat k` and `vec b = hat i- 6 hat j+ 4 hat k ` is

Answer

C

Explanation

a. Multiply the given expression by 2 and rewrite it as
`(2x-3y)^(2)+(3y-4z)^(2)+(4z-2x)^(2)=0`
`rArr2x=3y=4z`
So, `1/x,1/y,1/z` are in A.P.
Therefore,x,y,z are in H.P..
b. Since, `21=1^(2)+2^(2)+4^(2),` we have
`(1^(2)+2^(2)+4^(2))(x^(2)+y^(2)+z^(2))=(x+2y+4z)^(2)`
`rArr(1^(2)+2^(2)+4^(2))(x^(2)+y^(2)+z^(2))-(1-xxx+2xxy+4xxz)^(2)=0`
`rArr(y-2x)^(2)+(2z-4y)^(2)+(4z-x)^(2)=0`
This is possible only when
y-2=0,2z-4y=0 and 4x-z=0
or `y/x=2,z/y=2,z/x=4`
`rArry/x=z/yrArry^(2)=zx`
Therefore,x,y,z are in G.P..
c. We have `216x^(3)+64y^(3)+27z^(3)=72xyz`
`therefore (6x+4y+3z)(36x^(2)+16y^(2)+9z^(2)-24xy-12yz-18xz)=0`
Since x,y,z `gt0`
`36x^(2)+16y^(2)+9z^(2)-24xy-12yz-18xz=0`
`therefore72x^(2)+32y^(2)+18z^(2)-48xy-24yz-36xz=0`
`rArr(6x-4y)^(2)+(4y-3z)^(2)+(3z-6x)^(2)=0`
`rArr6x=4y=3z=lamda` (say)
`rArrx=lamda/6,y=lamda/4,z=lamda/3`
`rArrx+z=lamda/6+lamda/3=lamda/2=2y`
Therefore,x,y,z are in A.P..
d. `ax^(2)+2px+b=0` has root -1.
`therefore a-2p+b=0`
`rArra+b=2p`
So,a,p,b are in A.P..
`ax^(2)+2qx+b=0` has equal roots.
`therefore4q^(2)-4ab=0`
`rArrq^(2)=ab`
Therefore a,q,b are in G.P..
Line `x/a+y/b=2/r` passes through the point (1,1).
`therefore1/a+1/b=2/r`
Therefore,a,r,b are in H.P..
Thereforep,q,r are in G.P...