Statement 1: Let `veca, vecb, vecc` be three coterminous edges of a parallelopiped of volume 2 cubic units and `vecr` is any vector in space then
`|(vecr.veca)(vecbxxvecc)+(vecr.vecb)(veccxxveca)+(vecc.vecc)(vecaxxvecb|=2|vecr|`
Statement 2: Any vector in space can be written as a linear combination of three non-coplanar vectors.

Options

A. 1
B. 2
C. 3
D. 4

Answer

A

Explanation

Clearly, statement -2 is true
We have `|[(veca, vecb, vecc)]|=2`
`:.[(vecaxxvecb, vecbxxvecc, veccxxveca)]=[(veca,vecb, vecc)]^(2)=4!=0`
`implies vecaxxvecb, vecbxxvecc, veccxxveca` are non coplanar.
Using statement -2 we have
`vecr=x(vecaxxvecb)+y(vecbxxvecc)+z(veccxxveca)`...............i
Taking dot products with `veca,vecb` and `vecc` respectively, we get
`vecr.veca=y[(veca, vecb, vecc)],vecr.vecb=z[(veca, vecb, vecc)]` and `vecr.vecc=x[(veca,vecb,vecc)]`
Substituting the values of `x,y` in i we get
`vecr[(veca, vecb,vecc)]=(vecr.veca)(vecbxxvecc)+(vecr.vecb)(veccxxveca)+(vecr.vecc)(vecaxxvecb)`
`implies|(vecr.veca)(vecbxxvecc)+(vecr.vecb)(veccxxveca)+(vecr.veca)(vecaxxvecb)|`
`=|vecr||(veca, vecb,vecc)|`
`implies|(vecr.veca)(vecbxxvecc)+(vecr.vecb)(veccxxveca)+(vecr.vecc)(vecaxxvecb)|=2|vecr|`