Let `veca=hati+hatj and vecb=2hati-hatk.` Then the point of intersection of the lines `vecrxxveca=vecbxxveca and vecrxxvecb=vecaxxvecb` is (A) `(3,-1,10` (B) `(3,1,-1)` (C) `(-3,1,1)` (D) `(-3,-1,-10`

Options

A. `hati-hatj+hatk`
B. `3hati-hatj+hatk`
C. `3hati+hatj-hatk`
D. `hati-hatj-hatk`

Answer

c

Explanation

`vecrxxveca =vecbxxveca or (vecr-vecb) xxveca=0`
`vecrxxvecb = vecaxx vecb or (vecr-veca) xx vecb=0`
`if vecr=xhati + yhatj +zhatk`then
`|{:(hati,hatj,hatk),(x-2,y,z+1),(1,1,0):}|=0and|{:(hati,hatj,hatk),(x-1,y-1,z),(2,0,-1):}|=0`
`Rightarrow z+1=0,x-y=2`
`and y-1=0,x-1+2z=0`
`Rightarrow x=3,y=1,z=-1`