Let ` vec a= hat i+ hat j` and `vec b=2 hat i- hat k ,` then the point of intersection of the lines ` vec rxx vec a= vec bxx vec a` and ` vec rxx vec b= vec axx vec b` is a. `(3,-1,1)` b. `(3,1,-1)` c. `(-3,1,1)` d. `(-3,-1,-1)`

Options

A. (3,-1,1)
B. (3,1,-1)
C. (-3,1,1)
D. (-3,-1,-1)

Answer

b

Explanation

Let `vecrxxveca= vecbxxveca`
` rArr " "(vecr-vecb)xxveca= vec0 rArr vecr= vecb+ tveca`
Similarly, other line `vecr= veca+ kvecb`, where `t and k` are scalars.
Now `veca+ kvecb= vecb+ tveca`
`rArr" "t=1, k=1 " "` (equating the coefficients of `veca and vecb` )
`therefore " "vecr= veca+ vecb= hati+ hatj+ 2hati-hatk= 3hati+ hatj-hatk`
i.e., `(3, 1, -1)`.