The vector equation of the line passing through a point with position vector `2hati-hatj+hatk` and parallel to the line joining the points with position vectors `-hati+4hatj+hatk and hati+2hatj+2hatk ` is `r=(2hati-hatj+hatk)+lambda(xhati-2hatj+hatk)` where x is equal to

Options

A. 0
B. 1
C. 2
D. 4

Answer

C

Explanation

Let `a=2hati-hatj+hatk, b=-hati+4hatj+hatk`
`and c=hati+2hatj+2hatk`
Then the equation of the line will be `r=a+lambda(c-b)`
`=(2hati-hatj+hatk)+lamda{(hati+2hatj+2hatk)-(-hati+4hatj+hatk)}`
`=(2hati-hatj+hatk)+lambda(2hati-2hatj+hatk)`
But the original equation of a line is `r=(2hati-hatj+hatk)+lambda(2hati-2hatj+hatk)`
Hence, x=2.