Question

A line `L_(1)` passes through the point `3hati` and is parallel to the vector `-hati+hatj+hatk` and another line `L_(2)` passes through `hati+hatj` and is parallel to the vector `hati+hatk`, then point of intersection of the lines iss
A. `hati+2hatj+hatk`
B. `2hati+hatj+hatk`
C. `hati-2hatj-hatk`
D. `hati-2hatj+hatk`

Answer 1

Correct Answer - B
Equation of `L_(1)` is
`r=3hati+lambda(-hati+hatj+hatk)=(3-lambda)hati+lambdahatj+lambdahatk`
Equation of `L_(2)` is
`r=(hati+hatj)+mu(hati+hatk)=(1+mu)hati+hatj+muhatk`
For point of intersection, we get
`lambda=mu=1`
`therefore r=2hati+hatj+hatk`