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 issA. `hati+2hatj+hatk`B. `2hati+hatj+hatk`C. `hati-2hatj-hatk`D. `hati-2hatj+hatk`

A line `L_(1)` passes through the point `3hati` and is parallel to the

1.	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 issA. `hati+2hatj+hatk`B. `2hati+hatj+hatk`C. `hati-2hatj-hatk`D. `hati-2hatj+hatk`
Answer» 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`

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