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Question
Find the vector and the cartesian equations of the plane containing the point `hati + 2hatj - hatk` and parallel to the lines `vecr = (hati + 2hatj + 2hatk) + s(2hati - 3hatj + 2hatk)` and `vecr = (3hati + hatj - 2hatk) + t(hati - 3hatj + hatk)`
Solution
Since, the plane is parallel to the given lines, the cross product of the vectors `2hati - 3hatj + 2hatk` and `hati - 3hatj + hatk` will be a normal to the plane
`(2hati - 3hatj + 2hatk) xx (hati - 3hatj + hatk) = |(hati, hatj, hatk),(2, -3, 2),(1, -3, 1)| = 3hati - 3hatk`
The vector equation of the plane is `vecr.(3hati - 3hatk) = (hati + 2hatj - hatk).(3hati - 3hatk)` or `vecr.(hati - hatk)` = 2 and the cartesian equation of the plane is x – z – 2 = 0
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