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Consider the line L(t)=(3,4t−3,3). Perpendicular, Parallel, Neither to?
Consider the line L(t)=(3,4t-3,3). Then:
L is _____ to the plane 6y=6
L is _____ to the plane 8y=-16
L is _____ to the plane 8z=40
L is _____ to the plane 3x-3y-5z=-5
The answers can be either perpendicular, parallel, or neither for each question.
2 Answers
- JeroonkLv 68 years agoFavorite Answer
L(t) has a direction vector (0,4,0), as it can be written as: L(t) = (3,−3,3) + t∙(0,4,0).
* The plane 6y = 6 has a normal vector parallel to (0,6,0).
The normal is parallel to L, thus the plane is perpendicular to L.
* The plane 8y = −16 has a normal vector parallel to (0,8,0).
The normal is parallel to L, thus the plane is perpendicular to L.
* The plane 8z = 40 has a normal vector parallel to (0,0,8).
The normal is perpendicular to L, thus the plane is parallel to L.
* The plane 3x − 3y − 5z = −5 has a normal vector parallel to (3,−3,−5).
The normal is neither parallel nor perpendicular to the line, thus the plane isn't also.
