Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 3 (b)
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Question B 3 (b) 

Line and plane shown in the figure are given in parametric form as where

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! 
Hint 

The form given in the question is actually vector form of a plane: given any point in the plane, and two nonparallel vectors in the plane, then any other point can be written as .
Here, are parameters. We already know one direction in the plane, i.e., the direction of line , the target is to find one point and another direction in the plane. 
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

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Please rate my easiness! It's quick and helps everyone guide their studies. According to definition of parametric form of plane, we will write as Our purpose is to find . As hinted, since contains , we can set . Now let's look for , in fact, we know is perpendicular to , it not hard to see that the normal vector of is on the plane , this normal vector is going to be . We are already given the parametric form of , by the standard method of finding normal vector, we need to apply cross product to two nonparallel vectors in the plane . The cross product is given by the formal determinant so we set It remains to compute . Notice that the origin is in , so it is in since contains . Hence we can take . It follows that a parametric form of the equation for is 