Science:Math Exam Resources/Courses/MATH152/April 2012/Question 01 (a)
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q2 (a) • Q2 (b) • Q2 (c) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 (a) • Q4 (b) • Q4 (c) • Q4 (d) • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q6 (c) • Q6 (d) • Q6 (e) • Q7 (a) • Q7 (b) • Q7 (c) • Q7 (d) • Q8 (a) • Q8 (b) • Q8 (c) • Q8 (d) •
Question 01 (a) 

Let be the plane perpendicular to and through the point P=[1,1,0]. Let be the plane through the points A=[1,0,0], B=[1,2,1], C=[0,1,1]. By L, we denote the line of intersection of and . Determine the planes and in equation form. 
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 

Recall that in order to define a plane, we need a normal vector and a point on 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 

Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. For the first plane we know that its normal is as this is provided. We’re also told that the point is on the plane. Using this information we can easily get the equation for via We have to work a little harder to get the equation for the plane of . However, notice we are given three points that are on the plane. Let be the vector connecting A to B and be the vector connecting A to C. We therefore have that, We know that the normal to the plane is orthogonal to all vectors on the plane. Since and are on the plane, if we take their cross product, we will get a vector orthogonal to both and and thus be orthogonal to the plane. Call this vector , We can now use this normal vector along with any point on the plane to get the equation. We will use , Therefore, we get that the equation for is while the equation for is . 