Did you read my tips on how to study and learn Math 1300?  If not, here is a link to those important suggestions:

Did you see my tips for Assignment 1? Click here.

Here is a link to the actual assignment, in case you don't have it handy:

You need to study Lesson 10 (Lines and Planes) from my Linear Algebra & Vector Geometry book to prepare for this assignment.  I think you should find this assignment fairly straightforward if you do thoroughly study and do all the Practise Problems I give you in Lesson 10.

Don't have my book?  You can download a free sample of my book and audio lectures containing Lessons 1, 2 and 9:

Similar to my Lesson 10, question 3.  Also look at my question 7 for an example of how to write point-normal form.

Similar to my Lesson 10, question 4.  By the two-point vector form in part (b), they mean write the vector form of the line.  You also have the option of leaving the vector v unsimplified.  Which is to say, you can just leave v as p-q or q-p (whichever you chose to use), showing them the two points that are being subtracted to compute the vector.

The dihedral angle is simply the angle made by the two planes.  It is also the angle between the two planes' normal vectors.  Use your formula to find the sine or cosine of the angle between two vectors (cosine is quicker to find).  You will then have to use your calculator to compute the actual angle.  Make sure your calculator is in Degree mode (you should see Deg written in small type somewhere in the top of the screen), and use Inverse Cosine or Inverse Sine, depending on whether you chose to compute cosine or sine.  That usually means pressing "2nd F" or "Shift", then "Cos" or "Sin".

Similar to the previous question, but now you are computing the angle between the two parallel vectors from the given lines.  Note that an acute angle is less than 90 degrees.  If you get an angle greater than 90, simply subtract your answer from 180.  You can understand why if you visualize a horizontal line.  Now draw a line starting from that horizontal line but heading up to the northeast at an angle of say 45 degrees.  Those two lines are making an angle of 45 degrees on the right side, but that also means, if you measure from the left side of the northeast line, you have an angle of 180-45=135 degrees.  So, depending on which side you are measuring the angle between two lines, you can get either an acute angle or an obtuse angle.

Part (a) is very similar to my question 10 (a).

Part (b) is quite challenging.  The key here is to find a vector which takes you from point P to point R on the line, but which is orthogonal to the line.  Since you know that point R is on the line, let R=(x, y, z) be  point on the line, and replace x, y and z with the parametric equations for the line.  That way, there is only one unknown in R, the parameter t.  But, you know that arrow PR is orthogonal to the line, so you can set up a dot product equation to solve t.  Once you have solved t, you can compute the arrow PR exactly.  Since arrow PR is orthogonal to the line and takes you exactly from point P to point R on the line, it follows that the length of arrow PR is equal to the distance from P to the line.

Do not ask for any additional assistance on this problem, as I fear that I have already told you too much.

Again, quite a challenging question.

Part (a) is best solved by finding the plane that contains both lines.  That is similar to my question 7.

Part (b) is a repeat of question 4 above.

Part (c) is a repeat of question 5 (b) above.

Part (a) is very similar to my Practise Problem 19(a).

Part (b) is very similar to my Lecture Problem 5.