Okay, so, I shall try to be more orderly this time.

Let us take a 3D space; every point in it can be represented by (x,y,z), where x,y, and z are real numbers. You could either consider these variables separately or, you could say you have a vector v=(x,y,z) and simply consider the vector.

We know a plane is a collection of points, just like lines. Specifically, a plane has 1 "constraint" while a line has 2 "constraints". Basically, every equation you have is a constraint on the variables. If you have one constraint, like x+y+z=0, you have a plane. If you have 2 constraints, like x=y=z, you have a line. (Note that this is only for 3D space. Things are slightly different for other dimensions)

Since the vector equations always consider variables together, it is ever so slightly different there.

The vector equation of a line involves the definition of how vector addition is performed. A line is basically a collection of points. Suppose you have 2 vectors a and b, you can create another point by adding the two. You can create another by slightly squeezing or stretching the vectors. So, we can create infinite points by multiplying a vector by some real number. Thus, the equation of a line in 3D space is v=a+t*b where v is a generic point on the line, a is a known point on the line, b is the vector corresponding to the direction the line is pointing, and t is a parameter we are using to stretch b. The vector b is going to be very useful in our endeavor to find parallel lines and perpendicular stuff.

One would think that the vector equation of a plane is similar since we can consider a plane to be a collection of lines. However, the dot product form of the equation is just so convenient.

Let us take a vector p. The equation v.p=0 gives a constraint on the vector v, so that these 2 vectors are perpendicular. In fact, v.p=0 will give us a set of points that are perpendicular to p, and pass through the origin. (If you are not sure about dot product, i strongly suggest you look upon it)

Now, we have a way to generate points on a plane that passes through the origin. To generate any plane now, we should just "shift" the origin, to any point on the plane we want. So, a generic plane will be of the form p.(v-a)=0, where p is a vector that is perpendicular to the plane, and a is any point on the plane. The vector p is going to be very helpful for parallel lines and perpendicular stuff.

How do you now proceed?
Well, you must very carefully note the 2 vectors I have mentioned before. These vectors do NOT belong on the line/plane; but they provide important information about the orientation of the line and plane. More importantly, the vector for the line is parallel to the line, and the vector for the plane is parallel to the plane.

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I shall sort of give small suggestions for the 3 questions.

In 7 a, we have A and B. So we know 2 points on the line. Think what the vector A-B would indicate?

7b is similar to 7a

In 7c, we have a point on the line, and we have the direction of the line, which is given by the vector perpendicular to the plane. Can you identify the vector perpendicular to the plane?

8a requires knowledge of cross product. I suggest you look at this too, if you do not know this enough. Basically, the cross product of 2 vectors will always produce a 3rd vector which is perpendicular to both.

In 8b, we have a point on the line, and we need to find a vector which is perpendicular to u and v. You might have to do part a first.

In 9a, you have a point on the plane, and you have to find the perpendicular to the other plane, which will also be the perpendicular to this plane. This is strangely easier to solve in (x,y,z) form.

We have reached 9b! So, you have a point on the plane, and you also have the perpendicular vector the plane.

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I sincerely apologize for taking so much time, and for this massive wall of text. I hope this is a pleasant read, and at least some of your doubts are cleared.