LINES AND PLANES CHRIS JOHNSON

LINES AND PLANES CHRIS JOHNSON Abstract. In this lecture we derive the equations for lines and planes living in 3-space, as well as define the angle between two non-parallel planes, and determine the distance from a point to a plane using properties of vector projections. 1. Lines 1.1. Vector, Parametric, and Symmetric Equations. Recall that in two dimensions, to specify a line you need two pieces of information: a point the line passes through, and the slope of the line. The slope of the line really just tells us the direction the line points in. In three dimensions we also need two pieces of information to determine a line; if we want to give the equation of a line, then we need to know a point on the line and the direction of the line. While in two dimensions we could use a single number (the slope) to determine the direction of the line, in three dimensions we ll use vectors. (Really, the slope in two dimensions determines a vector: a slope of m is the same as a vector 1, m. Since the first component of this vector is always 1, it s only the second component that matters.) So let s say, to keep things easy, we want the equation of a line through the origin. Let s suppose the vector v = a, b, c points along the line. This means that if a point P = (x, y, z) lives on the line, we could perform the scalar multiplication t v to stretch v out enough so that its tip was at P. That is, the displacement vector r = OP = x, y, z is related to the direction vector v = a, b, c by the following equation: r = t v This is the vector equation of the line through the origin in direction v. Now let s suppose we want our line to pass through some point P 0 = (x 0, y 0, z 0 ) instead of the origin. Let r 0 denote the displacement vector r 0 = OP 0 = x 0, y 0, z 0. All we need to do is take our line through the origin, and translate it along the vector r 0 so that the line passes through P 0. So, if another point P = (x, y, z) is to be on the line, 1

2 CHRIS JOHNSON letting r denote the vector x, y, z, we have the following equation: r = r 0 + t v. This is the vector equation of the line through P 0 in the direction of v. If we write this out in terms of components we have x, y, z = x 0, y 0, z 0 + ta, tb, tc = x 0 + ta, y 0 + ta, z 0 + ta. Two vectors are equal if and only if their components are equal, so by equating components we actually have three equations: x =x 0 + ta y =y 0 + tb z =z 0 + tc. Notice that on the right-hand side the values a, b, c, x 0, y 0, z 0 are all fixed: they don t change once we say what the point P 0 is and where the vector v points. The t is the only thing on the right-hand side that can change. Thus the values x, y, z on the left-hand side are functions of t. So really the above equations should be written as x(t) =x 0 + ta y(t) =y 0 + tb z(t) =z 0 + tc. These three equations form the parametric equations of the line through P 0 = (x 0, y 0, z 0 ) in the direction of v = a, b, c. Example 1.1. Find the vector and parametric equations for the line through the point (3, 4, 1) in the direction of 1 /2, 1, 3. The vector equation is x, y, z = 3 + t2, 4 t, 1 + 3t. To get the parametric equations, just equate the components to get: x(t) =3 + t 2 y(t) = 4 t z(t) =1 + 3t.

LINES AND PLANES 3 Now, assuming none of a, b, or c is zero, we could solve each of our equations above for t to get t = x x 0 a t = y y 0 b t = z z 0. c Since each of the things on the right-hand side equals t, these quantities are all equal. Thus we have x x 0 a = y y 0 b = z z 0. c These are the symmetric equations of the line through P 0 = (x 0, y 0, z 0 ) in the direction of v = a, b, c. Again, notice these equations only make sense if none of a, b, or c equals zero. (If one of them did equal zero, we d have division by zero.) In our example above, the symmetric equations are x 3 1/2 = y + 4 1 = z 1 3. These symmetric equations are nice because they give us an easy way to determine if a point is on the line or not. Example 1.2. Are the points (4, 6, 7) and ( 5, 0, 14) on the line through (3, 4, 1) in the direction of 1 /2, 1, 3? This is our line from the last example, so the symmetric equations are given above. We plug in the coordinates of each point to see if we have equality or not. In the case of (4, 2, 7) we have: 4 3 1 /2 = 6 + 4 1 = 7 1 3. Each of these expressions is 2, so we have equality, and the point (4, 2, 7) is on the line. Plugging in ( 5, 0, 14) we have 5 3 1 /2 = 8 + 0 1 12 1. 3 The first two expressions equal 4, while the last one equals 5. Thus the point ( 5, 0, 14) is not on the line.

4 CHRIS JOHNSON 1.2. Parallel and Skew Lines. Just as in two dimensions, two lines are parallel if they point in the same direction. In the case of two dimensions this meant that the two lines had the same slope. In the case of three dimensions it means that the two direction vectors are scalar multiples of one another. For example, the lines x 2 3 = y + 1 1 = z 1 2 and x + 1 9 = y 3 = z 2 6 are parallel. These lines have respective direction vectors 3, 1, 2 and 9, 3, 6. The second one is 2 times the first. In the case of two dimensions, two lines are parallel if and only if they never touch. This is not the case in three dimensions. Consider the lines with the following parametric equations. x 1 (t) =3 t y 1 (t) =1 + 2t z 1 (t) = 1 x 1 (t) =4 t y 1 (t) =2 + t z 1 (t) =1 Let s call these lines L 1 and L 2. From the equations we can already tell that the first line passes through the point (3, 1, 1) in the direction of v 1 = 1, 2, 0, and the second line passes through the point (4, 2, 1) in the direction of v 2 = 1, 1, 0. Notice that L 1 is contained in the plane z = 1, and L 2 is in the plane z = 1. Thus there s no possible way these lines can ever intersect. However, their direction vectors are not scalar multiples of one another: there is no λ R that makes v 1 = λ v 2. So these lines are not parallel, nor do they intersect. When this happens we say that the lines are skew. 2. Planes 2.1. Equations of Planes. Now we consider planes. Of course we ve already seen several examples of equations of planes in previous lectures, but everything we looked at before was a very special case (e.g., the coordinate planes). We want to be able to determine the equation

6 CHRIS JOHNSON where d = (n 1 x 0 + n 2 y 0 + n 3 z 0 ). This is a linear equation, and just like the symmetric equations for lines, they give us a really easy way to determine if a point is on a plane or not. Example 2.1. Find the linear equation of the plane which contains the point ( 2, 3, 5) and whose normal vector is 7, 2, 3. Is the point (1, 3, 4) on this plane? To get the linear equation: 7(x + 2) + 2(y 3) + 3(z + 5) = 0 = 7x + 14 + 2y 6 + 3z + 15 = 0 = 7x + 2y + 3z + 23 = 0 Now we check if (1, 3, 4) is on the plane or not: So the point is not on the plane. 7 1 + 2 3 + 3 ( 4) + 23 = 24 0 2.2. Distance to a Plane. Recall that if P = (x, y, z) is some point in 3-space, we can measure the distance from that point to each of the coordinate planes by first projecting onto the plane, and then measuring the distance from the projection to our initial point. We can do the exact same thing but for other planes. Let s say that we re given a point P 1 = (x 1, y 1, z 1 ) and we want to measure the distance from this point to the plane ax + by + cz = d. What we want to do is find the point P 0 = (x 0, y 0, z 0 ) on the plane so that the displacement vector P 0 P 1 sticks orhotogonally out of the plane. That is, P 0 P 1 = λ n where n = a, b, c is the normal vector of the plane. We then need to figure out how long this displacement vector is. Let s suppose that P = (x, y, z) is any other point in this plane. Then if r = OP and r 0 = OP 0, we know n ( r r 0 ) = 0. Let s = P P 1. What we want to do is project s onto n, and measure the length of that vector that is, we want the absolute value of comp n s. In components, s = x 1 x, y 1 y, z 1 z. Thus the distance from

LINES AND PLANES 7 P 1 = (x 1, y 1, z 1 ) to the plane ax + by + cz = d is given by distance = comp n s = s n n = x 1 x, y 1 y, z 1 z a, b, c a2 + b 2 + c 2 = a(x 1 x) + b(y 1 y) + c(z 1 z) a2 + b 2 + c 2 = ax 1 + by 1 + cz 1 (ax + by + cz) a2 + b 2 + c 2 Now, since our point P = (x, y, z) lives on the plane, it satisfies the equation ax + by + cz = d, so the above becomes distance = ax 1 + by 1 + cz 1 d a2 + b 2 + c 2. Caution: The formula that appears in your book is slightly different than what s written here because your book assumes the equation of the plane is written as ax+by+cz+d = 0, which becomes ax+by+cz = d: our d is the negative of the d in the book. So you have to be a slightly cautious when using these formulas. Example 2.2. Find the distance from the point (1, 2, 4) to the plane 3x y + 5z = 6. Here the normal vector to plane is n = 3, 1, 5, d = 6, and (x 1, y 1, z 1 ) = (1, 2, 4). Plugging these values into our formula above, distance = ax 1 + by 1 + cz 1 (ax + by + cz) a2 + b 2 + c 2 3 2 20 6 = 9 + 1 + 25 = 25 35 4.23. 2.3. The Angle Between Two Planes. We ll say that two planes are parallel if their normal vectors are scalar multiples of one another.

8 CHRIS JOHNSON Example 2.3. Are the two planes below parallel? 3x + 6y 7z =13 x 2y + 7 3 z =0 From the equations, we can easily pull of the normal vectors: 3, 6, 7 and 1, 2, 7 /3. These vectors are obviously scalar multiples: Thus the planes are parallel. 3, 6, 7 = 3 1, 2, 7 /3. Example 2.4. Are the two planes below parallel? 2x y + z =3 4x y + 2x =6 Again, we pull off the normal vectors: 2, 1, 1 and 4, 1, 2. Notice these vectors can t be scalar multiples of one another: we d need to multiply the first vector by 2 to get a 4 in the first component of the second vector, but then this would give a 2 in the second component. Hence the planes are not parallel. If two planes are not parallel, we can talk about the angle at which those planes meet. Of course, geometrically, this is exactly what you think it should be, but how do you go about calculating this angle? It s relatively clear that the normal vectors to the planes meet at the same angle as the planes themselves, and we can calculate the angle between the vectors by using dot products. That is, if our planes have normal vectors n = n 1, n 2, n 3 and m = m 1, m 2, m 3, then the angle between the planes is the same as the angle between these vectors, which we know is just ( ) m n θ = cos 1. m n Example 2.5. What s the angle between the planes given by the equations below? 2x + 3y z =0 3x + y + 4z =12 Our normal vectors are 2, 3, 1 and 3, 1, 4. Notice these vectors aren t scalar multiples of one another, so the planes aren t parallel and must intersect. The angle between the planes is the same as the angle

LINES AND PLANES 9 between these two vectors: ( ) 2, 3, 1 3, 1, 4 θ = cos 1 2, 3, 1 3, 1, 4 ( ) = cos 1 6 + 3 12 4 + 9 + 1 9 + 1 + 16 ( ) 3 = cos 1 364 99.05