Solutions to Homework 8

Transcription

1 Solutions to Homework 8. Find an orthonormal basis for the plane 4x x + x = in R. Answer: First, we choose two linearly independent vectors that are in the plane. These vectors from a basis for the plane, so we can then use Gram-Schmidt to find an orthonormal basis. Two vectors in the plane are x = and x = 4. These vectors are not multiples of each other, so they are linearly independent and form a basis. Now, following the Gram-Schmidt procedure, we make the first vector have unit length. The norm of the first vector is + + =. u = x x = Next, we project x onto u : proj =< x, u > u = We create a vector perpendicular to u by subtracting the projection from x : x proj = Finally, we make this into a unit vector by dividing by its norm, which is =.

2 u = x proj x proj = Thus, an orthonormal basis for the plane is,. Consider the vector space C[, ] of all continuous functions defined on the real line from to, and consider the following inner product: < f(x), g(x) >= f(x)g(x)e x dx Let S be the subspace of this inner product space spanned by the functions, x, and x. Use Gram-Schmidt to find an orthonormal basis for this subspace. Answer: We use Gram-Schmidt on the function f (x) =, f (x) = x, and f (x) = x. First, we need to make the first function have unit length. We compute its norm: f =<, >= e x dx = Thus, the first function in our orthonormal set is u (x) = Next, we project the second function onto u (x): proj =< x, > u (x) = xe x dx = f (x) f (x) =.

3 Thus, a function orthogonal to u (x) is x. We need to make this function have unit length, so we compute its norm: x =< x, x >= (x ) e x dx = Thus, the second function in our orthonormal set is u (x) = x. Next, we project the third function onto the subspace spanned by u (x) and u (x): proj =< x, > u (x)+ < x, x > u (x) We can compute the two necessary inner products: < x, >= x e x dx = Thus, < x, x >= x (x )e x dx = 4 proj =< x, > u (x)+ < x, x > u (x) = + 4(x ) = 4x Thus, a function orthogonal to both u (x) and u (x) is x 4x +. We need to make this function have unit length, so we compute its norm: x 4x+ =< x 4x+, x 4x+ >= (x 4x+) e x dx = 4 Thus, x 4x + = 4 =, and the third function in our orthonormal set is u (x) = x x +. Thus, we get the orthonormal basis: {, x, } x x +

4 . Consider the three planes, x + y z =, x + 4y z = 5, and x 5y +z =. Do these three planes have a nonempty intersection? If their intersection is nonempty, do they intersect in a plane, a line, or a point? Answer: To find the intersection of the three planes, we find all solutions to the three equations by row reducing Thus, the solutions are x y z = + t This is a parametric equation for a line. Thus, the solutions to the three equations form a line. Thus, the intersection of the three planes is nonempty, and the three planes intersect in a line. 4. Consider the four points A(,, ), B(,, ), C(,, 5), and D(,, 4). Do these 4 points lie on a plane? Answer: Yes, the 4 points lie on a plane. There are several methods to determine the answer to this problem. Here are three methods. (a) Method : We can find the equation for the plane containing points A, B, and C. If D lies on this plane, then there is a plane containing all four points, and if D does not lie on this plane, then there is not such a plane. To find the equation for the plane containing points A, B, and C, first we find the normal vector to the plane by taking the cross product of AB and AC: 4

5 AB AC = Thus, the vector 8 6 î ĵ ˆk = 8î 6ĵ ˆk is normal to the plane containing the points A, B, and C. Since, any multiple of this vector is also normal to the plane, we can simplify it by dividing by. Thus, 4 the vector is normal to the plane. We can use the point A to find the equation for the plane: 4(x ) + (y ) + (z ) = Now, we can check whether the point D(,, 4) lies on this plane: 4( ) + ( ) + (4 ) = =. Thus, D does lie on the plane, so the 4 points all lie on the same plane. (b) Method : We want to determine whether the four points all lie on a plane. We can assume that the plane has equation Ax+By+ Cz + D = (where these A, B, C, and D, and are different from the names of the points). We can substitute the points into the equation for the plane, which will give us four linear equations for A, B, C, and D: A + B + C + D = A B + D = A + 5C + D = A + B + 4C + D = Now, we can solve the equations. If there is a solution with at least one of A, B, C, or D nonzero, then there is a plane containing the four points (and the solution gives the equation for the plane). If the only solution is A = B = C = D =, then the four points 5

6 do not lie on a plane (note that A = B = C = D = is always a solution to the given equations). We can create the augmented matrix and row reduce: 5 4 Thus, the solutions are A B C D = t Thus, since there are solutions other than A = B = C = D =, there is a plane containing the four points. An equation for the plane is 4x y z + 9 =. (c) Method : The absolute value of the triple scalar product of the vectors AB, AC, and AD equals the volume of the parallelepiped determined by those three vectors (see Example 5 in Section.4 for more information on the triple scalar product). If this number is, then the three vectors must all lie on a plane. If this number is nonzero, then the three vectors do not all lie on a plane. We compute the triple scalar product: ( AB AC) AD = Thus, the four points all lie on a plane. = 6

7 Additionally, in this example, the points A, B, and D all lie on a line. Any method which determines this (for example, noting that AB and AD are multiples of each other), also works to show that the four points lie on a plane. 7