MATH 2050 Assignment 4 Fall Due: Thursday. u v v 2 v = P roj v ( u) = P roj u ( v) =

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1 MATH 5 Assignment 4 Fall 8 Due: Thursday [5]. Let u = and v =. Find the projection of u onto v; and the projection of v onto u respectively. ANS: The projection of u onto v is P roj v ( u) = u v v. Note that = ( ) + ( v ) ( ) = and v = + + =. Hence, P roj v ( u) = v v =. Similarly, as u = = 7, one has, u u =. 7 []. (a) Find two orthogonal vectors in the plane x + y z =. ANS: First of all, let z =, one gets x + y =. Let y = and then x =. That is, vector u = is in the plane x + y z =. Second, let y =, one gets x z =. Let z = and then x =. That is, vector v = is in the plane x + y z =. We can check that =, and hence u is not orthogonal to v. We now calculate the projection of v onto u as follows: u u = =, as u = + =. Let f = v and e = u =. One can check that e and f are in the plane x + y z = and are orthogonal to each other.

2 (b) Find the projection of u = onto the plane x + y z =. ANS: Method : The projection of u onto the plane π : x + y z = is 5/ u e u f P roj π ( u) = e e + f f = /, where e =, e u =, f u = 6 and f =. Method : The plane π has normal vector n = projection of u onto n as follows: P roj n ( u) = / u n n n = = /, 6. We now calculate the as n = = 6 and u n =. So, the projection of u onto the plane π is / 5/ P roj π ( u) = u P roj n ( u) = / = / x + t [5]. Calculate the distance from point P (,, ) to the line y = t. z t ANS: The line has the direction u = and passes one point, say Q = (,, ) (by letting t = ). Let v = P Q = =. Now let us calculate u u = 4/9 8/9 & v 8/9 4/9 8/9 /9 x So the distance from point P (,, ) to the line y = + t t is the length of v P roj u ( v), which is, 8/9 = /. z t.

3 [5] 4. Calculate the distance from point P (,, ) to the plane x y + z =. ANS: A normal vector for the plane x y + z = is n = and one point in the plane is, say Q = (,, ). Let v = P Q = =. Now let us calculate P roj n ( v) = /9 n v n n = /9. /9 So the distance from point P (,, ) to the plane x y + z = is the length of P roj n ( v), which is, /. [5] 5. Are vectors u =, v =, w =, and x = linearly independent? ANS: Let c, c, c and c 4 be such that c u + c v + c w + c 4 x =. That is, c, c, c and c 4 satisfy the following equations: c c c 4 = ; c + c + c = ; c + c + c + c 4 =. Solve the above equations by elimination method: the first equation plus the second equation yields c = c 4 ; put this into the third equation to have c + c + c 4 =. Combining this equation with the first equation (first subtracting, then adding them) implies that c + c 4 = and c =. Hence, one gets c = and c = c 4. (Note that this system has equations with 4 variables, and hence there is at least one free variable, say c 4.) In conclusion, the solution for the above equations are c 4 = c = c and c =. In particular, one can let c 4 = and we get the following v + w + x =, and hence vectors u, v, w, and x are linearly dependent.

4 [] 6. Let u =, v = and w =. (a) Show that u, v and w are linearly independent. ANS: Let c, c and c be such that c u + c v + c w =. That is, c, c and c satisfy the following equations: c + c + c = ; c + c = ; c + c =. Solve the above equations by elimination method: the first equation minus the second equation yields c =. the first equation minus the third equation yields c =. By the first equation, one has c =. So the equation c u + c v + c w = only has solution c = c = c =, and hence vectors u, v and w are linearly independent. (b) Show that any -dimensional vector can be written as a linear combination of u, v and w. ANS: Let x = x x x be any -dimensional vector. Suppose that x = c u + c v + c w, and hence one has the following equations (treat x, x, x as constant numbers): c + c + c = x ; c + c = x ; c + c = x. Solve the above equations by elimination method: the first equation minus the second equation yields c = x x ; the first equation minus the third equation yields c = x x. The first equation then implies c = x + x x. In conclusion, the vector x has the following form x = (x x ) u + (x x ) v + (x + x x ) w, a linear combination of of u, v and w.

5 [] 7. Let u =, v =, and w =. (a) Are u, v and w linearly independent? ANS: Let c, c and c be such that c u + c v + c w =. That is, c, c and c satisfy the following equations: c + c c = ; c c c = ; c c + c =. Note the the first equation is identical to the third one. So in fact, we have only two equations with three variables: { c c c = ; c c + c =. Solve the above equations by elimination method: the second equation plus the first equation yields c c =. The first equation then implies c = c. By letting c =, one gets c = and c = is a solution for the desired system of linear equations. Hence, u + v w = and they are linearly dependent. (b) Can any -dimensional vector be written as a linear combination of u, v and w? If yes, please show your reason; if no, please provide one example. ANS: NO. For example, the cross product of u and v is not the linear combination of u, v and w. That is, u v = can not be written as a linear combination of u, v and w. Suppose yes, then there are constants c, c, c such that u v = c u + c v + c w, i.e., the following equation c + c c = ; c c c = ; c c + c =. The first equation plus the third equation yields = 6 which is not possible for any c, c and c. [5]