MATH ASSIGNMENT 5. v = 6

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1 () Let MATH - ASSIGNMENT 5 u = v = w = (a) Find a vector that is perpendicular to both v and w. (b) Find the angle between u and v. (c) Find dist(u, v). (d) Find proj u v. (e) Find the area of the parallelogram spanned by v and w. (f) Find the volume of the parallelepiped defined by u, v, w. (g) Do the three vectors lie in a plane? Explain. (a) v w = is perpendicular to both v and w. ( ) u v (b) θ = arccos u v = arccos(4/ 7) =.99 rad. (c) dist(u, v) = u v = 5 = + + ( 5) = 5 / (d) proj u v = u v u u u = 4 = 4/ / (e) The parallelogram spanned by v and w has area v w = ( ) + + = 8 (f) The volume of the parallelepiped defined by u, v, w is u (v w) = ( ) + () + () = (g) No, if the three vectors did lie in one plane then they would form a parallelepiped of volume (i.e. the shape would have no height). () (a) Find the distance from the line y = x + to the point (, ). (b) Find the distance from the point (,, ) to the plane x y + z = 5. Both of these problems can be solved using projections, but we will use the distance formulas here. (a) We write the line as x y + =, to get the distance (b) The distance is d = d = () () + + ( ) = / 5 () () + () 5 + () + = /

2 () Let u = and v =. (a) Find the projection of u onto v. (b) Find the portion of u that is perpendicular to v. (c) What is the distance from u to v? (d) What is the distance from the point given by u to the line passing through the origin and the point given by v? (a) proj v u = u v v v v = 4 7 = u v to v. (b) The portion of u that is perpendicular to v is u v = u u v = /7 8/7 = 8/7 (c) dist(u, v) = u v = 4 the portion of u that is parallel = 9 = 4.59 /7 9/7 4/7 (d) This distance is given by u v = 7 8 =.. Notice that this is shorter than dist(u, v). (4) Consider the lines in R y(x) = (,, )x + (,, ) and y(x) = (,, )x + (,, ) (a) Are the lines parallel? Explain. (b) Find a vector n, that is perpendicular to both lines (hint: you may find the cross product useful for this). (c) Find the minimum distance between these two lines (hint: start by finding some vector that starts on one line and ends on the other; then think about projections). (d) Do theses lines intersect? Explain. (a) No, their slope vectors are not parallel (i.e. (,, ) α(,, )) (b) n = (,, ) (,, ) = (5,, ) is perpendicular to both lines. (c) First we choose a point on each line, the most obvious point on a line is the point for when x =. We will let P = (,, ) and Q = (,, ). w = P Q = (,, ) joins the two lines. The projection proj n w = 8 n

3 represents the shortest line segment joining the two lines (it is perp to both lines). So the distance between the two lines is proj n w = 8 n = / 8. (5) (a) Findthe equation of the plane that passes through the point P = and is perpendicular to n =. (b) Find the distance from the line y = from part (a). + t to the plane (a) By n, we know that the the plane is of the form x + y + z = d, and by plugging P into this equation, we are able to solve for d; d = () + () + () =. So the plane has the equation x + y + z = (b) Let Q = (,, ), a point on the line. So w = P Q = Q P = (,, ) joins the line to the plane. Notice that n is perpendicular to both the plane and the slope of the line, so proj n w = (,, ) represents the shortest line segment joining the line to the plane. So proj n w = (,, ) = / is the distance between the line and the plane. () Find the distance from the point P to the line L. P = L(x) = x + First join the point to the line: Let Q = (,,, ), this is a point on the line. Let w = P Q = Q P = (,,, ), this joins the point to the line. Next project w onto the slope of the line: Let v = (,,, ) the slope of the line. proj v w = 4 (,,, ) = w v This is the part of w that is parallel to the line.

4 4 Find w v : w v = w w v = ( /4, 7/4, /4, 5/4) represents the shortest line joining the point to the line. So w v = ( /4, 7/4, /4, 5/4) = 4 7 is the distance from P to L. (7) Let T (x, x ) = (x x, x + x, x x ) (a) Find the domain and codomain of T (b) Find the standard matrix for T (c) Determine if T is a linear transformation (d) Find a vector x such that T ( x) = (, 4, 9) (a) T defines a transformation from R to R. That is, T : R R so the domain is R and the codomain is R. (b) Rewriting the transformation in matrix notation, w w w = Then the standard matrix is A = x x x. (c) For u = (u, u ), v = (v, v ) R, T ( u + v) = ((u +v )(u +v ), (u +v )+(u +v ), (u +v )(u +v ) = (u u, u + u, u u ) + (v v, v + v, v v ) = T ( u) + T ( v). And for c a scalar, T (c u) = ((cu ) (cu ), (cu ) + (cu ), (cu ) (cu )) = (c(u u ), c( u + u ), c(u u )) = c(x x, x + x, x x ) = ct ( u) Thus T is a linear transformation. (d) For x = (5, ) then T ( x) = (, 4, 9) (8) Given the vectors u = ( 7,,,, ), v = (5, 9,, 4, ) (a) Find the Euclidean inner product u v (b) Find the Euclidean distance between u and v (c) Verify the Cauchy-Schwarz inequality holds (a) u v = ( 7)(5) + ()( 9) + ()() + ( )(4) + ( )( ) = (b) d( u, v) = ( 7 5) + ( + 9) + ( + ) + ( 4) + ( + 4) = 5

5 5 (c) u v = u = ( 7) ( ) + ( ) = v = 5 + ( 9) + () ( ) = 5 Then u v 5.54 u v (9) Find the standard matrix of the linear transformation T : R R given by orthogonal projection onto the plane x y + z =. (It is helpful to note that this plane contains the origin). Let a = a a be a point in R. We want to find the standard matrix corresponding to T ( a). First we find a vector from a to point in the plane. Since (,, ) is a point in the plane, we have the vector a a v = a = a Now, the plane has the normal vector n = along n is given by proj n v = a a +., so the projection of v Then the orthogonal projection of v onto the plane (the transformation we re looking for) is given by v proj n v = = a a = aa+a 5a +a a 5a + a +a +5 5/ / / / 5/ / / / 5/ a a Thus the standard matrix of is T a a 5/ / / / 5/ / / / 5/

6 () Determine the standard matrix for the linear transformation T : R R that first dilates a vector by a factor k =, then reflects about the line y = x, then projects onto the y-axis. Dilation by a factor can be given as multiplication by the vector [ ] Reflection about the line y = x can be given as multiplication by the vector [ ] And projection onto the y-axis can be given as multiplication by the vector [ ] Then the transformation T ( u) for u = (u, u ) R is given by [ ] [ ] [ ] [ ] [ ] [ ] u u T ( u) = = u u and the standard matrix is [ ]