Thursday January 30 Solutions Projections distances and planes 1 Let a i j and b 2i 1j b a a Calculate projb a b b b and draw a picture of it together with a and b SOLUTION projb a 2 5 1 5 This is drawn below b b The orthogonal complement of a with respect to b is the vector orthb a a projb a Find orthb a and orthb a and draw two copies of it in your picture from part a one based at 0 and the other at projb a SOLUTION orthb a 3 5 1 5 1 0 orth b a a 0 5 orth b a 0 5 1 0 1 5 2 0 projb a 0 5 b 1 0 c Check that orthb a calculated in b is orthogonal to projb a calculated in a SOLUTION 2 5 1 5 3 5 6 5 6 25 6 25 0 so orthb a and projb a are orthogonal d Find the distance of the point 1 1 from the line x y t 2 1 SOLUTION This is the length of orthb a or p p 3 5 2 6 5 2 3 5 5 2 Let a and b be vectors in Rn Use the definitions of projb a and orthb a to show that orthb a is always orthogonal to projb a SOLUTION Since projb a points in the same direction as b it is equivalent to show that b is orthogonal to orthb a We take the dot product a b a b b orthb a b a b b a b b b a a b 0 b b b b Since the dot product of b and orthb a is 0 they are orthogonal 3 Find the distance between the point P 3 4 1 and the line l t 2 3 1 t 1 1 1 SOLUTION Let Q 2 3 1 a 3 4 1 2 3 1 1 1 1 and b 1 1 1 The distance from P to l t is given by the magnitude of orthb a as shown below P orth b a b a Q projb a 1 3 p 1 3 1 3 and orthb a a projb a 2 3 4 3 2 3 So the distance from P to l t is orthb a 2 3 6 4 Consider the equation of the plane x 2y 3z 12 a Find a normal vector to the plane SOLUTION A normal vector is n 1 2 3 b Find where the x y and z axes intersect the plane Sketch the portion of the plane in the first octant where x 0 y 0 z 0 SOLUTION The plane intersects the x y and z axes respectively at 12 0 0 0 6 0 and 0 0 4 The sketch is shown below 0 x 5 10 4 3 z 2 1 0 0 2 y 4 6 c Using the points in part b find two non parallel vectors that are parallel to the plane SOLUTION The vectors a 12 0 4 and b 0 6 4 work These vectors start at the intersection of the plane with the z axis and end at the intersections with the x and y axes respectively d Using the dot product to check that the vectors you found in c are really parallel to the plane SOLUTION A vector v is parallel to the plane if and only if it is orthogonal to a normal vector for the plane that is v n 0 So we check a n 12 0 4 1 2 3 12 0 12 0 b n 0 6 4 1 2 3 0 12 12 0 e Pick another normal vector n0 to the plane and one of the points from b Use these to find an alternative equation for the plane Compare this new equation to x 2y 3z 12 How are these two equations related Is it clear that they describe the same set of points x y z in R3 SOLUTION We use the point 0 0 4 and normal vector n0 2n 2 4 6 The plane consists of all points x y z such that the vector x y z 4 is orthogonal to the vector n0 This is expressed by n0 x y z 4 0 or 2x 4y 6 z 4 0 which is the same as 2x 4y 6z 24 If we divide both sides by 2 we obtain the equation x 2y 3z 12 which is the original equation These describe the same set of points because multiplying both sides of the original equation by any nonzero constant does not affect the solution set 5 The Triangle Inequality Let a and b be any vectors in Rn The triangle inequality states that a b a b a Give a geometric interpretation of the triangle inequality SOLUTION Fit a b and a b into a triangle as below The triangle inequality says the sum of the lengths of the sides of the triangle corresponding to a and b is less than the length of the side corresponding to a b a b b a b Use what we know about the dot product to explain why a b a b This is called the Cauchy Schwartz inequality SOLUTION a b a b cos where is the angle between a and b So a b a b cos a b since cos 1 c Use part b to justify the triangle inequality SOLUTION It is equivalent to show a b 2 a b 2 a 2 2 a b b 2 We begin with the equality a b 2 a b a b Since the dot product is distributive a b a b a a 2a b b b a 2 2a b b 2 a 2 2 a b b 2 where the last inequality follows from part b So this justifies the triangle inequality
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