Tuesday, September 4 ∗ Solutions ∗ Projections, distances, and planes.1. Let a =i +j and b =2i −1j(a) Calculate projba =³b·ab·b´b and draw a picture of it together with a and b.SOLUTION:projba =〈2/5,−1/5〉. This is drawn below (b).(b) The orthogonal complement of a with respect to b is the vectororthba =a −projba.Find orthba and orthba and draw two copies of it in your picture from part (a), one basedat 0 and the other at projba.SOLUTION:orthba =〈3/5,−1/5〉projbaorthbaorthbaab0.51.01.52.0-1.0-0.50.51.0(c) Check that orthb(a) calculated in (b) is orthogonal to projba calculated in (a).SOLUTION:〈2/5,−1/5〉·〈3/5,6/5〉=6/25 −6/25 =0, so orthb(a) and projba 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), orp(3/5)2+(6/5)2=3p5/5.2. Let a and b be vectors in Rn. Use the definitions of projba and orthba to show that orthba isalways orthogonal to projba.SOLUTION:Since projba points in the same direction as b, it is equivalent to show that b is orthogonal toorthba. We take the dot product:b ·orthba =b ·µa −a ·bb ·bb¶=b ·a −µa ·bb ·b¶b ·b =b ·a −a ·b =0Since the dot product of b and orthba 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 orthba as shown below.PaborthbaQprojba =〈1/3,−1/3,1/3〉and orthba = a −projba =〈2/3,4/3,2/3〉. So the distance from P to l(t)is |orthba|=2p63.4. 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 thefirst 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). Thesketch is shown below.0510x0246y01234z(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 ofthe 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 theplane.SOLUTION:A vector v is parallel to the plane if and only if it is orthogonal to a normal vector for theplane, that is v·n =0. So we check:a ·n =〈12,0,−4〉·〈1,2,3〉=12 +0 −12 =0b ·n =〈0,6,−4〉·〈1,2,3〉=0 +12 −12 =0(e) Pick another normal vector n0to the plane and one of the points from (b). Use these tofind 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 allpoints (x, y, z) such that the vectorx, y, z −4®is orthogonal to the vector n0. This is ex-pressed byn0·x, y, z −4®=0or2x +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 originalequation. These describe the same set of points because multiplying both sides of theoriginal 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 thelengths of the sides of the triangle corresponding to a and b is less than the length of theside corresponding to a+b.aba+b(b) Use what we know about the dot product to explain why |a ·b| ≤ |a||b|. This is called theCauchy-Schwartz inequality.SOLUTION:a ·b =|a||b|cosθ, where θ is the angle between a and b. So|a ·b|=|a||b||cosθ|≤|a||b|, since |cosθ|≤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|2We 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|2where the last inequality follows from part (b). So this justifies the triangle
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