Math 115 Lab 1 Solutions

1. Find the point 15 of the way from P (2,1, 5) to Q(3, 0, 4).

Solution: Let ~p = [2 1 5]T and ~q = [3 0 4]T be the position vectors of the points.The position vector of the point 15 the way from ~p to ~q is

~t = ~p+ 15 (~q ~p) = 45~p+ 15~q = 15 [11 4 24]T .Thus, the point 15 of the way from P to Q is (

115 ,

45 ,

245 ).

2. Given the vectors ~v = [3 1 2]T and ~d = [1 2 1]T , write the vector ~v as a sum ~v = ~v1 + ~v2 where ~v1 isparallel to ~d and ~v2 is orthogonal to ~d.

Solution: Let ~v1 = proj~d(~v) =~v~d~d2

~d = 12 ~d =12 [1 2 1]

T .

Then ~v2 = ~v ~v1 = 12 [5 4 3]T .

3. Find the shortest distance from the point P (1, 0, 2) to the line [x y z]T = [1 1 0]T + t[2 1 1]T .

Solution: Write P0 = P0(1,1, 0), ~v = P0P = [0 1 2]T and ~d = [2 1 1]T .Compute ~v1 = proj~d(~v) =

12~d.

Then the shortest distance is ~v ~v1 = 12 [2 1 3]T

= 1214.4. Find the vector equation of the line through the points P1(1, 0,2) and P2(2, 1,1).

Solution: Now ~d =P1P2 = [1 1 1]T , so the line is [x y z]T = [1 0 2]T + t[1 1 1]T .

5. Determine the parametric equations of the line through the point P (1,1, 0) which is perpendicular tothe plane x+ y 2z = 3.

Solution: The normal ~n = [1 1 2]T will serve as direction vector of the line (it is perpendicular tothe plane).

As P (1,1, 0) is in the line, the parametric equations are

x = 1 + t

y = 1 + t

z = 2t

6. If ~v and ~w are orthogonal vectors, show that ~v2 + ~w2 = ~v + ~w2.

Solution: We have ~v + ~w2 = (~v + ~w) (~v + ~w) = ~v2 + 2~v ~w + ~w2 .But ~v and ~w are orthogonal, so ~v ~w = 0.Hence ~v2 + ~w2 = ~v + ~w2.

7. Find the point of intersection of the line [x y z]T = [2 1 3]T + t[1 1 4]T

and the plane 3x+ y 2z = 4.

Solution: Every point on the line has the form [x y z]T = [2 + t 1 t 3 4t]T .This point lies on the plane if 3(2 + t) + (1 t) 2(3 4t) = 4, which gives t = 12 .Hence the point is [x y z]T = [52 32 1]T .

8. Find vector equation of the line through (2, 5, 0) that is parallel to the planes 2x + y 4z = 0 andx+ 2y + 1 = 0.

Solution: We need the direction vector ~d = [d1 d2 d3]T . Since ~d is parallel to the plane 2x+y4z = 0,it must be orthogonal to the normal vector of the plane [2 1 4]T .ie. [d1 d2 d3]T [2 1 4]T = 0.Also, since ~d is parallel to the plane x+ 2y + 1 = 0, we have [d1 d2 d3]T [1 2 0]T = 0.This gives us the equations 2d1 + d2 4d3 = 0 and d1 + 2d2 = 0. Substituting d1 = 2d2 into thefirst equation, we obtain d3 = 54d2. There are many possible direction vectors (all parallel). We will let

d2 = 4 and so d1 = 8 and d3 = 5. So ~d = [8 4 5]T and the line passes through (2, 5, 0), so the vectorequation of the line is [x y z]T = (2, 5, 0) + [8 4 5]T .

9. Find the area of the triangle with vertices P (2, 1, 0), Q(3,1, 1) and R(1, 0, 1).

Solution: The area of a triangle is 12base height.We will take ~QR as the base and ~QP proj ~QR ~QP as the height.Now, ~QR = [2 1 0]T =(2)2 + 12 + 02 = 5 andproj ~QR

~QP =~QP ~QR ~QR2

~QR =[1 2 1]T [2 1 0]T

52 [2 1 0]T =

45[2 1 0]T .

So ~QP proj ~QR ~QP = [1 2 1]T 45 [2 1 0]T = [ 35 65 1]T =( 35 )

2 + ( 65 )2 + 12 =

705 .

Thus the area of the triangle is 125705 =

142 .

10. Consider the points A(2, 2, 1), B(1, 1, 0) and C(2, 3,3).

(a) Are these points the vertices of a right-angled triangle? Justify your answer.

Solution: We haveAB = [1 1 1]T , AC = [0 1 4]T and BC = [1 2 3]T .

HenceAB BC = 0 so the angle at B is a right angle.

(b) Find the cosine of the interior angle of the triangle at vertex C.

Solution: If is the internal angle at C then, sinceCA = [0 1 4]T and CB = [1 2 3]T ,

we have cos =CA CBCACB = 141714 =

1417

17.