Parametric Equations? ):

Question

Q1a) Let x = p + td be a parametric equation of the line through the points A(4,4,1)

i) What is the vector p, given that its second component is 4?

ii) Give the vector d that has the first component 3

 

b) Let x = q + se be a parametric equation of the line through the point C(-3,1,4) and in the direction (-6,-2,5)

i) What is the vector q, given that its third component is 4?

ii) Give the vector e that has second component -2

 

c) Are these two lines the same?

Yes or No

 

d) In the line with parametric equation x = (1,0,1) + r(-1,-3,9) perpendicular to the line AB?

 

Q2a) Same sort of question just different numbers

If x = p + td  is a parametric equation of the line through  (-4,-3,5) and in the direction (-2,-1,2) then:

i) What is the vector p, given that its third component is 5?

ii) Give the vector d that has first component -2

 

b) If x = p + td is a parametric equation of the line through the points A(0,-1,1) and B(6,2,-5), then:

i) What is the vector p, given that its second component is -1?

ii) Give the vector d that has the third component -6

 

c) Are the lines in a) and b) the same?

 

d) Find the point D distance 3 from A along the line towards B

 

e) Is the line with parametric equation x = (3 + s, 3 + 2s, 2 + 2s) perpendicular to the line AB?

Yes or No?

Answer 1

1.(b).                

The vectors are p = (- 3, 1, 4) and d = (- 12, - 2, 23/2).

Comments

Please mention the point B in the question number 1 to find the solutions for the remaining questions [ numbers 1(a), 1(c) and 1(d) ].

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Sorry! Point B is (7,0,0)

Answer 2

2(a).

The vectors are p = (- 4, - 3, 5) and d = (- 2, - 2, 3).

Answer 3

(2).b

The vectors are p = (0, - 1, 1) and d = (6, 3, - 6).

Answer 4

(2).c

The two lines are not same.

Comments

(2)C.

To decide if the lines are coincident, we check to see whether a point on one of the lines satisfies the vector equation of the other. The point (0, - 1, 1) is on the second line. If it is also on the first line, then (0, - 1, 1) = (- 4, - 3, 5) + t(- 2, - 1, 2).

Then 0 = - 4 - 2t                -1 = - 3 - t                1 = 5 + 2t

              t = - 2                          t = - 2                     t = - 2

Since the same parameter value t = - 2 obtained each equation, the point (0, - 1, 1) from the second line does lie on the first line, and the lines are coincident.

So, the two lines are same.

Answer 5

2.(e).

So, the two lines are perpendicular.

Answer 6

2.(d).

The point D = (2, 0, - 1).

Answer 7

1(a).

The vectors are p = (4, 4, 1) and d = (3, - 4, - 1).