Exercise 1
Write the parametric and Cartesian equations of the line passing through the point A(3, −5) and parallel to the line x + 2y − 4 = 0.
Answer. A vector parallel to the line is (2, −1). Consequently, parametric equations of the line are
x = 2t + 3 y = −t − 5 A Cartesian equation of the line is
x + 2y + a = 0
where a is to be determined so that point A belongs to the line. This gives
3 + 2(−5) + a = 0 whence
a = 7 Thus the required equation is
x + 2y + 7 = 0
Exercise 1
Write the parametric and Cartesian equations of the line passing through the point A(3, −5) and parallel to the line x + 2y − 4 = 0.
Answer. A vector parallel to the line is (2, −1). Consequently, parametric equations of the line are
x = 2t + 3 y = −t − 5 A Cartesian equation of the line is
x + 2y + a = 0
where a is to be determined so that point A belongs to the line. This gives
3 + 2(−5) + a = 0 whence
a = 7 Thus the required equation is
x + 2y + 7 = 0
Exercise 2
Given the lines r : x + ky + k = 0 and s : kx + y + k = 0 determine the values of k such that:
the lines are parallel
their common point belongs to the line t : x + y + 3 = 0
Exercise 2, answer
The lines are parallel if and only if the vector (1, k) and (k, 1) are parallel, if and only if
1 k k 1
= 1 − k2= 0 if and only if k = ±1.
Exercise 2, answer (cont.)
If the lines are not parallel, that is k 6= ±1, the coordinates of their common point are the solutions of the linear system
x + ky = −k kx + y = −k that is
x =
−k k
−k 1
1 − k2 =k2− k
1 − k2 = − k 1 + k
y =
1 −k k −k
1 − k2 =k2− k
1 − k2 = − k 1 + k Such a point belongs to line t if and only if
− k
1 + k − k
1 + k + 3 = 0
which means 1+k2k = 3, that is 2k = 3 + 3k, and finally k = −3.
Exercise 3
Find the equation of the following planes:
(1) the plane through the point A = (1, 1, 0) and parallel to the vectors u = (1, 0, −1) and v = (0, 2, 3)
(2) the plane through the points B = (0, 1, −1), C = (3, 2, 1) and parallel to the vector w = (0, 0, 5)
(3) the plane through the point D = (1, 1, −1) and orthogonal to the vector n = (1, −1, 2)
(4) the plane through the points E = (0, 1, 0), F = (2, −1, 0) and G = (1, 2, 2)
Exercise 3, answer
(1)
x − 1 y − 1 z
1 0 −1
0 2 3
= 0
− 2(x − 1) − 3(y − 1) + 2z = 0 2x + 3y − 2z − 4 = 0
(2) This plane is parallel to the vector (C − B) = (3, 1, 2), so
x y − 1 z + 1
3 1 2
0 0 5
= 0 5x − 15(y − 1) = 0 x − 3y + 3 = 0
Exercise 3, answer (cont.)
(3)
x − 1 − (y − 1) + 2(z + 1) = 0 x − y + 2z + 2 = 0
(4) This plane is parallel to the vectors
(F − E ) = (2, −2, 0), (F − G ) = (1, −3, −2) thus also to the vector
1
2(F − E ) = (1, −1, 0) so
x y − 1 z
1 −1 0
1 −3 −2
= 0 2x + 2(y − 1) − 2z = 0 x + y − z − 1 = 0
Exercise 4
Discuss the mutual position of the plane and the line in the following cases:
(1) π : 3x − y + z = 1, and r : x = 1 + t, y = −2 + t, z = 2 − t (2) π : 3x − y + z = 1, and r : x − 1 = 0, z − y = 0
Exercise 4, answer
(1) Plane π is orthogonal to the vector (3, −1, 1), while line r is parallel to the vector (1, 1, −1), thus π, r are not parallel (nor orthogonal), and they meet in a point, given be the solution to the equation
3(1 + t) − (−2 + t) + 2 − t = 1 t = −7
thus π ∩ r = {(−6, −9, 5)}.
(2) Parametric equations for r are:
x = 1
y = t
z = t
Consequently, r is parallel to the vector (0, 1, 1). Since π is orthogonal to the vector (3, −1, 1) and (0, 1, 1) · (3, −1, 1) = 0, it turns out that π, r are parallel.
As the equation
3 − t + t = 1 has no solution, it follows that π ∩ r = ∅.
Exercise 5
Given the lines r : 4x + 3y − 10 = 0, s : x + 5y + 6 = 0 and
t : 3x − 2y + 1 = 0 compute the perimeter and the area of the triangle they determine.
Exercise 5, answer
r ∩ s :
4x + 3y = 10 x + 5y = −6
x =
10 3
−6 5
4 3 1 5
= 68 17 = 4
y =
4 10 1 −6
4 3 1 5
=−34 17 = −2
that is r ∩ s = {(4, −2)}.
Exercise 5, answer (cont.)
r ∩ t :
4x + 3y = 10 3x − 2y = −1
x =
10 3
−1 −2
4 3
3 −2
= −17
−17= 1
y =
4 10 3 −1
4 3
3 −2
=−34
−17 = 2
that is r ∩ t = {(1, 2)}.
Exercise 5, answer (cont.)
s ∩ t :
x + 5y = −6 3x − 2y = −1
x =
−6 5
−1 −2
1 5
3 −2
= 17
−17= −1
y =
1 −6 3 −1
1 5
3 −2
= 17
−17 = −1
that is s ∩ t = {(−1, −1)}
Exercise 5, answer (cont.)
The edges of the triangle are the vectors
(4, −2) − (1, 2) =(3, −4) (4, −2) − (−1, −1) =(5, −1) (1, 2) − (−1, −1) =(2, 3) The perimeter of the triangle is thus
||(3, −4)||+||(5, −1)||+||(2, 3)|| =√
9 + 16+√
25 + 1+√
4 + 9 = 5+√ 26+√
13 The area of the triangle is
1
2||(2, 3, 0) ∧ (3, 4, 0)|| = 1
2||(0, 0, 2 · 4 − 3 · 4)|| = 1 24 = 2
