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MATH1002 Quizzes

Quiz 4: Lines and planes
Question 1 Questions
Find the equation of the line joining P(2,1,1) and Q(0,3,1) in parametric vector form. Exactly one option must be correct)
a)
r = 2i + j k + t(3j + k)
b)
r = 3j + k + t(2i + j k)
c)
r = 2i + j k + t(2i + 2j + 2k)
d)
r = 2i + 2j + 2k + t(2i + 2j + 2k)

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
A vector parallel to the line is the vector PQ = OQ OP = 2i + 2j + 2k.
PIC
Hence the line can be represented by the parametric vector equation r = 2i + j k + t(2i + 2j + 2k). Note that there are many other ways of giving a vector parametric equation for this line. For example, instead of using the position vector of the known point P in the equation, we could have used the position vector of Q, to give another version of the equation, r = 3j + k + t(2i + 2j + 2k).
Choice (d) is incorrect
A line has Cartesian equations x 2 3 = y 1 2 = z + 6 1 . A vector parallel to the line is: Exactly one option must be correct)
a)
3i + 2j k
b)
2i j + 6k
c)
2 3i + 1 2j + 6k
d)
i + j 7k

Choice (a) is correct!
When the cartesian equations of a line are given in the form x α a = y β b = z γ c , we can identify the coordinates of one point on the line (namely (α,β,γ)) and a vector parallel to the line, namely ai + bj + ck. In this case, a = 3,b = 2,c = 1.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Given the parametric scalar equations of a line, x = 3 + 5t,y = 2 t,z = 3 + 2t, find the cartesian equations of the same line. Exactly one option must be correct)
a)
x + 3 5 = y + 2 1 = z + 3 2
b)
x 5 3 = y + 1 2 = z 2 3
c)
x 15 = y 2 = z 6
d)
x 3 5 = y 2 1 = z 3 2

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
In the three parametric scalar equations, the coefficients 5, 1,2 of t are the components of a vector parallel to the line. (Therefore the line has direction parallel to 5i j + 2k.) The numbers 5, 1,2 become the denominators in the cartesian equations of the line. The constants 3,2,3 in the scalar parametric equations are (in that order) the coordinates of a particular point (3,2,3) on the line. This information allows us to form the numerators in the cartesian form, x 3,y 2,z 3.
Find the equation of the line through (1,2,1) parallel to the line
x 2 3 = y 1 1 = z 5 4 .
Exactly one option must be correct)
a)
r = i + 2j + k + t(2i + j + 5k)
b)
r = i + 2j + k + t(6i + 2j + 8k)
c)
r = 2i + j + 5k + t(3i + j + 4k)
d)
r = i + 2j + k + t(3i j + 4k)

Choice (a) is incorrect
Choice (b) is correct!
A vector parallel to the line is 3i + j + 4k (from the denominators in the cartesian form). So the line is also parallel to 6i + 2j + 8k. Therefore since the line passes through the point (1,2,1), it has vector parametric equation r = i + 2j + k + t(6i + 2j + 8k).
Choice (c) is incorrect
Choice (d) is incorrect
Find the parametric vector equation of the line through (1,1,1) which is perpendicular to the two lines (1) and (2) below.
(1)r = i j + tk(2)x 4 = y + 1 = z 2 3
Exactly one option must be correct)
a)
r = i + j + k + ti
b)
r = i + j + k + t(i + 4j)
c)
r = i + j + k + t(3i + 4k)
d)
r = i + j + k + t(i + 4j)

Choice (a) is incorrect
Choice (b) is correct!
Line (1) is parallel to k and line (2) is parallel to 4i + j + 3k. Hence the required line, which is perpendicular to both, must be parallel to the cross product of these two vectors, k × (4i + j + 3k) = i + 4j. Its parametric vector equation is therefore r = i + j + k + t(i + 4j).
Choice (c) is incorrect
Choice (d) is incorrect
Suppose that P and Q are two distinct points in 3-dimensional space. How many planes are there which contain both P and Q? Exactly one option must be correct)
a)
None.
b)
One only.
c)
Two only.
d)
None of the above.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
There are infinitely many planes containing any two given points. To see this, visualise the line joining the points as the spine of a book, and the infinitely many planes as pages of the book.
Find the cartesian equation of the plane which goes through the point (3,1,0) and is perpendicular to the vector i j + 2k. Exactly one option must be correct)
a)
x y + 2z = 0
b)
x y + 2z = 4
c)
x y + 2z = 2
d)
x 3y = 0

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The equation of the plane through the point (p,q,r) perpendicular to ai + bj + ck is a(x p) + b(y q) + c(z r) = 0. Hence in this case, the cartesian equation is 1(x 3) 1(y 1) + 2(z 0) = 0, that is, x y + 2z = 2.
Choice (d) is incorrect
Find the cartesian equation of the plane which contains the three points A = (1,3,1), B = (1,1,5) and C = (2,1,0). Exactly one option must be correct)
a)
10x + 2y z = 17
b)
10x + 21y 6z = 1
c)
x + 21y 6z = 19
d)
10x 18y 6z = 38

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
In order to know the equation of a plane, we must know a point on the plane and a vector perpendicular to the plane. We know three points on the plane, so we’re OK there. We can find a vector perpendicular to the plane by using the vector cross product. Notice that the vectors AB and AC are both parallel to the plane, so their cross product is normal to the plane. Since AB = 2j + 6k and AC = 3i 2j + k, we calculate AB ×AC = 10i 18j 6k. This is a normal vector to the plane. The cartesian equation is therefore 10(x 1) 18(y 3) 6(z + 1) = 0, or 10x 18y 6z = 38.
Find the cartesian equation of the plane which contains the point (1,1,1) and is parallel to the two vectors i + j k and 2i 5k. Exactly one option must be correct)
a)
3x + y 6z = 2
b)
x y 4z = 4
c)
5x + 3y + 2z = 10
d)
5x 3y + 2z = 4

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
A vector perpendicular to the plane is
(i+jk)×(2i5k) = ij k 11 12 0 5 = 5i+3j2k.
So the cartesian equation of the plane is 5(x 1) + 3(y 1) 2(z 1) = 0 i.e 5x 3y + 2z = 4.
Find the acute angle between the planes 3x + y + z = 0 and x 2y + z = 3. Exactly one option must be correct)
a)
1.55 radians
b)
1.32 radians
c)
0.87 radians
d)
0.69 radians

Choice (a) is incorrect
Hint: Find the angles between the normals to the two planes.
Choice (b) is correct!
The two planes are not parallel, since the first plane has normal vector n = 3i + j + k and the second plane has normal m = i 2j + k; so n is not parallel to m. Denote the two planes by ABCD and EBCF, respectively, so that BC is the line of intersection.
PIC
The diagram above shows that the angle between two planes is the same as the angle between the normals to the planes, so we need to find the angle between the vectors n and m. This angle θ is given by the now well-known formula cosθ = n m |n||m| = 2 116. So, the angle between the two planes is θ 1.32 radians.
Choice (c) is incorrect
Hint: Find the angles between the normals to the two planes.
Choice (d) is incorrect
Hint: Find the angles between the normals to the two planes.