Lesson F1-F2
F1-F2 Vector equations of lines and planes Quiz: AQA Further Maths, Unit 2
20 questions
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Lesson F1-F2, Vector equations of lines and planes: 20 multiple choice questions for the AQA Further Maths (7367), Unit 2: Compulsory content, written with Revision Ninja.
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The 20 questions
-
In the vector equation r = a + lambda b of a straight line, what does the vector b represent?
- The position vector of the origin
- A normal to the line
- A point on the line
- A direction vector of the line
-
In the Cartesian equation ax + by + cz = d of a plane, which vector is normal to the plane?
- (d, a, b)
- (d, d, d)
- (x, y, z)
- (a, b, c)
-
If a plane has vector equation r = a + s b + t c, which vector is normal to the plane?
- b . c
- b + c
- b x c
- a x b
-
Which scalar product form gives the equation of a plane with normal n passing through a point with position vector a?
- r x n = a x n
- r . n = 0
- r + n = a
- r . n = a . n
-
Which point lies on the plane x + 2y - z = 2?
- (1, 1, 0)
- (1, 1, 1)
- (2, 0, 1)
- (0, 0, 0)
-
Find the Cartesian equation of the plane through the points (1, 0, 0), (0, 1, 0) and (0, 0, 1).
- x + y + z = 1
- x + y - z = 1
- x + y + z = 0
- x - y + z = 1
-
A line passes through (1, 2, 3) and (4, 0, -1). Which vector equation describes this line?
- r = (1, 2, 3) + t(4, 0, -1)
- r = (3, -2, -4) + t(1, 2, 3)
- r = (1, 2, 3) + t(3, 2, 4)
- r = (1, 2, 3) + t(3, -2, -4)
-
Find the Cartesian equation of the plane through (2, 1, -1) with normal vector (3, -1, 2).
- 3x + y + 2z = 3
- 3x - y + 2z = 6
- 3x - y + 2z = 3
- 3x - y - 2z = 3
-
Find the Cartesian equation of the plane through (1, 0, 0), (0, 2, 0) and (0, 0, 3).
- 6x + 3y + 2z = 6
- 6x + 3y + 2z = 0
- 3x + 2y + z = 6
- x + 2y + 3z = 6
-
The point (1, 2, k) lies on the plane 2x + 3y - z = 10. What is the value of k?
- -6
- -2
- 18
- 2
-
Write the line r = (1, 2, 3) + t(2, -1, 4) in Cartesian form.
- (x - 2)/1 = (y + 1)/(-2) = (z - 3)/4
- (x + 1)/2 = (y + 2)/(-1) = (z + 3)/4
- (x - 1)/2 = (y - 2)/(-1) = (z - 3)/4
- (x - 1)/2 = (y - 2)/1 = (z - 3)/4
-
Find the Cartesian equation of the plane through (1, 2, 3) with normal vector (2, -1, 1).
- 2x + y + z = 3
- 2x - y + z = 3
- 2x - y - z = 3
- 2x - y + z = 1
-
Find the direction vector of the line (x - 3)/2 = (y + 1)/(-4) = z/5.
- (2, -4, 5)
- (3, 1, 0)
- (3, -1, 0)
- (2, 4, 5)
-
What point is reached on the line r = (2, 0, 1) + t(1, 1, 0) when t = 3?
- (5, 3, 0)
- (3, 3, 0)
- (5, 3, 1)
- (2, 0, 1)
-
A plane has normal (1, -2, 2) and passes through the origin. Which point also lies on this plane?
- (0, 0, 1)
- (1, 1, 1)
- (2, 2, 0)
- (2, 1, 0)
-
The plane with vector equation r = (1, 0, 0) + s(1, 1, 0) + t(0, 1, 2) has which Cartesian form?
- 2x - 2y + z = 2
- 2x + 2y + z = 2
- x + y + 2z = 1
- 2x - 2y + z = 0
-
Find the Cartesian equation of the plane r = (1, 2, 0) + s(1, 0, 1) + t(0, 1, -1).
- x + y + z = 3
- -x + y + z = 3
- -x + y + z = 1
- x - y - z = 1
-
A line is given by x = 1 + 2t, y = 2 - t, z = 3 + 4t. What is the point on the line where x = 5?
- (5, 0, 7)
- (5, 1, 11)
- (5, 0, 11)
- (5, 2, 3)
-
The point (k, 1, 2) lies on the plane x + 2y - 3z = 4. What is k?
- 4
- 8
- -8
- 2
-
Find the vector equation of the line through the points P(2, -1, 3) and Q(5, 1, 0).
- r = (2, -1, 3) + t(3, -2, 3)
- r = (2, -1, 3) + t(3, 2, -3)
- r = (2, -1, 3) + t(2, 1, 0)
- r = (3, 2, -3) + t(2, -1, 3)
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