3-1 The 3D orthogonal coordinate system: practice questions
Pure Mathematics — First: Algebra · Mathematics. These are 10 questions from the Magmo3 bank on this lesson, from easy to hard. Solve each one, then open "Show answer and explanation". Magmo3 has 37 questions on this lesson, each with an explanation.
Find the point on the z-axis that is equidistant from A(3, 1, 2) and B(1, −1, −2).
A(0, 0, −1)
B(0, 0, 2)
C(0, 0, 1)
D(0, 1, 0)
Show answer and explanation
Correct answer: (C) (0, 0, 1)
Explanation: Step 1: A point on the z-axis has the form P = (0, 0, z).
Step 2: PA² = 9 + 1 + (z − 2)² and PB² = 1 + 1 + (z + 2)².
Step 3: PA = PB gives 10 + z² − 4z + 4 = 2 + z² + 4z + 4, so 8z = 8 and z = 1.
Step 4: Check: PA² = 9 + 1 + 1 = 11 and PB² = 1 + 1 + 9 = 11.
∴ P = (0, 0, 1)
Question 5Medium
C(1, −2, 3) is the midpoint of AB, where A(x, 4, −1) and B(5, y, z). Find the value of x + y + z.
A−10
B−4
C4
D12
Show answer and explanation
Correct answer: (B) −4
Explanation: Step 1: (x + 5)/2 = 1, so x = −3.
Step 2: (4 + y)/2 = −2, so y = −8.
Step 3: (−1 + z)/2 = 3, so z = 7.
Step 4: x + y + z = −3 − 8 + 7 = −4.
∴ −4
Question 6Medium
The distance between the point P(k, 3, 2k) and the y-axis is 5√5 length units. Find the positive value of k.
A10
B5
C25
D√5
Show answer and explanation
Correct answer: (B) 5
Explanation: Step 1: The foot of the perpendicular from P to the y-axis is (0, 3, 0).
Step 2: Distance = √(k² + (2k)²) = √(5k²) = √5 k for k positive.
Step 3: √5 k = 5√5, so k = 5.
∴ k = 5
Question 7Medium
Find the perimeter of the triangle whose vertices are O(0, 0, 0), A(2, 3, 6) and B(2, 3, 0).
The points A(−1, 2, 0), B(1, 3, 2) and C(5, 5, 6):
Aare collinear, and A lies between B and C
Bare collinear, and B lies between A and C
Care the vertices of a right-angled triangle
Dare the vertices of an isosceles triangle
Show answer and explanation
Correct answer: (B) are collinear, and B lies between A and C
Explanation: Step 1: AB = √(4 + 1 + 4) = 3.
Step 2: BC = √(16 + 4 + 16) = 6.
Step 3: AC = √(36 + 9 + 36) = 9.
Step 4: AB + BC = 3 + 6 = 9 = AC, so the three points lie on one straight line with B between A and C.
∴ collinear, B between A and C
Question 9Hard
The points A(4, 1, 1), B(1, 4, 1) and C(1, 1, k) are the vertices of a triangle. Find the value(s) of k that make the triangle ABC equilateral.
Ak = 3 or k = −3
Bk = −4 or k = 2
Ck = 4 or k = −2
Dk = 4 only
Show answer and explanation
Correct answer: (C) k = 4 or k = −2
Explanation: Step 1: AB² = (1 − 4)² + (4 − 1)² + 0² = 18.
Step 2: AC² = 9 + 0 + (k − 1)² and BC² = 0 + 9 + (k − 1)², so AC = BC for every k.
Step 3: Equilateral needs AC² = AB²: 9 + (k − 1)² = 18, so (k − 1)² = 9 and k − 1 = ±3.
∴ k = 4 or k = −2
Question 10Hard
A lies on the x-axis, B on the y-axis and C on the z-axis. If (2, 3, 0) is the midpoint of AB and (0, 3, 4) is the midpoint of BC, find the length of AC.
A4√5
B10
C2√13
D2√5
Show answer and explanation
Correct answer: (A) 4√5
Explanation: Step 1: Let A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c).
Step 2: Midpoint of AB = (a/2, b/2, 0) = (2, 3, 0), so a = 4 and b = 6.
Step 3: Midpoint of BC = (0, b/2, c/2) = (0, 3, 4), so c = 8.
Step 4: AC = √((0 − 4)² + 0² + (8 − 0)²) = √80 = 4√5.
∴ AC = 4√5 length units