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4-1 The equation of a straight line in space: 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.

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Question 1Easy

Find a direction vector of the straight line passing through the points A(2, 1, 3) and B(5, 4, 1).

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Correct answer: (A) (3, 3, −2)

Explanation: Step 1: A direction vector of the line = B − A. Step 2: B − A = (5−2, 4−1, 1−3) = (3, 3, −2). ∴ direction vector = (3, 3, −2)

Question 2Easy

If L₁ and L₂ are two straight lines in space whose direction vectors d₁ and d₂ are NOT parallel, then L₁ and L₂ are ..........

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Correct answer: (B) intersecting or skew

Explanation: Step 1: Parallel (or coincident) lines need d₁ ∥ d₂, which is not the case here. Step 2: Two non-parallel lines either meet at one point or never meet; lines that are neither parallel nor intersecting are called skew. ∴ the two lines are intersecting or skew.

Question 3Easy

A line is given by (x − 1)/4 = (y + 3)/(−2) = z/5. Find its direction vector.

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Correct answer: (A) (4, −2, 5)

Explanation: Step 1: In (x − x₀)/a = (y − y₀)/b = (z − z₀)/c the direction vector is (a, b, c). Step 2: Here a = 4, b = −2, c = 5. ∴ direction vector = (4, −2, 5)

Question 4Medium

Find the measure of the angle between the two straight lines L₁: 2x = y = −2z and L₂: (x − 1)/2 = y + 3 = z.

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Correct answer: (D) 60°

Explanation: Step 1: Write L₁ as x/1 = y/2 = z/(−1) (divide 2x = y = −2z by 2): d₁ = (1, 2, −1). Step 2: L₂: d₂ = (2, 1, 1). Step 3: cos θ = |d₁·d₂|/(‖d₁‖‖d₂‖) = |2 + 2 − 1|/(√6 × √6) = 3/6 = 1/2. Step 4: θ = 60°.

Question 5Medium

Two lines have direction vectors (1, 2, −1) and (2, 4, −2). The two lines are:

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Correct answer: (D) parallel

Explanation: Step 1: Check if one direction is a scalar multiple of the other. Step 2: (2, 4, −2) = 2×(1, 2, −1), so the directions are proportional. ∴ the lines are parallel

Question 6Medium

The length of the perpendicular drawn from the point C(5, 3, 2) to the straight line passing through the two points A(1, 2, 0) and B(3, 0, 1) equals .......... length unit.

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Correct answer: (C) 5√5/3

Explanation: Step 1: Direction vector d = AB = (2, −2, 1), ‖d‖ = 3; AC = (4, 1, 2), ‖AC‖² = 21. Step 2: Projection of AC on the line = |AC · d|/‖d‖ = |8 − 2 + 2|/3 = 8/3. Step 3: Length = √(21 − 64/9) = √(125/9) = 5√5/3. ∴ 5√5/3 (8/3 is the projection; √253/3 adds instead of subtracting).

Question 7Medium

The two straight lines r₁ = (1, 0, 2) + t₁(1, 1, 0) and r₂ = (0, 1, 0) + t₂(0, 1, 1) are ..........

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Correct answer: (C) skew

Explanation: Step 1: d₁ = (1, 1, 0), d₂ = (0, 1, 1) are not parallel, so the lines intersect or are skew. Step 2: Equate r₁ = r₂: 1 + t₁ = 0, t₁ = 1 + t₂, 2 = t₂. Step 3: The first gives t₁ = −1, then t₂ = −2, which contradicts t₂ = 2. ∴ no common point: the lines are skew.

Question 8Hard

The lines L₁: (x, y, z) = (2, 0, 3) + t(1, 1, 1) and L₂: (x, y, z) = (4, 0, 4) + s(1, −1, 0). These two lines:

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Correct answer: (B) intersect at the point (3, 1, 4)

Explanation: Step 1: Directions (1, 1, 1) and (1, −1, 0) are not proportional, so not parallel. Step 2: From L₂, z = 4 ⇒ L₁: 3 + t = 4 ⇒ t = 1, giving L₁ point (3, 1, 4). Step 3: Check L₂: x = 4 + s = 3 ⇒ s = −1, then y = −s = 1 ✓, z = 4 ✓. ∴ they intersect at (3, 1, 4)

Question 9Hard

For which value of k do the lines (x, y, z) = (1, 0, 0) + t(1, 1, 0) and (x, y, z) = (0, 0, k) + s(0, 1, 1) intersect?

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Correct answer: (A) k = 1

Explanation: Step 1: Equate: 1 + t = 0, t = s, 0 = k + s ⇒ t = −1, s = −1. Step 2: The third equation gives k − 1 = 0 ⇒ k = 1 (for any other k the lines are skew, since the directions are not parallel); they meet at (0, −1, 0). ∴ k = 1

Question 10Hard

Which of the following straight lines is skew to the straight line r = t(1, 1, 0)?

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Correct answer: (D) r = (0, 0, 1) + s(1, −1, 0)

Explanation: Step 1: (2, 2, 0) = 2(1, 1, 0), and (1, 0, 0) is not on the given line, so that line is parallel to it. Step 2: r = (1, 1, 1) + s(0, 0, 1) passes through (1, 1, 0) at s = −1, and r = (0, 0, 3) + s(1, 1, 3) passes through (−1, −1, 0) at s = −1; both points are on r = t(1, 1, 0), so both intersect it. Step 3: r = (0, 0, 1) + s(1, −1, 0) always has z = 1 while the given line has z = 0, and (1, −1, 0) is not parallel to (1, 1, 0). ∴ r = (0, 0, 1) + s(1, −1, 0) is skew to it.

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