Vector equation of a line (2D/3D) practice questions.

A line has vector equation r = (1, 2) + t(3, -1). Determine the direction vector of the line.

• (3, -1)
• (1, 2)
• (4, 1)
• (-3, 1)

Answer: (3, -1)

In r = a + t·d, the coefficient of the parameter t is the direction vector, here (3, -1).

Determine a vector equation of the line passing through the point A(4, 1) with direction vector (2, 5).

• r = (4, 1) + t(2, 5)
• r = (2, 5) + t(4, 1)
• r = (4, 1) + t(5, 2)
• r = (2, 5) + t(1, 4)

Answer: r = (4, 1) + t(2, 5)

The vector equation is r = a + t·d, where a is the position vector of a point on the line and d is the direction: r = (4, 1) + t(2, 5).

A line has vector equation r = (0, 3) + t(4, 3). Calculate the gradient of this line.

• 3/4
• 4/3
• -3/4
• 3

Answer: 3/4

Gradient = (change in y)/(change in x) = 3/4 from the direction vector (4, 3).

A line in three dimensions is given by r = (1, 0, 2) + t(2, -1, 3). Determine the coordinates of the point when t = 1.

• (3, -1, 5)
• (2, -1, 3)
• (1, 0, 2)
• (3, 1, 5)

Answer: (3, -1, 5)

r = (1 + 2, 0 - 1, 2 + 3) = (3, -1, 5). Add the direction vector once to the base point.

Determine whether the point (7, 4) lies on the line r = (1, -2) + t(2, 2).

• Yes, when t = 3
• Yes, when t = 2
• No, it does not lie on the line
• Yes, when t = 4

Answer: Yes, when t = 3

x: 1 + 2t = 7 gives t = 3; y: -2 + 2(3) = 4. Both hold, so the point lies on the line at t = 3.

A particle moves along r = (1, 0) + t(6, 8), where t is in seconds and distance in metres. Calculate its speed.

• 10 m/s
• 14 m/s
• 48 m/s
• √14 m/s

Answer: 10 m/s

Speed = magnitude of the velocity vector = √(6² + 8²) = √100 = 10 m/s.