Multivariable-Calculus · Unit 5 · Video 1 · Interactive Practice

A Line Is a Moving Point: Parametric Equations From Two Points in Space

IKey Formulas

Formula	Name	What you need
$\overrightarrow{Q_0Q(t)} = t\,\overrightarrow{Q_0Q_1}$	The one observation	Constant speed, with $Q(0) = Q_0$ and $Q(1) = Q_1$
$\mathbf{v} = \overrightarrow{Q_0Q_1} = Q_1 - Q_0$	Velocity (direction) vector	Two points: subtract the coordinates of $Q_0$ from those of $Q_1$ , one component at a time
$Q(t) = Q_0 + t\,\mathbf{v}$	Vector form of the line	One point and one direction
$x = x_0 + at$ $y = y_0 + bt$ $z = z_0 + ct$	Parametric equations	Constants: the position at $t = 0$ ; coefficients: $\mathbf{v} = \langle a, b, c\rangle$

Key Insight: A parametric line carries six numbers — the constants are the point at $t = 0$ , the coefficients of $t$ are the velocity — and finding a point on it means substituting one value of $t$ , never solving a system.

IIA Clock on the Line

At constant velocity the displacement from $Q_0$ is exactly $t$ times $\overrightarrow{Q_0Q_1}$ , for every real $t$ .

t 0.5

IIIOne Vector Equation, Three Coordinate Formulas

Matching components turns $\overrightarrow{Q_0Q(t)} = t\,\mathbf{v}$ into three formulas, each a constant plus a constant times $t$ .

Step 1

Step 1 — Subtract the coordinates

\mathbf{v} = \overrightarrow{Q_0Q_1} = \langle 1-(-1),\ 3-2,\ -1-2\rangle

\mathbf{v} = \langle 2,\ 1,\ -3\rangle

Step 2 — The arrow to the moving point

\overrightarrow{Q_0Q(t)} = \langle x-(-1),\ y-2,\ z-2\rangle

\overrightarrow{Q_0Q(t)} = \langle x+1,\ y-2,\ z-2\rangle

Step 3 — Match the three components

\langle x+1,\ y-2,\ z-2\rangle = t\,\langle 2,\ 1,\ -3\rangle

\begin{aligned} x+1 &= 2t\\ y-2 &= t\\ z-2 &= -3t \end{aligned}

Step 4 — Move each constant across

\boxed{\begin{aligned} x &= -1 + 2t\\ y &= 2 + t\\ z &= 2 - 3t \end{aligned}}

The coefficients

2, 1, -3

are

\mathbf{v}

; the constants

-1, 2, 2

are

Q_0

Step 5 — Check both clock settings

t = 0

\ (-1,\ 2,\ 2) = Q_0\ \checkmark

t = 1

\ (-1+2,\ 2+1,\ 2-3) = (1,\ 3,\ -1) = Q_1\ \checkmark

Dropping the constant in

y = 2 + t

gives

y = t

, which sits at

(-1, 0, 2) \ne Q_0

when

t = 0

: a parallel line through neither point.

IVReading the Six Numbers Backwards

The constants are the position at $t = 0$ ; the coefficients of $t$ are the velocity along the line.

x(t): constant −1 coefficient of t 2

y(t): constant 2 coefficient of t 1

z(t): constant 2 coefficient of t −3

💡 Challenge: change all six numbers and still describe the video's line.

VQuiz Questions

Problem 1 · Two Points, Three Formulas

Given: $A = (2, -1, 3)$ and $B = (5, 1, 2)$ — find parametric equations of the line through them with $Q(0) = A$ and $Q(1) = B$ .

✅ Correct!

\overrightarrow{AB} = \langle 3, 2, -1\rangle

supplies the coefficients of

t

and

A = (2, -1, 3)

supplies the constants.

❌ The roles are swapped. Here the constants are

\langle 3, 2, -1\rangle

and the coefficients are the coordinates of

A

. It does reach

\mathbf{v} + A = B

t = 1

, but at

t = 0

it sits at

(3, 2, -1)

, not

A

, and it moves along

\langle 2, -1, 3\rangle

instead of

\overrightarrow{AB}

❌ Check the middle constant.

y - (-1) = 2t

gives

y = -1 + 2t

. With

y = 1 + 2t

the point sits at

(2, 1, 3) \ne A

when

t = 0

— a parallel line through neither point.

❌ That velocity is $A - B$ . It traces the same line, but with the clock running backwards: it starts at

A

and heads away from

B

, reaching

(-1, -3, 4)

t = 1

and passing

B

only at

t = -1

. Subtract the coordinates of

A

from those of

B

❌ Not quite. The coefficients of

t

come from

B - A

and the constants from

A

Show solution

Subtract the coordinates of $A$ from those of $B$ , one component at a time:

\mathbf{v} = \overrightarrow{AB} = \langle 5-2,\ 1-(-1),\ 2-3\rangle = \langle 3,\ 2,\ -1\rangle

The arrow from $A$ to the moving point is $t$ times $\mathbf{v}$ :

\langle x-2,\ y+1,\ z-3\rangle = \langle 3t,\ 2t,\ -t\rangle

Move each constant across:

\boxed{x = 2 + 3t,\quad y = -1 + 2t,\quad z = 3 - t}

Check: $t = 0$ gives $(2, -1, 3) = A$ , and $t = 1$ gives $(2+3,\ -1+2,\ 3-1) = (5, 1, 2) = B\ \checkmark$

Problem 2 · Which Point Is on the Line?

Given: the video's line $x = -1 + 2t$ , $y = 2 + t$ , $z = 2 - 3t$ — which of these points lies on it?

✅ Correct! All three coordinates give the same time:

-1 + 2t = -3

2 + t = 1

and

2 - 3t = 5

each give

t = -1

, one unit of time before the point reaches

Q_0

❌ That is the velocity, not a position. Treated as a point,

x = 2

needs

t = \tfrac{3}{2}

while

y = 1

needs

t = -1

❌ Two coordinates out of three is not enough.

x

and

y

both give

t = 2

, but at

t = 2

the third coordinate is

2 - 6 = -4

, not

-3

❌ That point is on the parallel line $y = t$ — the video's line with its constant

2

dropped, at

t = 1

. On the actual line

x = 1

needs

t = 1

but

y = 1

needs

t = -1

❌ Not quite. Solve each coordinate equation for

t

; the point is on the line only if all three give the same

t

Show solution

A point is on the line exactly when one value of $t$ produces all three of its coordinates. Solve each coordinate equation for $t$ :

$(2, 1, -3)$ : $x$ gives $t = \tfrac{3}{2}$ , $y$ gives $t = -1$ — no.
$(3, 4, -3)$ : $x$ gives $t = 2$ , $y$ gives $t = 2$ , $z$ gives $2 - 3t = -3$ , so $t = \tfrac{5}{3}$ — no.
$(1, 1, -1)$ : $x$ gives $t = 1$ , $y$ gives $t = -1$ — no.
$(-3, 1, 5)$ : $x$ gives $t = -1$ , $y$ gives $t = -1$ , $z$ gives $2 - 3t = 5$ , so $t = -1$ — yes.

Q(-1) = (-1 - 2,\ 2 - 1,\ 2 + 3) = (-3,\ 1,\ 5) = Q_0 - \mathbf{v}

Problem 3 · A Different Clock

Given: a point moves along a line at constant velocity; at $t = 0$ it is at $A = (2, -1, 0)$ and at $t = 3$ it is at $B = (8, 5, -6)$ — find its velocity and its position at $t = 1$ .

What is the velocity vector?

Where is the point one unit of time after leaving A?

✅ Correct! The displacement

\overrightarrow{AB} = \langle 6, 6, -6\rangle

takes

3

units of time, so

\mathbf{v} = \langle 2, 2, -2\rangle

and

Q(1) = A + \mathbf{v} = (4, 1, -2)

❌ That is the whole displacement $\overrightarrow{AB}$ . It is covered in

3

units of time, not one, so the velocity is a third of it.

❌ Only the coordinates of $B$ were divided by $3$ . The velocity is the displacement

B - A

divided by the elapsed time

3

❌ That adds the coordinates of $A$ and $B$ . The displacement is the difference

B - A = \langle 6, 6, -6\rangle

❌ Check the velocity. Velocity

=

displacement

\div

elapsed time

= (B - A) / 3

❌ That is $B$ , where the point is at $t = 3$ . Arriving at the second point at

t = 1

was the clock chosen in the video; this clock reaches

B

t = 3

❌ That is the midpoint of $AB$ . At constant velocity the point is halfway at

t = \tfrac{3}{2}

, not at

t = 1

❌ That is the velocity, not a position. The moving point starts from

A

Q(1) = A + \mathbf{v}

❌ Check the position. Start at

A

and add one velocity:

Q(1) = A + \mathbf{v}

Show solution

The displacement from $A$ to $B$ :

\overrightarrow{AB} = \langle 8-2,\ 5-(-1),\ -6-0\rangle = \langle 6,\ 6,\ -6\rangle

It is covered between $t = 0$ and $t = 3$ , so each unit of time covers a third of it:

\mathbf{v} = \tfrac{1}{3}\langle 6,\ 6,\ -6\rangle = \langle 2,\ 2,\ -2\rangle

With $A$ as the constants and $\mathbf{v}$ as the coefficients (the $z$ -constant is $0$ because $A$ lies in the $xy$ -plane):

x = 2 + 2t,\quad y = -1 + 2t,\quad z = -2t

Q(1) = (2 + 2,\ -1 + 2,\ -2) = (4,\ 1,\ -2)

Check: $Q(3) = (2 + 6,\ -1 + 6,\ -6) = (8, 5, -6) = B\ \checkmark$

Problem 4 · Same Line, Different Numbers

Given: the video's line $x = -1 + 2t$ , $y = 2 + t$ , $z = 2 - 3t$ — which of these describes the same line?

✅ Correct! Its constants

(1, 3, -1)

are

Q_1

, a point of the line, and its coefficients

\langle 4, 2, -6\rangle = 2\mathbf{v}

point along it: the same line, travelled twice as fast with the clock started at

Q_1

❌ The constant of $y$ is missing. The direction is right, but at

t = 0

this line is at

(-1, 0, 2)

— two units from

Q_0

in the

-y

direction, on a parallel line through neither

Q_0

nor

Q_1

❌ Constants and coefficients have traded places. This line starts at

(2, 1, -3)

and moves along

\langle -1, 2, 2\rangle

, which is not a multiple of

\langle 2, 1, -3\rangle

❌ One coefficient changed sign. It starts at

Q_0

but moves along

\langle 2, 1, 3\rangle

, not a multiple of

\langle 2, 1, -3\rangle

, so the two lines meet at

Q_0

and nowhere else.

❌ Not quite. Same line means a parallel velocity AND a starting point that lies on the video's line.

Show solution

Read each candidate backwards: the constants are its point at $t = 0$ , the coefficients its velocity. It is the video's line exactly when (1) the velocity is a multiple of $\mathbf{v} = \langle 2, 1, -3\rangle$ and (2) the starting point lies on the video's line.

(a) velocity $\langle 2, 1, -3\rangle$ $\checkmark$ , but $(-1, 0, 2)$ : $x$ gives $t = 0$ , $y$ gives $t = -2$ — not on the line. Parallel, different.
(b) velocity $\langle 4, 2, -6\rangle = 2\langle 2, 1, -3\rangle$ $\checkmark$ , and $(1, 3, -1)$ gives $t = 1$ in all three coordinates — it is $Q_1$ . Same line.
(c) velocity $\langle -1, 2, 2\rangle$ — not a multiple of $\langle 2, 1, -3\rangle$ .
(d) velocity $\langle 2, 1, 3\rangle$ — not a multiple of $\langle 2, 1, -3\rangle$ .

For (b), the point it reaches at time $s$ is the video's point at time $t = 1 + 2s$ :

\big(-1 + 2(1+2s),\ 2 + (1+2s),\ 2 - 3(1+2s)\big) = (1 + 4s,\ 3 + 2s,\ -1 - 6s)

Solved: 0 / 4