Multivariable-Calculus · Unit 5 · Video 2 · Interactive Practice

Plug the Line Into the Plane: the Coefficient of t Decides Whether It Crosses

IKey Formulas

Formula	Name	What you need
$x_0 + 2y_0 + 4z_0$ against $7$	Side test for $x + 2y + 4z = 7$	The point's coordinates: $> 7$ and $< 7$ are the two half-spaces, $= 7$ is the plane itself
$x(t) + 2y(t) + 4z(t) = 11 - 8t$	The video's line substituted into the plane	$x = -1 + 2t,\ y = 2 + t,\ z = 2 - 3t$
$a\,x(t) + b\,y(t) + c\,z(t) = (\mathbf{N} \cdot \mathbf{v})\,t + \mathbf{N} \cdot \overrightarrow{OQ_0}$	Any line into any plane $ax + by + cz = d$	$\mathbf{N} = \langle a, b, c \rangle$ and the line $\overrightarrow{OQ_0} + t\,\mathbf{v}$
$\mathbf{N} \cdot \mathbf{v} = 0$	Line parallel to the plane or inside it	The constant decides: $\mathbf{N} \cdot \overrightarrow{OQ_0} = d$ puts the whole line in the plane, $\neq d$ means it never meets it

Key Insight: The coefficient of $t$ is $\mathbf{N} \cdot \mathbf{v} = 1(2) + 2(1) + 4(-3) = -8 \neq 0$ , so $11 - 8t = 7$ has exactly one solution, $t = \tfrac{1}{2}$ , and the line crosses the plane once, at $Q(\tfrac{1}{2}) = (0, \tfrac{5}{2}, \tfrac{1}{2})$ .

IITwo Points, Two Sides

Along any path $x + 2y + 4z$ changes continuously, so values on opposite sides of $7$ force a crossing.

Second point

IIIWhere the Moving Point Meets the Plane

Along the line $x + 2y + 4z = 11 - 8t$ , so the moving point is on the plane exactly when $11 - 8t = 7$ .

IVOne Coefficient Decides: $\mathbf{N} \cdot \mathbf{v}$

Substituting any line leaves $(\mathbf{N} \cdot \mathbf{v})\,t$ plus a constant; when the coefficient vanishes, the constant alone settles parallel versus contained.

v₁ 2 v₂ 1 v₃ −3 Base point

VQuiz Questions

Problem 1 · Which Side?

Given: the plane $x + 2y + 4z = 7$ and the points $A = (2, 1, 1)$ and $B = (3, -2, 1)$ — find where $A$ and $B$ lie relative to the plane.

✅ Correct!

A

gives

8 > 7

and

B

gives

3 < 7

: the two values fall on opposite sides of

7

❌ Watch the sign of $y$ in $B$ .

2y

y = -2

-4

, so

B

gives

3 - 4 + 4 = 3

, not

3 + 4 + 4 = 11

— and

3

is below

7

while

A

8

is above.

❌ The $y$ -coefficient is $2$ .

2 + 1 + 4 = 7

drops it; with it,

A

gives

2 + 2(1) + 4(1) = 8

. Only a value of exactly

7

puts a point in the plane, and neither point has one.

❌ No picture is needed. One line of arithmetic per point decides it: compare

x + 2y + 4z

with

7

❌ Not quite. Put each point into

x + 2y + 4z

and compare the result with

7

Show solution

Evaluate the left side of the plane's equation at each point and compare with $7$ :

A:\ 2 + 2(1) + 4(1) = 2 + 2 + 4 = 8 > 7

B:\ 3 + 2(-2) + 4(1) = 3 - 4 + 4 = 3 < 7

Neither value equals $7$ , so neither point is in the plane, and they miss it in opposite directions: opposite sides. Along the segment from $A$ to $B$ the value falls steadily from $8$ to $3$ , reaching $7$ one fifth of the way along, at $A + \tfrac{1}{5}(B - A) = (\tfrac{11}{5}, \tfrac{2}{5}, 1)$ ; check: $\tfrac{11}{5} + \tfrac{4}{5} + 4 = 7$ .

Problem 2 · The Crossing Point

Given: the line $x = -1 + 2t,\ y = 1 - 2t,\ z = 1 + t$ and the plane $x + 2y + 4z = 7$ — find the point where the line meets the plane.

✅ Correct!

5 + 2t = 7

gives

t = 1

, and

Q(1) = (1, -1, 2)

satisfies

1 - 2 + 8 = 7

❌ That is $t = -1$ . From

5 + 2t = 7

, subtract

5

2t = +2

, so

t = 1

. Check your point:

-3 + 6 + 0 = 3

, not

7

❌ That is $t = 2$ — the division by the coefficient was skipped.

2t = 2

gives

t = 1

. Check your point:

3 - 6 + 12 = 9

, not

7

❌ The $4$ must multiply both terms of $z$ .

4(1 + t) = 4 + 4t

, not

4 + t

; that slip turns the sum into

5 - t

and gives

t = -2

. Check your point:

-5 + 10 - 4 = 1

, not

7

❌ Not quite. Substitute

x(t), y(t), z(t)

into

x + 2y + 4z

, solve for

t

, then put

t

back into the line.

Show solution

Substitute the line into the left side of the plane's equation:

(-1 + 2t) + 2(1 - 2t) + 4(1 + t) = -1 + 2t + 2 - 4t + 4 + 4t = 5 + 2t

The coefficient of $t$ is $\mathbf{N} \cdot \mathbf{v} = 1(2) + 2(-2) + 4(1) = 2 \neq 0$ , so there is exactly one crossing. Set the result equal to $7$ :

5 + 2t = 7 \quad\Longrightarrow\quad 2t = 2 \quad\Longrightarrow\quad t = 1

Put $t = 1$ back into the line:

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

Check: $1 + 2(-1) + 4(2) = 1 - 2 + 8 = 7$ ✓

Problem 3 · Tuning the Line to Run Parallel

Given: the plane $x + 2y + 4z = 7$ and the family of lines $x = 1 + kt,\ y = 2 - t,\ z = 3 + t$ — find the value of $k$ that makes the coefficient of $t$ vanish, and what the line then does.

Which $k$ makes the coefficient of $t$ zero?

With that $k$ , the line

✅ Correct! At

k = -2

the substitution collapses to the constant

17

, and

17 = 7

is never true: the line is parallel to the plane and misses it.

❌ Sign slip. The coefficient is

k - 2 + 4 = k + 2

; at

k = 2

it is

4

, not

0

❌ That dots $\mathbf{v}$ with the base point.

\langle 1, 2, 3 \rangle \cdot \langle k, -1, 1 \rangle = k + 1

is not the coefficient of

t

; the coefficient is

\mathbf{N} \cdot \mathbf{v}

with

\mathbf{N} = \langle 1, 2, 4 \rangle

❌ The coefficient must be $0$ , not $7$ .

k + 2 = 7

makes the

t

-term

7t

, and then

17 + 7t = 7

has the solution

t = -\tfrac{10}{7}

❌ Read the constant. With the

t

-terms gone, the left side is

1 + 2(2) + 4(3) = 17

at every point of the line, and

17 \neq 7

❌ There is no $t$ left to solve for. A zero coefficient leaves

17 = 7

, which no value of

t

can make true.

❌ Check $k$ . Collect the

t

-terms of

(1 + kt) + 2(2 - t) + 4(3 + t)

and set their coefficient to

0

❌ Check the case. With the coefficient zero, compare the leftover constant with

7

Show solution

Substitute the line into $x + 2y + 4z$ :

(1 + kt) + 2(2 - t) + 4(3 + t) = 17 + (k - 2 + 4)\,t = 17 + (k + 2)\,t

The coefficient is $\mathbf{N} \cdot \mathbf{v} = 1(k) + 2(-1) + 4(1) = k + 2$ , which is zero exactly when $k = -2$ : then $\mathbf{v} = \langle -2, -1, 1 \rangle$ is perpendicular to $\mathbf{N} = \langle 1, 2, 4 \rangle$ and the line runs along the plane's direction.

With $k = -2$ the equation to solve is

17 = 7,

which is false for every $t$ . The constant $17 = 1 + 2(2) + 4(3)$ is the side-test value of the base point $(1, 2, 3)$ — it sits above $7$ , and so does every point of the line. The line is parallel to the plane and never meets it.

Problem 4 · A New Plane

Given: the plane $2x - y + 3z = 6$ and the line $x = 1 + t,\ y = -1 + 5t,\ z = 1 + t$ — find how the line and the plane meet.

✅ Correct!

\mathbf{N} \cdot \mathbf{v} = 2 - 5 + 3 = 0

and the constant is

6

, the plane's own constant:

6 = 6

for every

t

❌ The base point is on the plane, but it is not the only such point. The coefficient of

t

\mathbf{N} \cdot \mathbf{v} = 0

, so the value

6

t = 0

is the value at every

t

— the whole line is in the plane.

❌ $\mathbf{N} \cdot \mathbf{v} = 0$ allows two cases — read the constant. The substitution gives

6

, which equals the plane's

6

, so every point of the line satisfies the equation.

❌ $\mathbf{N} \cdot \mathbf{v} = 0$ means $\mathbf{v}$ is perpendicular to the normal. A direction perpendicular to

\mathbf{N}

runs along the plane; a line perpendicular to the plane would need

\mathbf{v}

parallel to

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

❌ Not quite. Substitute the line into

2x - y + 3z

and look at both the coefficient of

t

and the constant.

Show solution

Read off $\mathbf{N} = \langle 2, -1, 3 \rangle$ from the plane and $\mathbf{v} = \langle 1, 5, 1 \rangle$ from the line. Substitute:

2(1 + t) - (-1 + 5t) + 3(1 + t) = 2 + 2t + 1 - 5t + 3 + 3t = 6

The $t$ -terms cancel because $\mathbf{N} \cdot \mathbf{v} = 2(1) + (-1)(5) + 3(1) = 0$ , so the line runs along the plane's direction. The constant left over is $6$ , and the equation to solve is

6 = 6,

true for every $t$ . Every point of the line satisfies the plane's equation: the line lies in the plane. (Had the constant come out anything other than $6$ , the same zero coefficient would have meant parallel and never meeting.)

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Plug the Line Into the Plane: the Coefficient of t Decides Whether It Crosses

IKey Formulas

IITwo Points, Two Sides

IIIWhere the Moving Point Meets the Plane

IVOne Coefficient Decides: N⋅v\mathbf{N} \cdot \mathbf{v}N⋅v

VQuiz Questions

IVOne Coefficient Decides: $\mathbf{N} \cdot \mathbf{v}$