Session 04 - Lines and Planes in Space

Here is the step-by-step solution to the problem presented in the image.

Given Data

• Point $P$: $(-4, -6, -2)$

• Point $Q$: $(10, 2, 1)$

1. Vector Equation of the Line

To find the vector equation, we first need a position vector $\vec{r}_0$ for a point on the line and a direction vector $\vec{v}$ that is parallel to the line.

1. Find the direction vector $\vec{v}$ by finding the vector from $P$ to $Q$:

2. Use point $P$ as the initial position vector $\vec{r}_0$:

3. Write the vector equation $\vec{r}(t) = \vec{r}_0 + t\vec{v}$:

Evaluating at a few values of $t$:

• For $t = 0$:

• For $t = 1$:

• For $t = -1$:

2. Parametric Equations of the Line

The parametric equations break the vector equation down into separate components for $x$, $y$, and $z$ using the format:

• $x = x_0 + at$

• $y = y_0 + bt$

• $z = z_0 + ct$

Substituting our point $P(-4, -6, -2)$ and direction vector components $\langle 14, 8, 3 \rangle$:

• $$x = -4 + 14t$$

• $$y = -6 + 8t$$

• $$z = -2 + 3t$$

3. Symmetric Equation of the Line

To convert the parametric equations into a symmetric equation, we solve each individual equation for $t$ and set them equal to each other:

1. Solve for $t$ in the $x$ equation:

2. Solve for $t$ in the $y$ equation:

3. Solve for $t$ in the $z$ equation:

Equating all three expressions for $t$ gives the symmetric equation:

Here is the complete solution for Problem S1.1 from the activity:

Given Data

• Point $A$: $(1, 4, -2)$

• Point $B$: $(-3, 5, 0)$

1. Vector Equation of the Line

To construct the vector equation, we need an initial position vector $\vec{r}_0$ and a direction vector $\vec{v}$.

1. Find the direction vector $\vec{v}$ by finding the displacement vector from $A$ to $B$:

2. Set point $A$ as your initial position vector $\vec{r}_0$:

3. Assemble the vector equation $\vec{r}(t) = \vec{r}_0 + t\vec{v}$:

Evaluating at a few values of $t$:

• For $t = 0$:

• For $t = 1$:

• For $t = 2$:

2. Parametric Equations of the Line

By breaking the vector equation down into its distinct scalar components ($x$, $y$, and $z$) using the format $x = x_0 + at$, $y = y_0 + bt$, and $z = z_0 + ct$, we get:

• $$x = 1 - 4t$$

• $$y = 4 + t$$

• $$z = -2 + 2t$$

3. Symmetric Equation of the Line

To find the symmetric equation, we isolate $t$ in each parametric component and set them equal to one another:

1. Solve for $t$ in $x$:

2. Solve for $t$ in $y$:

3. Solve for $t$ in $z$:

Equating these gives the symmetric equation:

Here is the step-by-step solution for Skill 2 Demo using the distance formula provided in the activity.

Given Data

• Point $M$ (anywhere in space): $(1, 1, 3)$

• Symmetric equation of the line: $\frac{x-3}{4} = \frac{y+1}{2} = z-3$

1. Extract Information from the Line

From the symmetric equation form $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$, we can extract:

1. A point $P$ on the line: $(3, -1, 3)$

2. The direction vector $\vec{v}$ along the line: $\langle 4, 2, 1 \rangle$ (note that $z-3$ is equivalent to $\frac{z-3}{1}$)

2. Find the Vector $\vec{PM}$

Now, find the vector pointing from the line point $P$ to the external point $M$:

3. Compute the Cross Product $\vec{PM} \times \vec{v}$

The formula requires the cross product of $\vec{PM}$ and the direction vector $\vec{v}$:

Expand the determinant by the top row:

4. Calculate the Magnitudes

Now find the magnitudes (lengths) of both the cross product vector and the direction vector $\vec{v}$:

• Magnitude of $\vec{PM} \times \vec{v}$:

• Magnitude of $\vec{v}$:

5. Calculate the Distance $d$

Substitute these magnitudes into the distance formula $d = \frac{||\vec{PM} \times \vec{v}||}{||\vec{v}||}$:

To write it under a single radical or rationalize it:

The exact distance between the point and the line is $\frac{2\sqrt{798}}{21}$ (or approximately $2.69$ units).

Given Data

• Point $M$ (external point): $(0, 3, 6)$

• Parametric equations of the line: $x = 1 - t$, $y = 1 + 2t$, $z = 5 + 3t$

1. Extract Information from the Line

From the parametric equations form $x = x_0 + at$, $y = y_0 + bt$, and $z = z_0 + ct$ , we can easily read off a point on the line and its direction vector:

1. A point $P$ on the line (by looking at the constant terms where $t = 0$):

2. The direction vector $\vec{v}$ along the line (the coefficients of $t$):

2. Find the Vector $\vec{PM}$

Now, find the vector pointing from the line's point $P$ to the external point $M$:

3. Compute the Cross Product $\vec{PM} \times \vec{v}$

Next, find the cross product required by the formula:

Expand the determinant along the top row:

4. Calculate the Magnitudes

Now compute the magnitudes (lengths) of our cross product vector and the line's direction vector $\vec{v}$:

• Magnitude of $\vec{PM} \times \vec{v}$:

• Magnitude of $\vec{v}$:

5. Calculate the Distance $d$

Substitute these values into the distance formula $d = \frac{||\vec{PM} \times \vec{v}||}{||\vec{v}||}$:

To rationalize the denominator:

The exact distance between the point and the line is $\frac{\sqrt{70}}{7}$ (or approximately $1.20$ units).

Given Data

• Line 1 ($L_1$): $x = 5t, \quad y = -13 - 6t, \quad z = 19 - t$

• Line 2 ($L_2$): $x = -1 - 6s, \quad y = -12 - 12s, \quad z = 19 - 18s$

Step 1: Extract and Compare Direction Vectors

First, we extract the direction vectors ($\vec{u}$ and $\vec{v}$) from the coefficients of the parameters $t$ and $s$:

• Direction vector of $L_1$ ($\vec{u}$): $\langle 5, -6, -1 \rangle$

• Direction vector of $L_2$ ($\vec{v}$): $\langle -6, -12, -18 \rangle$

To check if the lines are parallel, we see if one vector is a scalar multiple of the other ($\vec{u} = a\vec{v}$):

Because the components do not scale proportionally, the direction vectors are not parallel. Looking at our relationship table, this leaves two possibilities: Intersecting or Skew.

Step 2: Check for an Intersection Point

To find out if they share a common point, we set the corresponding coordinate components equal to each other ($x_1 = x_2$, $y_1 = y_2$, $z_1 = z_2$):

1. $$5t = -1-6s$$

2. $$-13-6t = -12-12s$$

3. $$19-t = 19-18s$$

Solve the system using equations (1) and (3):

From equation (3), we can simplify by subtracting 19 from both sides:

Substitute $t = 18s$ into equation (1):

Now find $t$:

Test the values in the remaining equation (2):

We must check if these values for $t$ and $s$ satisfy equation (2) to see if the lines actually cross:

Since $\text{LS} = \text{RS}$, the system is consistent! The lines share a common intersection point.

Conclusion

Because the direction vectors are not parallel and the lines share a common point, the lines are Intersecting.

• Equal

• Parallel

• Skew (Incorrect option)

• Intersecting (Correct Answer)

Here is the step-by-step solution for Problem S3.1 to determine the relationship between line A and line B.

Given Data

• Line A passes through: $A_1 = (2, 0, 3)$ and $A_2 = (4, 7, 8)$

• Line B passes through: $B_1 = (-3, -1, 4)$ and $B_2 = (-1, 5, 9)$

Step 1: Find and Compare Direction Vectors

First, we calculate the direction vectors ($\vec{u}$ and $\vec{v}$) for both lines by subtracting their given coordinates:

• Direction vector for Line A ($\vec{u}$):

• Direction vector for Line B ($\vec{v}$):

Now we check if the direction vectors are parallel ($\vec{u} = a\vec{v}$): Comparing the components:

• $x$-component ratio: $\frac{2}{2} = 1$

• $y$-component ratio: $\frac{7}{6}$

Since $\frac{2}{2} \neq \frac{7}{6}$, the direction vectors are not parallel. Based on the classification table, the lines must be either Intersecting or Skew.

Step 2: Write Parametric Equations for Both Lines

To check if they share a common point, we first write out their parametric forms using $A_1$ and $B_1$ as our initial points:

• Line A (using parameter $t$):

• Line B (using parameter $s$):

Step 3: Check for an Intersection Point

We set the corresponding coordinate equations equal to one another ($x_A = x_B$, $y_A = y_B$, $z_A = z_B$):

1. $$2 + 2t = -3 + 2s$$

2. $$7t = -1 + 6s$$

3. $$3 + 5t = 4 + 5s$$

Solve the system using equations (1) and (3):

From equation (3), subtract $5s$ and 3 from both sides:

From equation (1), rearrange to group $t$ and $s$:

Notice the contradiction:

From equation (3) we get $t - s = 0.2$, but from equation (1) we get $t - s = -2.5$. Because $0.2 \neq -2.5$, it is mathematically impossible to find values for $t$ and $s$ that satisfy this system. The system is inconsistent, meaning the lines do not share a common point.

Conclusion

Since the direction vectors are not parallel and the lines do not share a common point, the lines are Skew.

• Equal

• Parallel

• Skew (Correct Answer)

• Intersecting