https://file.aiquickdraw.com/mathbot/file/1725999332439b4458iw5.jpg

To show that the given lines are coplanar and to find the equation of the plane containing them, we can follow these steps:

### Step 1: Represent the Lines in Vector Form

The two lines are given as:

\[
\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}
\]
\[
\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 6}{7}
\]

These can be written in vector form as:

**Line 1:**
\[
\mathbf{r}_1 = (-1, -3, -5) + \lambda(3, 5, 7)
\]

**Line 2:**
\[
\mathbf{r}_2 = (2, 4, 6) + \mu(1, 4, 7)
\]

Here, \(\lambda\) and \(\mu\) are parameters.

### Step 2: Find the Direction Vectors

The direction vector for Line 1 is:
\[
\mathbf{d}_1 = (3, 5, 7)
\]

The direction vector for Line 2 is:
\[
\mathbf{d}_2 = (1, 4, 7)
\]

### Step 3: Check for Coplanarity

To determine if the lines are coplanar, we can check if the vectors \(\mathbf{d}_1\), \(\mathbf{d}_2\), and the vector connecting a point on Line 1 to a point on Line 2 are coplanar. This is done by checking if the scalar triple product is zero.

Let \(\mathbf{a}\) be the vector connecting the point \((-1, -3, -5)\) on Line 1 to the point \((2, 4, 6)\) on Line 2:
\[
\mathbf{a} = (2 - (-1), 4 - (-3), 6 - (-5)) = (3, 7, 11)
\]

Now, compute the scalar triple product \(\mathbf{a} \cdot (\mathbf{d}_1 \times \mathbf{d}_2)\).

First, find the cross product \(\mathbf{d}_1 \times \mathbf{d}_2\):
\[
\mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
3 & 5 & 7 \\
1 & 4 & 7
\end{vmatrix}
= \mathbf{i}(5 \times 7 - 7 \times 4) - \mathbf{j}(3 \times 7 - 7 \times 1) + \mathbf{k}(3 \times 4 - 5 \times 1)
\]
\[
= \mathbf{i}(35 - 28) - \mathbf{j}(21 - 7) + \mathbf{k}(12 - 5)
\]
\[
= \mathbf{i}(7) - \mathbf{j}(14) + \mathbf{k}(7)
\]
\[
= (7, -14, 7)
\]

Next, compute the dot product \(\mathbf{a} \cdot (7, -14, 7)\):
\[
\mathbf{a} \cdot (7, -14, 7) = 3 \times 7 + 7 \times (-14) + 11 \times 7
\]
\[
= 21 - 98 + 77 = 0
\]

Since the scalar triple product is zero, the lines are coplanar.

### Step 4: Find the Equation of the Plane

The equation of the plane can be derived using the normal vector obtained from the cross product \(\mathbf{d}_1 \times \mathbf{d}_2 = (7, -14, 7)\) and any point on one of the lines, say \((-1, -3, -5)\).

The equation of the plane is:
\[
7(x + 1) - 14(y + 3) + 7(z + 5) = 0
\]

Expanding and simplifying:
\[
7x + 7 - 14y - 42 + 7z + 35 = 0
\]
\[
7x - 14y + 7z = 0
\]

Divide by 7:
\[
x - 2y + z = 0
\]

### Final Answers
1. The lines are coplanar.
2. The equation of the plane containing these lines is \(x - 2y + z = 0\).

---

Would you like any more details on this solution, or do you have any questions?

**Further questions:**
1. How would you find if two lines are parallel?
2. What are the possible outcomes when checking the intersection of two lines in 3D space?
3. What is the significance of the scalar triple product in vector analysis?
4. How can you verify if two planes are parallel?
5. Can two skew lines be made to lie on the same plane?

**Tip:** Remember that if the scalar triple product of three vectors is zero, it implies that the vectors are coplanar.

Show that the lines (x+1)/3 = (y+3)/5 = (z+5)/7 and (x-2)/1 = (y-4)/4 = (z-6)/7 are coplanar. Also, find the equation of the plane containing these lines.

Learn how to verify the coplanarity of two lines and find the equation of the plane containing them. This problem covers key vector algebra concepts, including the scalar triple product, cross product, and vector form of lines in 3D space.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation