https://file.aiquickdraw.com/mathbot/file/1730164186427vef88jgj.jpg

To solve question 21, we need to find the shortest distance between the two given lines. The lines are provided in vector form as follows:

\[
\vec{r_1} = \vec{a_1} + \lambda \vec{b_1} = -\hat{i} + 4\hat{j} + 3\hat{k} + \lambda (4\hat{i} + \hat{j} + 4\hat{k})
\]
\[
\vec{r_2} = \vec{a_2} + \mu \vec{b_2} = -4\hat{i} + \hat{j} + 4\hat{k} + \mu (5\hat{i} + 2\hat{j} + \hat{k})
\]

To find the shortest distance \(d\) between these skew lines, we use the formula:

\[
d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}
\]

### Step 1: Define \( \vec{a_1} \), \( \vec{a_2} \), \( \vec{b_1} \), and \( \vec{b_2} \)
From the given equations:
- \( \vec{a_1} = -\hat{i} + 4\hat{j} + 3\hat{k} \)
- \( \vec{a_2} = -4\hat{i} + \hat{j} + 4\hat{k} \)
- \( \vec{b_1} = 4\hat{i} + \hat{j} + 4\hat{k} \)
- \( \vec{b_2} = 5\hat{i} + 2\hat{j} + \hat{k} \)

### Step 2: Find \( \vec{a_2} - \vec{a_1} \)
\[
\vec{a_2} - \vec{a_1} = (-4\hat{i} + \hat{j} + 4\hat{k}) - (-\hat{i} + 4\hat{j} + 3\hat{k})
\]
\[
= (-4 + 1)\hat{i} + (1 - 4)\hat{j} + (4 - 3)\hat{k} = -3\hat{i} - 3\hat{j} + \hat{k}
\]

### Step 3: Compute \( \vec{b_1} \times \vec{b_2} \)
\[
\vec{b_1} = 4\hat{i} + \hat{j} + 4\hat{k}
\]
\[
\vec{b_2} = 5\hat{i} + 2\hat{j} + \hat{k}
\]

Using the cross product formula:
\[
\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 4 \\ 5 & 2 & 1 \end{vmatrix}
\]

Expanding the determinant:
\[
= \hat{i}(1 \cdot 1 - 4 \cdot 2) - \hat{j}(4 \cdot 1 - 4 \cdot 5) + \hat{k}(4 \cdot 2 - 1 \cdot 5)
\]
\[
= \hat{i}(1 - 8) - \hat{j}(4 - 20) + \hat{k}(8 - 5)
\]
\[
= -7\hat{i} + 16\hat{j} + 3\hat{k}
\]

So, \( \vec{b_1} \times \vec{b_2} = -7\hat{i} + 16\hat{j} + 3\hat{k} \).

### Step 4: Calculate \( |(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})| \)
\[
\vec{a_2} - \vec{a_1} = -3\hat{i} - 3\hat{j} + \hat{k}
\]
\[
\vec{b_1} \times \vec{b_2} = -7\hat{i} + 16\hat{j} + 3\hat{k}
\]

Dot product:
\[
(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-3)(-7) + (-3)(16) + (1)(3)
\]
\[
= 21 - 48 + 3 = -24
\]

So, \( |(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})| = 24 \).

### Step 5: Calculate \( |\vec{b_1} \times \vec{b_2}| \)
\[
|\vec{b_1} \times \vec{b_2}| = \sqrt{(-7)^2 + 16^2 + 3^2}
\]
\[
= \sqrt{49 + 256 + 9} = \sqrt{314}
\]

### Step 6: Calculate the shortest distance \(d\)
\[
d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}
\]
\[
= \frac{24}{\sqrt{314}}
\]

Thus, the shortest distance between the two lines is:

\[
d = \frac{24}{\sqrt{314}}
\]

Would you like a more detailed explanation of each step or have any questions?

---

### Related Questions

1. How can the formula for the shortest distance between skew lines be derived?
2. What are other methods to find distances between skew lines?
3. How does the cross product help in finding perpendicular distances in 3D geometry?
4. What are practical applications of finding distances between lines in 3D space?
5. How can vector projections be used to solve similar problems?

### Tip
When calculating distances in 3D, always ensure vectors are correctly defined and cross products are accurately computed for precise results.

Solve question 21: Show that the shortest distance between the lines \(\vec{r} = -\hat{i} + 4\hat{j} + 3\hat{k} + \lambda (4\hat{i} + \hat{j} + 4\hat{k})\) and \(\vec{r} = -4\hat{i} + \hat{j} + 4\hat{k} + \mu (5\hat{i} + 2\hat{j} + \hat{k})\) is 24/√314.

Learn to find the shortest distance between two skew lines given in vector form using cross and dot products. This solution walks through a step-by-step process, perfect for Grade 11-12 students studying vector algebra.

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