Find the distance between the skew lines $$ \vec{r}(t)=\langle-1,4,-5\rangle t+\langle 6,0,-6\rangle $$ anc $$ \vec{p}(s)=\langle 0,2,5\rangle s+\langle-4,-3,-7\rangle $$.

Question

UpStudy AI · Accepted Answer

To find the distance between the skew lines given by the vector equations

$$
\vec{r}(t) = \langle -1, 4, -5 \rangle t + \langle 6, 0, -6 \rangle
$$

and

$$
\vec{p}(s) = \langle 0, 2, 5 \rangle s + \langle -4, -3, -7 \rangle,
$$

we can follow these steps:

1. **Identify direction vectors and points on the lines:**
   - The direction vector of line $$ \vec{r}(t) $$ is $$ \vec{d_1} = \langle -1, 4, -5 \rangle $$.
   - The direction vector of line $$ \vec{p}(s) $$ is $$ \vec{d_2} = \langle 0, 2, 5 \rangle $$.
   - A point on line $$ \vec{r}(t) $$ is $$ \vec{a} = \langle 6, 0, -6 \rangle $$.
   - A point on line $$ \vec{p}(s) $$ is $$ \vec{b} = \langle -4, -3, -7 \rangle $$.

2. **Find the vector connecting the two points:**
   $$
   \vec{b} - \vec{a} = \langle -4, -3, -7 \rangle - \langle 6, 0, -6 \rangle = \langle -10, -3, -1 \rangle.
   $$

3. **Calculate the cross product of the direction vectors:**
   $$
   \vec{d_1} \times \vec{d_2} = \begin{vmatrix}
   \hat{i} & \hat{j} & \hat{k} \
   -1 & 4 & -5 \
   0 & 2 & 5
   \end{vmatrix}.
   $$

Calculating the determinant:
   $$
   \vec{d_1} \times \vec{d_2} = \hat{i} \begin{vmatrix} 4 & -5 \ 2 & 5 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & -5 \ 0 & 5 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 4 \ 0 & 2 \end{vmatrix}.
   $$

Evaluating the determinants:
   $$
   = \hat{i} (4 \cdot 5 - (-5) \cdot 2) - \hat{j} (-1 \cdot 5 - 0) + \hat{k} (-1 \cdot 2 - 0).
   $$
   $$
   = \hat{i} (20 + 10) - \hat{j} (-5) + \hat{k} (-2).
   $$
   $$
   = \hat{i} (30) + \hat{j} (5) - \hat{k} (2) = \langle 30, 5, -2 \rangle.
   $$

4. **Calculate the magnitude of the cross product:**
   $$
   |\vec{d_1} \times \vec{d_2}| = \sqrt{30^2 + 5^2 + (-2)^2} = \sqrt{900 + 25 + 4} = \sqrt{929}.
   $$

5. **Calculate the distance between the skew lines:**
   The distance $$ D $$ between the skew lines is given by the formula:
   $$
   D = \frac{|(\vec{b} - \vec{a}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}.
   $$

First, calculate the dot product:
   $$
   (\vec{b} - \vec{a}) \cdot (\vec{d_1} \times \vec{d_2}) = \langle -10, -3, -1 \rangle \cdot \langle 30, 5, -2 \rangle.
   $$
   $$
   = -10 \cdot 30 + (-3) \cdot 5 + (-1) \cdot (-2) = -300 - 15 + 2 = -313.
   $$

Now, calculate the distance:
   $$
   D = \frac{|-313|}{\sqrt{929}} = \frac{313}{\sqrt{929}}.
   $$

Thus, the distance between the skew lines is

$$
\frac{313}{\sqrt{929}}.
$$
The distance between the skew lines is $$ \frac{313}{\sqrt{929}} $$.

Find the distance between the skew lines anc
.

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Step 1: Determine Direction Vectors

Step 2: Find the Vector Between Two Points on Each Line

Step 3: Construct the Vector Between Points A and B

Step 4: Use the Cross Product to Find the Direction of Minimum Distance

Step 5: Magnitude of the Cross Product

Step 6: Using the Formula for the Distance Between Skew Lines

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Find the distance between the skew lines anc .