Questions: Help Entering Answers (1 point) The distance between a point P and a line: r(t)=r0+tv can be given by: d=PR x v / v Where R is any point on the line r(t) (in particular it can be r0 ) Find the distances between the line r(t)=<-1-5t,-3-t, 3+4t> and points: 1. (5,-4,-5): 2. (2,-5,-10): 3. (4,-2,-8):

Help Entering Answers
(1 point) The distance between a point P and a line: r(t)=r0+tv can be given by:
d=PR x v / v

Where R is any point on the line r(t) (in particular it can be r0 )
Find the distances between the line r(t)=<-1-5t,-3-t, 3+4t> and points:
1. (5,-4,-5):
2. (2,-5,-10):
3. (4,-2,-8):

Transcript text: Help Entering Answers (1 point) The distance between a point $P$ and a line: $\mathbf{r}(t)=\mathbf{r}_{0}+t \mathrm{v}$ can be given by: \[ d=\frac{|\overrightarrow{P R} \times \mathrm{v}|}{|\mathrm{v}|} \] Where $R$ is any point on the line $\mathbf{r}(t)$ (in particular it can be $\mathbf{r}_{0}$ ) Find the distances between the line $\mathbf{r}(t)=\langle-1-5 t,-3-t, 3+4 t\rangle$ and points: 1. $(5,-4,-5):$ $\square$ 2. $(2,-5,-10):$ $\square$ 3. $(4,-2,-8):$ $\square$

Solution

Solution Steps

To find the distance between a point $ P $ and a line given by the vector equation $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}$, we use the formula:

\[ d = \frac{|\overrightarrow{PR} \times \mathbf{v}|}{|\mathbf{v}|} \]

Identify $\mathbf{r}_0$ and $\mathbf{v}$ from the line equation.
For each point $ P $, calculate the vector $\overrightarrow{PR}$ where $ R = \mathbf{r}_0 $.
Compute the cross product $\overrightarrow{PR} \times \mathbf{v}$.
Calculate the magnitude of the cross product and the magnitude of $\mathbf{v}$.
Use the formula to find the distance.

Step 1: Identify Line Parameters

The line is given by the vector equation $\mathbf{r}(t) = \langle -1 - 5t, -3 - t, 3 + 4t \rangle$. From this, we identify:

A point on the line, $\mathbf{r}_0 = \langle -1, -3, 3 \rangle$.
The direction vector, $\mathbf{v} = \langle -5, -1, 4 \rangle$.

Step 2: Calculate Distance for Each Point

For each point $ P $, we calculate the distance to the line using the formula:

\[ d = \frac{|\overrightarrow{PR} \times \mathbf{v}|}{|\mathbf{v}|} \]

where $ R = \mathbf{r}_0 $.

Point 1: $ (5, -4, -5) $

Calculate $\overrightarrow{PR} = \langle 5 - (-1), -4 - (-3), -5 - 3 \rangle = \langle 6, -1, -8 \rangle$.
Compute the cross product $\overrightarrow{PR} \times \mathbf{v}$.
Calculate the magnitudes and use the formula to find $ d \approx 3.522 $.

Point 2: $ (2, -5, -10) $

Calculate $\overrightarrow{PR} = \langle 2 - (-1), -5 - (-3), -10 - 3 \rangle = \langle 3, -2, -13 \rangle$.
Compute the cross product $\overrightarrow{PR} \times \mathbf{v}$.
Calculate the magnitudes and use the formula to find $ d \approx 9.022 $.

Point 3: $ (4, -2, -8) $

Calculate $\overrightarrow{PR} = \langle 4 - (-1), -2 - (-3), -8 - 3 \rangle = \langle 5, 1, -11 \rangle$.
Compute the cross product $\overrightarrow{PR} \times \mathbf{v}$.
Calculate the magnitudes and use the formula to find $ d \approx 5.508 $.