Find the distance of the point (4,4,-4)
 from the line r(t)=<-1+4t,-1+3t,3-3t>

To find the distance of the point \((4, 4, -4)\) from the line represented by the vector equation:

\[
r(t) = \langle -1 + 4t, -1 + 3t, 3 - 3t \rangle
\]

we use the formula for the distance \(d\) from a point \(P(x_1, y_1, z_1)\) to a line:

\[
d = \frac{|\vec{PQ} \times \vec{v}|}{|\vec{v}|}
\]

Where:
- \(P(x_1, y_1, z_1)\) is the point (4, 4, -4).
- \(\vec{v}\) is the direction vector of the line, which we get from the parametric form of the line: \( \vec{v} = \langle 4, 3, -3 \rangle \).
- \(Q(x_0, y_0, z_0)\) is any point on the line. We can take \(t = 0\) to get the point \(Q = (-1, -1, 3)\).

Now we compute the distance step by step.

### Step 1: Find vector \( \vec{PQ} \)
The vector \( \vec{PQ} \) is the vector from the point \(Q(-1, -1, 3)\) to the point \(P(4, 4, -4)\).

\[
\vec{PQ} = \langle 4 - (-1), 4 - (-1), -4 - 3 \rangle = \langle 5, 5, -7 \rangle
\]

### Step 2: Compute the cross product \( \vec{PQ} \times \vec{v} \)

We now compute the cross product of \( \vec{PQ} = \langle 5, 5, -7 \rangle \) and \( \vec{v} = \langle 4, 3, -3 \rangle \).

\[
\vec{PQ} \times \vec{v} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
5 & 5 & -7 \\
4 & 3 & -3 \\
\end{vmatrix}
\]

This determinant gives:

\[
\vec{PQ} \times \vec{v} = \mathbf{i}(5(-3) - (-7)(3)) - \mathbf{j}(5(-3) - (-7)(4)) + \mathbf{k}(5(3) - 5(4))
\]

\[
= \mathbf{i}(-15 + 21) - \mathbf{j}(-15 + 28) + \mathbf{k}(15 - 20)
\]

\[
= \mathbf{i}(6) - \mathbf{j}(13) + \mathbf{k}(-5)
\]

\[
= \langle 6, -13, -5 \rangle
\]

### Step 3: Find the magnitude of the cross product \( |\vec{PQ} \times \vec{v}| \)

The magnitude of the vector \( \vec{PQ} \times \vec{v} = \langle 6, -13, -5 \rangle \) is:

\[
|\vec{PQ} \times \vec{v}| = \sqrt{6^2 + (-13)^2 + (-5)^2} = \sqrt{36 + 169 + 25} = \sqrt{230}
\]

### Step 4: Find the magnitude of \( \vec{v} \)

The magnitude of \( \vec{v} = \langle 4, 3, -3 \rangle \) is:

\[
|\vec{v}| = \sqrt{4^2 + 3^2 + (-3)^2} = \sqrt{16 + 9 + 9} = \sqrt{34}
\]

### Step 5: Compute the distance \(d\)

Finally, the distance is:

\[
d = \frac{|\vec{PQ} \times \vec{v}|}{|\vec{v}|} = \frac{\sqrt{230}}{\sqrt{34}} = \sqrt{\frac{230}{34}} = \sqrt{6.76} \approx 2.6
\]

Thus, the distance from the point \((4, 4, -4)\) to the line is approximately \(2.6\) units.

Would you like more details or have any questions? Here are some related questions for further exploration:

1. How do you find the equation of a line given two points in 3D?
2. What is the formula for the shortest distance between two parallel lines?
3. How do you calculate the angle between two vectors in 3D?
4. Can we find the closest point on a line to a given point?
5. What are other methods to compute the distance between a point and a line in 3D?

**Tip**: In problems involving distance, cross products are especially useful when the perpendicular distance from a point to a line or plane is needed.

Find the distance of the point (4, 4, -4) from the line r(t)=<-1+4t,-1+3t,3-3t>

Learn how to calculate the distance from the point (4, 4, -4) to the line r(t) = <-1+4t, -1+3t, 3-3t> in 3D space. This solution explores vector algebra, cross products, and the distance formula. Suitable for high school and college students.

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