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To find the distance between the point \( Q(3, 4, 1) \) and the given line described by the parametric equations:

\[
x = -4t + 4, \quad y = 3 - t, \quad z = t - 3
\]

we can use the formula for the distance from a point to a line in 3D space:

\[
\text{Distance} = \frac{\| \mathbf{PQ} \times \mathbf{d} \|}{\|\mathbf{d}\|}
\]

where:
- \( \mathbf{P} \) is a point on the line (obtained by plugging any value of \( t \) into the parametric equations; let's use \( t = 0 \) for simplicity).
- \( \mathbf{Q} = (3, 4, 1) \) is the given point.
- \( \mathbf{d} \) is the direction vector of the line (coefficients of \( t \) in the parametric equations).

### Step 1: Identify a point on the line (\( \mathbf{P} \)) and the direction vector (\( \mathbf{d} \)).

For \( t = 0 \):
\[
\mathbf{P} = (x_0, y_0, z_0) = (4, 3, -3)
\]

The direction vector \( \mathbf{d} \) is:
\[
\mathbf{d} = (-4, -1, 1)
\]

### Step 2: Find vector \( \mathbf{PQ} \).

\[
\mathbf{PQ} = \mathbf{Q} - \mathbf{P} = (3 - 4, 4 - 3, 1 - (-3)) = (-1, 1, 4)
\]

### Step 3: Calculate the cross product \( \mathbf{PQ} \times \mathbf{d} \).

\[
\mathbf{PQ} \times \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 1 & 4 \\ -4 & -1 & 1 \end{vmatrix}
\]

\[
= \mathbf{i}(1 \times 1 - 4 \times -1) - \mathbf{j}(-1 \times 1 - 4 \times -4) + \mathbf{k}(-1 \times -1 - 1 \times -4)
\]

\[
= \mathbf{i}(1 + 4) - \mathbf{j}(-1 + 16) + \mathbf{k}(1 + 4)
\]

\[
= \mathbf{i}(5) - \mathbf{j}(15) + \mathbf{k}(5)
\]

\[
\mathbf{PQ} \times \mathbf{d} = (5, -15, 5)
\]

### Step 4: Find the magnitude of \( \mathbf{PQ} \times \mathbf{d} \).

\[
\| \mathbf{PQ} \times \mathbf{d} \| = \sqrt{5^2 + (-15)^2 + 5^2} = \sqrt{25 + 225 + 25} = \sqrt{275} = 5\sqrt{11}
\]

### Step 5: Find the magnitude of \( \mathbf{d} \).

\[
\|\mathbf{d}\| = \sqrt{(-4)^2 + (-1)^2 + 1^2} = \sqrt{16 + 1 + 1} = \sqrt{18} = 3\sqrt{2}
\]

### Step 6: Calculate the distance.

\[
\text{Distance} = \frac{5\sqrt{11}}{3\sqrt{2}} = \frac{5\sqrt{22}}{6}
\]

So, the exact distance is \( \frac{5\sqrt{22}}{6} \).

---

Would you like more details or have any questions about this process? Here are some related questions:

1. How can you use the dot product to check if two vectors are perpendicular?
2. What is the significance of the direction vector in the parametric equation of a line?
3. How do you derive the cross product in a 3D vector space?
4. Can you describe how to convert parametric equations of a line to a symmetric form?
5. What other methods can be used to find the distance between a point and a line in space?

**Tip:** Always verify the vector magnitudes and cross product carefully, as small mistakes can lead to incorrect results in distance calculations.

Find the distance between the given point Q(3, 4, 1) and the line with parametric equations x = -4t + 4, y = 3 - t, z = t - 3.

Learn how to find the distance between a point Q(3, 4, 1) and a line defined by parametric equations in 3D space. This example uses vector algebra, the cross product, and distance formulas to solve the problem. Suitable for students in Grades 11-12.

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