find the distance from the point (5, 0, 0) to the line : x = 5 + t, y = 2t, z = 24(sqroot5) + 2t

To find the distance from a point \((x_1, y_1, z_1)\) to a line defined parametrically by:

\[
x = x_0 + at, \quad y = y_0 + bt, \quad z = z_0 + ct
\]

we can use the formula for the distance between a point and a line in 3D space. The formula is:

\[
d = \frac{| \mathbf{v} \times \mathbf{p_0p_1} |}{|\mathbf{v}|}
\]

Where:
- \(\mathbf{v} = \langle a, b, c \rangle\) is the direction vector of the line.
- \(\mathbf{p_0} = (x_1, y_1, z_1)\) is the given point.
- \(\mathbf{p_1} = (x_0, y_0, z_0)\) is any point on the line (found by using \(t = 0\)).

### Step 1: Determine the direction vector of the line
The line is given as:

\[
x = 5 + t, \quad y = 2t, \quad z = 24\sqrt{5} + 2t
\]

From these parametric equations, the direction vector \(\mathbf{v}\) is:

\[
\mathbf{v} = \langle 1, 2, 2 \rangle
\]

### Step 2: Pick a point on the line
For \(t = 0\), the point \(\mathbf{p_1}\) on the line is:

\[
\mathbf{p_1} = (5, 0, 24\sqrt{5})
\]

### Step 3: Compute the vector \(\mathbf{p_0p_1}\)
Now, the vector from the point \((5, 0, 0)\) to \(\mathbf{p_1} = (5, 0, 24\sqrt{5})\) is:

\[
\mathbf{p_0p_1} = \langle 5 - 5, 0 - 0, 0 - 24\sqrt{5} \rangle = \langle 0, 0, -24\sqrt{5} \rangle
\]

### Step 4: Compute the cross product \(\mathbf{v} \times \mathbf{p_0p_1}\)
The cross product of the vectors \(\mathbf{v} = \langle 1, 2, 2 \rangle\) and \(\mathbf{p_0p_1} = \langle 0, 0, -24\sqrt{5} \rangle\) is:

\[
\mathbf{v} \times \mathbf{p_0p_1} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & 2 \\
0 & 0 & -24\sqrt{5}
\end{vmatrix}
\]

This determinant evaluates to:

\[
\mathbf{v} \times \mathbf{p_0p_1} = \mathbf{i}(2 \cdot -24\sqrt{5} - 2 \cdot 0) - \mathbf{j}(1 \cdot -24\sqrt{5} - 2 \cdot 0) + \mathbf{k}(1 \cdot 0 - 0 \cdot 2)
\]

\[
= \mathbf{i}(-48\sqrt{5}) - \mathbf{j}(-24\sqrt{5}) + \mathbf{k}(0)
= \langle -48\sqrt{5}, 24\sqrt{5}, 0 \rangle
\]

### Step 5: Compute the magnitude of the cross product
The magnitude of \(\mathbf{v} \times \mathbf{p_0p_1}\) is:

\[
|\mathbf{v} \times \mathbf{p_0p_1}| = \sqrt{(-48\sqrt{5})^2 + (24\sqrt{5})^2 + 0^2}
\]

\[
= \sqrt{(48\sqrt{5})^2 + (24\sqrt{5})^2}
= \sqrt{23040 + 5760}
= \sqrt{28800}
= 120\sqrt{2}
\]

### Step 6: Compute the magnitude of \(\mathbf{v}\)
The magnitude of the direction vector \(\mathbf{v} = \langle 1, 2, 2 \rangle\) is:

\[
|\mathbf{v}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3
\]

### Step 7: Compute the distance
Finally, the distance is:

\[
d = \frac{|\mathbf{v} \times \mathbf{p_0p_1}|}{|\mathbf{v}|} = \frac{120\sqrt{2}}{3} = 40\sqrt{2}
\]

Thus, the distance from the point \((5, 0, 0)\) to the line is \(40\sqrt{2}\).

Would you like any further details or have any questions?

---

Here are 5 related questions for further practice:
1. How would the distance change if the direction vector of the line were \(\mathbf{v} = \langle 1, 3, 2 \rangle\)?
2. What is the general formula for the shortest distance between a point and a plane?
3. How can the parametric equation of a line be derived from two points in space?
4. What is the geometric significance of the cross product in this context?
5. How can the vector projection be used to find the distance between a point and a line?

**Tip**: When calculating distances in 3D, always make sure to correctly identify the direction vector and any point on the line for accurate cross-product calculations!

Find the distance from the point (5, 0, 0) to the line: x = 5 + t, y = 2t, z = 24√5 + 2t.

Learn how to find the distance from the point (5, 0, 0) to the parametric line x = 5 + t, y = 2t, z = 24√5 + 2t using vector cross product and geometry. This detailed solution is perfect for students in Grades 11-12 studying 3D geometry.

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