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

The formula for finding the distance \( D \) from a point \( P(x_1, y_1, z_1) \) to a line given by parametric equations:

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

is:

\[
D = \frac{| \mathbf{v} \times \mathbf{PQ} |}{|\mathbf{v}|}
\]

where:
- \( \mathbf{v} = \langle a, b, c \rangle \) is the direction vector of the line,
- \( \mathbf{PQ} = \langle x_1 - x_0, y_1 - y_0, z_1 - z_0 \rangle \) is the vector from a point \( Q(x_0, y_0, z_0) \) on the line to the point \( P \).

### Step 1: Identify relevant components.
- Point: \( P(5, 0, 0) \)
- Line parametric equations:
  \[
  x = 5 + t, \quad y = 2t, \quad z = 24\sqrt{5} + 2t
  \]
  Therefore, the point \( Q(5, 0, 24\sqrt{5}) \) is on the line when \( t = 0 \), and the direction vector \( \mathbf{v} \) is:

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

### Step 2: Compute the vector \( \mathbf{PQ} \).
\[
\mathbf{PQ} = \langle 5 - 5, 0 - 0, 0 - 24\sqrt{5} \rangle = \langle 0, 0, -24\sqrt{5} \rangle
\]

### Step 3: Calculate the cross product \( \mathbf{v} \times \mathbf{PQ} \).
\[
\mathbf{v} \times \mathbf{PQ} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 2 \\ 0 & 0 & -24\sqrt{5} \end{vmatrix}
\]

This results in:
\[
\mathbf{v} \times \mathbf{PQ} = \mathbf{i} \left(2(0) - 2(0)\right) - \mathbf{j} \left(1(0) - 2(0)\right) + \mathbf{k} \left(1(0) - 2(0)\right)
= \mathbf{i}(0) - \mathbf{j}(0) - \mathbf{k}(24\sqrt{5})
\]

So:
\[
\mathbf{v} \times \mathbf{PQ} = \langle 0, 0, -24\sqrt{5} \rangle
\]

### Step 4: Find the magnitude of \( \mathbf{v} \times \mathbf{PQ} \).
\[
|\mathbf{v} \times \mathbf{PQ}| = \sqrt{0^2 + 0^2 + (-24\sqrt{5})^2} = \sqrt{(24\sqrt{5})^2} = 24\sqrt{5}
\]

### Step 5: Find the magnitude of the direction vector \( \mathbf{v} \).
\[
|\mathbf{v}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3
\]

### Step 6: Calculate the distance.
\[
D = \frac{|\mathbf{v} \times \mathbf{PQ}|}{|\mathbf{v}|} = \frac{24\sqrt{5}}{3} = 8\sqrt{5}
\]

Thus, the distance from the point \( (5, 0, 0) \) to the line is \( \boxed{8\sqrt{5}} \).

---

Would you like more details or have any questions? Here are 5 related questions for deeper understanding:

1. How can you find the distance from a point to a plane instead of a line?
2. What is the geometric significance of the cross product in distance calculations?
3. How would you approach the problem if the line were given in vector form instead of parametric form?
4. Can this method be used for finding distances in higher dimensions?
5. How does the direction vector of the line affect the result?

**Tip:** Always remember that the cross product of two vectors gives a vector perpendicular to the plane containing the two vectors. This is key when calculating distances.

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 line with parametric equations x = 5 + t, y = 2t, z = 24√5 + 2t. This solution explores vector algebra, cross product calculations, and the application of 3D geometry formulas. Suitable for students in Grades 10-12.

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