Question

1. Find the distance of the point P1(−1, 3, 0) from the line with parametric equation
L : x = 1, y = 1 + 3t, z = −1 + 2t
2. Find the distance from the point P(1, 2, 3) to the plane π : 3x + 5y − 4z + 37 = 0
3. Find the distance between the planes π1 : x + 2y − 2z = 3 and π2 : 2x + 4y + −4z = 7

Answer 1

Let's solve each of the three problems step-by-step.

### Problem 1: Distance from Point to Line

The point \( P_1(-1, 3, 0) \) and the line \( L : x = 1, y = 1 + 3t, z = -1 + 2t \).

1. **Find a point on the line**: Let's take \( t = 0 \) to find a point on the line:
\[
P_L(1, 1 + 3(0), -1 + 2(0)) = P_L(1, 1, -1)
\]

2. **Calculate the direction vector of the line**: From the parametric equations, the direction vector \( \mathbf{v} \) of the line is:
\[
\mathbf{v} = (0, 3, 2)
\]

3. **Find the vector from the point \( P_1 \) to point \( P_L \)**:
\[
\mathbf{P_1P_L} = P_L - P_1 = (1 - (-1), 1 - 3, -1 - 0) = (2, -2, -1)
\]

4. **Calculate the cross product of \( \mathbf{P_1P_L} \) and \( \mathbf{v} \)**:
\[
\mathbf{P_1P_L} \times \mathbf{v} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & -2 & -1 \\
0 & 3 & 2
\end{vmatrix}
= \hat{i}((-2)(2) - (-1)(3)) - \hat{j}((2)(2) - (0)(-1)) + \hat{k}((2)(3) - (0)(-2))
\]
\[
= \hat{i}(-4 + 3) - \hat{j}(4 - 0) + \hat{k}(6 - 0) = \hat{i}(-1) - \hat{j}(4) + \hat{k}(6)
\]
\[
= (-1, -4, 6)
\]

5. **Find the magnitude of the cross product**:
\[
\|\mathbf{P_1P_L} \times \mathbf{v}\| = \sqrt{(-1)^2 + (-4)^2 + 6^2} = \sqrt{1 + 16 + 36} = \sqrt{53}
\]

6. **Find the magnitude of the direction vector** \( \mathbf{v} \):
\[
\|\mathbf{v}\| = \sqrt{0^2 + 3^2 + 2^2} = \sqrt{0 + 9 + 4} = \sqrt{13}
\]

7. **Calculate the distance from point to line**:
\[
d = \frac{\|\mathbf{P_1P_L} \times \mathbf{v}\|}{\|\mathbf{v}\|} = \frac{\sqrt{53}}{\sqrt{13}} = \sqrt{\frac{53}{13}} = \sqrt{4.0769} \approx 2.018
\]

### Problem 2: Distance from Point to Plane

The point \( P(1, 2, 3) \) and the plane \( \pi : 3x + 5y − 4z + 37 = 0 \).

The formula for the distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is:
\[
d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]

Here, \( A = 3, B = 5, C = -4, D = 37 \) and \( (x_0, y_0, z_0) = (1, 2, 3) \):
\[
d = \frac{|3(1) + 5(2) - 4(3) + 37|}{\sqrt{3^2 + 5^2 + (-4)^2}}
\]
\[
= \frac{|3 + 10 - 12 + 37|}{\sqrt{9 + 25 + 16}} = \frac{|38|}{\sqrt{50}} = \frac{38}{5\sqrt{2}} = \frac{19\sqrt{2}}{5} \approx 5.364
\]

### Problem 3: Distance Between Two Planes

The planes \( \pi_1 : x + 2y - 2z = 3 \) and \( \pi_2 : 2x + 4y - 4z = 7 \).

1. **Convert \( \pi_2 \) to a comparable form**:
Dividing \( \pi_2 \) by 2 gives:
\[
x + 2y - 2z = \frac{7}{2}
\]

2. **The normal vectors for both planes are the same**:
\[
\mathbf{n} = (1, 2, -2)
\]

3. **Use the distance formula for parallel planes**:
The distance \( d \) between the planes \( Ax + By + Cz = D_1 \) and \( Ax + By + Cz = D_2 \) is:
\[
d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}}
\]
Here, \( D_1 = 3 \) and \( D_2 = \frac{7}{2} \):
\[
d = \frac{\left| \frac{7}{2} - 3 \right|}{\sqrt{1^2 + 2^2 + (-2)^2}} = \frac{\left| \frac{7}{2} - \frac{6}{2} \right|}{\sqrt{1 + 4 + 4}} = \frac{\left| \frac{1}{2} \right|}{\sqrt{9}} = \frac{\frac{1}{2}}{3} = \frac{1}{6}
\]

So, the final results are:
1. Distance to line: \( \approx 2.018 \)
2. Distance to plane: \( \frac{19\sqrt{2}}{5} \approx 5.364 \)
3. Distance between planes: \( \frac{1}{6} \)