Let's solve each problem in turn.

Problem 71

Objective: Find the distance from the point $(1, -2, 4)$ to the plane $3x + 2y + 6z = 5$.

The formula for the distance $D$ from a point $(x_1, y_1, z_1)$ to a plane $Ax + By + Cz + D = 0$ is: $D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$

Steps:

1. Rewrite the plane equation $3x + 2y + 6z = 5$ in the form $Ax + By + Cz + D = 0$: $3x + 2y + 6z - 5 = 0$ Here, $A = 3$, $B = 2$, $C = 6$, and $D = -5$.

2. Substitute the point $(1, -2, 4)$ into the formula: $D = \frac{|3(1) + 2(-2) + 6(4) - 5|}{\sqrt{3^2 + 2^2 + 6^2}}$

3. Calculate the numerator: $3(1) + 2(-2) + 6(4) - 5 = 3 - 4 + 24 - 5 = 18$

4. Calculate the denominator: $\sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$

5. Now, find the distance: $D = \frac{18}{7}$

So, the distance from the point $(1, -2, 4)$ to the plane $3x + 2y + 6z = 5$ is $\frac{18}{7}$.

Problem 73

Objective: Find the distance between the parallel planes $2x - 3y + z = 4$ and $4x - 6y + 2z = 3$.

1. First, observe that the second plane equation can be simplified by dividing by 2: $4x - 6y + 2z = 3 \Rightarrow 2x - 3y + z = \frac{3}{2}$ So, the two planes in simplified form are:

   • $2x - 3y + z = 4$
   • $2x - 3y + z = \frac{3}{2}$

2. Since they are parallel, we can use the formula for the distance between two parallel planes $Ax + By + Cz = D_1$ and $Ax + By + Cz = D_2$: $D = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$ Here, $A = 2$, $B = -3$, $C = 1$, $D_1 = 4$, and $D_2 = \frac{3}{2}$.

3. Calculate the numerator: $|D_1 - D_2| = \left|4 - \frac{3}{2}\right| = \left|\frac{8 - 3}{2}\right| = \frac{5}{2}$

4. Calculate the denominator: $\sqrt{A^2 + B^2 + C^2} = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$

5. Now, find the distance: $D = \frac{\frac{5}{2}}{\sqrt{14}} = \frac{5}{2\sqrt{14}} = \frac{5\sqrt{14}}{28}$

Thus, the distance between the two parallel planes is $\frac{5\sqrt{14}}{28}$.

Would you like further details or have any questions on these steps?

Related Questions:

1. How would you find the distance from a point to a line in 3D space?
2. Can you explain why the formula for distance between two parallel planes works?
3. What happens if the planes in problem 73 are not parallel? How would we approach it?
4. How do you derive the formula for the distance from a point to a plane?
5. What are the different ways to check if two planes are parallel?

Tip: For problems involving distance in geometry, double-check that all planes and points are correctly placed in their simplified forms, as simplifications can make the math easier and more accurate.