Answer :
To show that the equation [tex]\( z = px + qy + \sqrt{p^2 + q^2 + 1} \)[/tex] represents a family of planes, each at a unit distance from the origin, we will follow these steps:
1. Understand the Given Equation:
The equation of the plane given is:
[tex]\[ z = px + qy + \sqrt{p^2 + q^2 + 1} \][/tex]
We need to rewrite this in the standard form of a plane in 3D space: [tex]\( Ax + By + Cz + D = 0 \)[/tex].
2. Rewrite the Equation:
To convert it into the standard form, we rearrange as follows:
[tex]\[ z - px - qy = \sqrt{p^2 + q^2 + 1} \][/tex]
Or equivalently,
[tex]\[ -px - qy + z - \sqrt{p^2 + q^2 + 1} = 0 \][/tex]
Here, we identify the coefficients [tex]\( A, B, C, \)[/tex] and [tex]\( D \)[/tex]:
[tex]\[ A = -p, \quad B = -q, \quad C = 1, \quad D = -\sqrt{p^2 + q^2 + 1} \][/tex]
3. Distance from the Origin:
To determine the distance of the plane from the origin, we use the formula for the distance [tex]\( D \)[/tex] from a point [tex]\( (x_0, y_0, z_0) \)[/tex] (in this case, the origin [tex]\((0, 0, 0)) \)[/tex] to the plane [tex]\( Ax + By + Cz + D = 0 \)[/tex]:
[tex]\[ \text{Distance} = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \][/tex]
Plugging in the values [tex]\( A = -p, B = -q, C = 1, \)[/tex] and [tex]\( D = -\sqrt{p^2 + q^2 + 1} \)[/tex], we get:
[tex]\[ \text{Distance} = \frac{|- \sqrt{p^2 + q^2 + 1}|}{\sqrt{(-p)^2 + (-q)^2 + 1^2}} \][/tex]
4. Simplify the Expression:
First, simplify the numerator:
[tex]\[ |-\sqrt{p^2 + q^2 + 1}| = \sqrt{p^2 + q^2 + 1} \][/tex]
Now, simplify the denominator:
[tex]\[ \sqrt{(-p)^2 + (-q)^2 + 1^2} = \sqrt{p^2 + q^2 + 1} \][/tex]
Therefore, the distance expression becomes:
[tex]\[ \text{Distance} = \frac{\sqrt{p^2 + q^2 + 1}}{\sqrt{p^2 + q^2 + 1}} \][/tex]
5. Calculate the Final Distance:
By simplifying the fraction, we get:
[tex]\[ \text{Distance} = \frac{\sqrt{p^2 + q^2 + 1}}{\sqrt{p^2 + q^2 + 1}} = 1 \][/tex]
Thus, the distance from the origin to the plane described by the equation [tex]\( z = px + qy + \sqrt{p^2 + q^2 + 1} \)[/tex] is always 1 unit. Hence, the given equation represents a family of planes, each of which is at a unit distance from the origin.
1. Understand the Given Equation:
The equation of the plane given is:
[tex]\[ z = px + qy + \sqrt{p^2 + q^2 + 1} \][/tex]
We need to rewrite this in the standard form of a plane in 3D space: [tex]\( Ax + By + Cz + D = 0 \)[/tex].
2. Rewrite the Equation:
To convert it into the standard form, we rearrange as follows:
[tex]\[ z - px - qy = \sqrt{p^2 + q^2 + 1} \][/tex]
Or equivalently,
[tex]\[ -px - qy + z - \sqrt{p^2 + q^2 + 1} = 0 \][/tex]
Here, we identify the coefficients [tex]\( A, B, C, \)[/tex] and [tex]\( D \)[/tex]:
[tex]\[ A = -p, \quad B = -q, \quad C = 1, \quad D = -\sqrt{p^2 + q^2 + 1} \][/tex]
3. Distance from the Origin:
To determine the distance of the plane from the origin, we use the formula for the distance [tex]\( D \)[/tex] from a point [tex]\( (x_0, y_0, z_0) \)[/tex] (in this case, the origin [tex]\((0, 0, 0)) \)[/tex] to the plane [tex]\( Ax + By + Cz + D = 0 \)[/tex]:
[tex]\[ \text{Distance} = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \][/tex]
Plugging in the values [tex]\( A = -p, B = -q, C = 1, \)[/tex] and [tex]\( D = -\sqrt{p^2 + q^2 + 1} \)[/tex], we get:
[tex]\[ \text{Distance} = \frac{|- \sqrt{p^2 + q^2 + 1}|}{\sqrt{(-p)^2 + (-q)^2 + 1^2}} \][/tex]
4. Simplify the Expression:
First, simplify the numerator:
[tex]\[ |-\sqrt{p^2 + q^2 + 1}| = \sqrt{p^2 + q^2 + 1} \][/tex]
Now, simplify the denominator:
[tex]\[ \sqrt{(-p)^2 + (-q)^2 + 1^2} = \sqrt{p^2 + q^2 + 1} \][/tex]
Therefore, the distance expression becomes:
[tex]\[ \text{Distance} = \frac{\sqrt{p^2 + q^2 + 1}}{\sqrt{p^2 + q^2 + 1}} \][/tex]
5. Calculate the Final Distance:
By simplifying the fraction, we get:
[tex]\[ \text{Distance} = \frac{\sqrt{p^2 + q^2 + 1}}{\sqrt{p^2 + q^2 + 1}} = 1 \][/tex]
Thus, the distance from the origin to the plane described by the equation [tex]\( z = px + qy + \sqrt{p^2 + q^2 + 1} \)[/tex] is always 1 unit. Hence, the given equation represents a family of planes, each of which is at a unit distance from the origin.