To find the distance from the point [tex]\( (2, 7, 8) \)[/tex] to the plane [tex]\( 2y + 2z = 0 \)[/tex], we use the formula for the distance from a point to a plane in three-dimensional space. The standard form of a plane equation is [tex]\( Ax + By + Cz + D = 0 \)[/tex].
First, we rewrite the given plane equation in standard form:
[tex]\[ 2y + 2z = 0 \][/tex]
This can be seen as:
[tex]\[ 0x + 2y + 2z + 0 = 0 \][/tex]
From this, we identify the coefficients:
- [tex]\( A = 0 \)[/tex]
- [tex]\( B = 2 \)[/tex]
- [tex]\( C = 2 \)[/tex]
- [tex]\( D = 0 \)[/tex]
The point provided is [tex]\( (x_1, y_1, z_1) = (2, 7, 8) \)[/tex].
The distance [tex]\( d \)[/tex] from a point [tex]\( (x_1, y_1, z_1) \)[/tex] to a plane given by [tex]\( Ax + By + Cz + D = 0 \)[/tex] is calculated using the formula:
[tex]\[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \][/tex]
Substitute the values into the formula:
1. Calculate the numerator:
[tex]\[ |A \cdot x_1 + B \cdot y_1 + C \cdot z_1 + D| = |0 \cdot 2 + 2 \cdot 7 + 2 \cdot 8 + 0| \][/tex]
[tex]\[ = |0 + 14 + 16 + 0| \][/tex]
[tex]\[ = |30| \][/tex]
[tex]\[ = 30 \][/tex]
2. Calculate the denominator:
[tex]\[ \sqrt{A^2 + B^2 + C^2} = \sqrt{0^2 + 2^2 + 2^2} \][/tex]
[tex]\[ = \sqrt{0 + 4 + 4} \][/tex]
[tex]\[ = \sqrt{8} \][/tex]
[tex]\[ = \sqrt{4 \cdot 2} \][/tex]
[tex]\[ = 2 \sqrt{2} \approx 2.8284271247461903 \][/tex]
3. Calculate the distance:
[tex]\[ d = \frac{30}{2.8284271247461903} \][/tex]
[tex]\[ \approx 10.606601717798211 \][/tex]
Therefore, the distance from the point [tex]\( (2, 7, 8) \)[/tex] to the plane [tex]\( 2y + 2z = 0 \)[/tex] is approximately [tex]\( 10.606601717798211 \)[/tex].
