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## Solution:
Given:
- A point [tex]\( P(-1, 3, 2) \)[/tex]
- The plane equation [tex]\( x + 2y + 2z = 3 \)[/tex]
We need to:
1. Find the equation of the line through [tex]\((-1, 3, 2)\)[/tex] and perpendicular to the plane.
2. Find the length (distance) of the perpendicular from the point to the plane.
3. Determine the coordinates of the foot of the perpendicular.
### 1. Equation of the Line
The normal to the given plane [tex]\( x + 2y + 2z = 3 \)[/tex] provides the direction vector for the line. The normal vector of the plane can be derived from its coefficients:
[tex]\[ \mathbf{n} = (1, 2, 2) \][/tex]
Thus, the direction vector [tex]\( \mathbf{d} \)[/tex] of the line perpendicular to the plane is:
[tex]\[ \mathbf{d} = (1, 2, 2) \][/tex]
Using the equation of a line in parametric form, through point [tex]\( P(-1, 3, 2) \)[/tex] and in the direction of [tex]\( \mathbf{d} \)[/tex]:
[tex]\[ \vec{r} = \vec{r_0} + t \mathbf{d} \][/tex]
where [tex]\( \vec{r_0} = (-1, 3, 2) \)[/tex] and [tex]\( \mathbf{d} = (1, 2, 2) \)[/tex]. Thus, the parametric equations of the line are:
[tex]\[ \begin{cases} x = -1 + t \\ y = 3 + 2t \\ z = 2 + 2t \end{cases} \][/tex]
### 2. Distance from the Point to the Plane
The formula to find the distance [tex]\( d \)[/tex] from a point [tex]\( (x_1, y_1, z_1) \)[/tex] to a plane [tex]\( Ax + By + Cz = D \)[/tex] is:
[tex]\[ d = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C^2}} \][/tex]
Substituting the given point [tex]\((-1, 3, 2)\)[/tex] and the plane equation [tex]\( x + 2y + 2z = 3 \)[/tex]:
- Coefficients [tex]\( A = 1 \)[/tex], [tex]\( B = 2 \)[/tex], [tex]\( C = 2 \)[/tex], and [tex]\( D = 3 \)[/tex]
- Point coordinates [tex]\( x_1 = -1 \)[/tex], [tex]\( y_1 = 3 \)[/tex], [tex]\( z_1 = 2 \)[/tex]
Calculate the numerator:
[tex]\[ |1(-1) + 2(3) + 2(2) - 3| = | -1 + 6 + 4 - 3 | = |6| = 6 \][/tex]
Calculate the denominator:
[tex]\[ \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \][/tex]
Thus, the distance [tex]\( d \)[/tex] is:
[tex]\[ d = \frac{6}{3} = 2 \][/tex]
### 3. Coordinates of the Foot of the Perpendicular
The foot of the perpendicular from point [tex]\( P(x_1, y_1, z_1) \)[/tex] to the plane [tex]\( Ax + By + Cz = D \)[/tex] has coordinates [tex]\( (x_0, y_0, z_0) \)[/tex] given by:
[tex]\[ \begin{aligned} x_0 &= x_1 - \frac{A(Ax_1 + By_1 + Cz_1 - D)}{A^2 + B^2 + C^2} \\ y_0 &= y_1 - \frac{B(Ax_1 + By_1 + Cz_1 - D)}{A^2 + B^2 + C^2} \\ z_0 &= z_1 - \frac{C(Ax_1 + By_1 + Cz_1 - D)}{A^2 + B^2 + C^2} \end{aligned} \][/tex]
Plugging in the values:
[tex]\[ \begin{aligned} x_0 &= -1 - \frac{1(6)}{9} = -1 - \frac{6}{9} = -1 - \frac{2}{3} = -\frac{5}{3} = -1.6667 \\ y_0 &= 3 - \frac{2(6)}{9} = 3 - \frac{12}{9} = 3 - \frac{4}{3} = 3 - 1.3333 = 1.6667 \\ z_0 &= 2 - \frac{2(6)}{9} = 2 - \frac{12}{9} = 2 - \frac{4}{3} = 2 - 1.3333 = 0.6667 \end{aligned} \][/tex]
Thus, the coordinates of the foot of the perpendicular are:
[tex]\[ \left( -1.6667, 1.6667, 0.6667 \right) \][/tex]
### Summary
1. Equation of the line through [tex]\( (-1, 3, 2) \)[/tex] and perpendicular to the plane:
[tex]\[ \begin{cases} x = -1 + t \\ y = 3 + 2t \\ z = 2 + 2t \end{cases} \][/tex]
2. Distance from the point to the plane: [tex]\( 2 \)[/tex] units
3. Coordinates of the foot of the perpendicular:
[tex]\[ \left( -1.6667, 1.6667, 0.6667 \right) \][/tex]
Given:
- A point [tex]\( P(-1, 3, 2) \)[/tex]
- The plane equation [tex]\( x + 2y + 2z = 3 \)[/tex]
We need to:
1. Find the equation of the line through [tex]\((-1, 3, 2)\)[/tex] and perpendicular to the plane.
2. Find the length (distance) of the perpendicular from the point to the plane.
3. Determine the coordinates of the foot of the perpendicular.
### 1. Equation of the Line
The normal to the given plane [tex]\( x + 2y + 2z = 3 \)[/tex] provides the direction vector for the line. The normal vector of the plane can be derived from its coefficients:
[tex]\[ \mathbf{n} = (1, 2, 2) \][/tex]
Thus, the direction vector [tex]\( \mathbf{d} \)[/tex] of the line perpendicular to the plane is:
[tex]\[ \mathbf{d} = (1, 2, 2) \][/tex]
Using the equation of a line in parametric form, through point [tex]\( P(-1, 3, 2) \)[/tex] and in the direction of [tex]\( \mathbf{d} \)[/tex]:
[tex]\[ \vec{r} = \vec{r_0} + t \mathbf{d} \][/tex]
where [tex]\( \vec{r_0} = (-1, 3, 2) \)[/tex] and [tex]\( \mathbf{d} = (1, 2, 2) \)[/tex]. Thus, the parametric equations of the line are:
[tex]\[ \begin{cases} x = -1 + t \\ y = 3 + 2t \\ z = 2 + 2t \end{cases} \][/tex]
### 2. Distance from the Point to the Plane
The formula to find the distance [tex]\( d \)[/tex] from a point [tex]\( (x_1, y_1, z_1) \)[/tex] to a plane [tex]\( Ax + By + Cz = D \)[/tex] is:
[tex]\[ d = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C^2}} \][/tex]
Substituting the given point [tex]\((-1, 3, 2)\)[/tex] and the plane equation [tex]\( x + 2y + 2z = 3 \)[/tex]:
- Coefficients [tex]\( A = 1 \)[/tex], [tex]\( B = 2 \)[/tex], [tex]\( C = 2 \)[/tex], and [tex]\( D = 3 \)[/tex]
- Point coordinates [tex]\( x_1 = -1 \)[/tex], [tex]\( y_1 = 3 \)[/tex], [tex]\( z_1 = 2 \)[/tex]
Calculate the numerator:
[tex]\[ |1(-1) + 2(3) + 2(2) - 3| = | -1 + 6 + 4 - 3 | = |6| = 6 \][/tex]
Calculate the denominator:
[tex]\[ \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \][/tex]
Thus, the distance [tex]\( d \)[/tex] is:
[tex]\[ d = \frac{6}{3} = 2 \][/tex]
### 3. Coordinates of the Foot of the Perpendicular
The foot of the perpendicular from point [tex]\( P(x_1, y_1, z_1) \)[/tex] to the plane [tex]\( Ax + By + Cz = D \)[/tex] has coordinates [tex]\( (x_0, y_0, z_0) \)[/tex] given by:
[tex]\[ \begin{aligned} x_0 &= x_1 - \frac{A(Ax_1 + By_1 + Cz_1 - D)}{A^2 + B^2 + C^2} \\ y_0 &= y_1 - \frac{B(Ax_1 + By_1 + Cz_1 - D)}{A^2 + B^2 + C^2} \\ z_0 &= z_1 - \frac{C(Ax_1 + By_1 + Cz_1 - D)}{A^2 + B^2 + C^2} \end{aligned} \][/tex]
Plugging in the values:
[tex]\[ \begin{aligned} x_0 &= -1 - \frac{1(6)}{9} = -1 - \frac{6}{9} = -1 - \frac{2}{3} = -\frac{5}{3} = -1.6667 \\ y_0 &= 3 - \frac{2(6)}{9} = 3 - \frac{12}{9} = 3 - \frac{4}{3} = 3 - 1.3333 = 1.6667 \\ z_0 &= 2 - \frac{2(6)}{9} = 2 - \frac{12}{9} = 2 - \frac{4}{3} = 2 - 1.3333 = 0.6667 \end{aligned} \][/tex]
Thus, the coordinates of the foot of the perpendicular are:
[tex]\[ \left( -1.6667, 1.6667, 0.6667 \right) \][/tex]
### Summary
1. Equation of the line through [tex]\( (-1, 3, 2) \)[/tex] and perpendicular to the plane:
[tex]\[ \begin{cases} x = -1 + t \\ y = 3 + 2t \\ z = 2 + 2t \end{cases} \][/tex]
2. Distance from the point to the plane: [tex]\( 2 \)[/tex] units
3. Coordinates of the foot of the perpendicular:
[tex]\[ \left( -1.6667, 1.6667, 0.6667 \right) \][/tex]
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