If the distance of the point P (1, - 2, 1) from the plane x + 2y - 2z = a, where a > 0, is 5, then the foot of the perpendicular from P to the plane is?

To find the foot of the perpendicular from the point P(1, -2, 1) to the plane defined by the equation x + 2y - 2z = a, we first need to determine the relationship between the point and the plane. The distance from a point to a plane can be calculated using a specific formula, and in this case, we know that the distance is 5. Let's break this down step by step.

Understanding the Distance Formula

The distance \( d \) from a point \( (x_0, y_0, z_0) \) to a plane given by the equation \( Ax + By + Cz + D = 0 \) is calculated using the formula:

\( d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \)

Identifying the Components

In our case, the plane equation is \( x + 2y - 2z = a \). We can rewrite this in the standard form:

\( x + 2y - 2z - a = 0 \)

Here, we have:

• A = 1
• B = 2
• C = -2
• D = -a

Substituting the Point Coordinates

Now, we substitute the coordinates of point P(1, -2, 1) into the distance formula:

\( d = \frac{|1(1) + 2(-2) - 2(1) - a|}{\sqrt{1^2 + 2^2 + (-2)^2}} \)

This simplifies to:

\( d = \frac{|1 - 4 - 2 - a|}{\sqrt{1 + 4 + 4}} = \frac{| -5 - a|}{3} \)

Setting Up the Equation

Since we know the distance \( d \) is equal to 5, we can set up the equation:

\( \frac{| -5 - a|}{3} = 5 \)

Multiplying both sides by 3 gives us:

\( | -5 - a| = 15 \)

Solving for 'a'

This absolute value equation leads to two cases:

• Case 1: \( -5 - a = 15 \) → \( a = -20 \) (not valid since \( a > 0 \))
• Case 2: \( -5 - a = -15 \) → \( a = 10 \) (valid)

Finding the Foot of the Perpendicular

Now that we have determined \( a = 10 \), we can find the foot of the perpendicular from point P to the plane. The foot of the perpendicular can be found using the direction ratios of the normal to the plane, which are given by the coefficients of x, y, and z in the plane equation.

Normal Vector and Direction Ratios

The normal vector \( \mathbf{n} \) to the plane is \( (1, 2, -2) \). The foot of the perpendicular \( F(x, y, z) \) can be expressed as:

\( F = P + t \cdot \mathbf{n} \)

Where \( t \) is a scalar that we will determine. The coordinates of point P are (1, -2, 1), so:

\( F = (1 + t, -2 + 2t, 1 - 2t) \)

Substituting into the Plane Equation

Now, we substitute the coordinates of F into the plane equation \( x + 2y - 2z = 10 \):

\( (1 + t) + 2(-2 + 2t) - 2(1 - 2t) = 10 \)

This simplifies to:

\( 1 + t - 4 + 4t - 2 + 4t = 10 \)

Combining like terms gives:

\( 9t - 5 = 10 \)

Solving for \( t \):

\( 9t = 15 \) → \( t = \frac{15}{9} = \frac{5}{3} \)

Calculating the Coordinates of the Foot

Now we can find the coordinates of the foot of the perpendicular:

• x-coordinate: \( 1 + \frac{5}{3} = \frac{8}{3} \)
• y-coordinate: \( -2 + 2 \cdot \frac{5}{3} = -2 + \frac{10}{3} = \frac{4}{3} \)
• z-coordinate: \( 1 - 2 \cdot \frac{5}{3} = 1 - \frac{10}{3} = -\frac{7}{3} \)

Final Result

The foot of the perpendicular from the point P(1, -2, 1) to the plane \( x + 2y - 2z = 10 \) is located at:

\( F\left(\frac{8}{3}, \frac{4}{3}, -\frac{7}{3}\right) \)