Let N be the foot of perpendicular from the point `P(1,-2,3)` on the line passing through the points `(4,5,8)` and `(1,-7,5)` . Then the distance of N from the plane `2x - 2y + z +5 = 0` is

Options

A. `7`
B. `9`
C. `6`
D. `8`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Find the direction ratios of the line passing through points \( A(4, 5, 8) \) and \( B(1, -7, 5) \).

The direction ratios can be calculated as follows:
\[ \text{Direction ratios} = B - A = (1 - 4, -7 - 5, 5 - 8) = (-3, -12, -3) \]

### Step 2: Write the parametric equations of the line.

The parametric equations of the line can be expressed as:
\[ x = 4 - 3t, \quad y = 5 - 12t, \quad z = 8 - 3t \]

### Step 3: Find the foot of the perpendicular \( N \) from point \( P(1, -2, 3) \) to the line.

Let \( N \) be the foot of the perpendicular. The coordinates of \( N \) can be expressed in terms of the parameter \( t \):
\[ N(t) = (4 - 3t, 5 - 12t, 8 - 3t) \]

### Step 4: Set up the condition for perpendicularity.

The vector \( \overrightarrow{PN} \) from \( P \) to \( N \) must be perpendicular to the direction vector of the line. Thus, we need to calculate:
\[ \overrightarrow{PN} = N(t) - P = (4 - 3t - 1, 5 - 12t + 2, 8 - 3t - 3) = (3 - 3t, 7 - 12t, 5 - 3t) \]
The direction ratios of the line are \( (-3, -12, -3) \). The dot product must equal zero:
\[ (3 - 3t)(-3) + (7 - 12t)(-12) + (5 - 3t)(-3) = 0 \]

### Step 5: Simplify the dot product equation.

Expanding the dot product:
\[ -9 + 9t - 84 + 144t - 15 + 9t = 0 \]
Combine like terms:
\[ 162t - 108 = 0 \]
Solving for \( t \):
\[ 162t = 108 \implies t = \frac{108}{162} = \frac{2}{3} \]

### Step 6: Substitute \( t \) back to find coordinates of \( N \).

Substituting \( t = \frac{2}{3} \) into the parametric equations:
\[ N\left(\frac{2}{3}\right) = \left(4 - 3\left(\frac{2}{3}\right), 5 - 12\left(\frac{2}{3}\right), 8 - 3\left(\frac{2}{3}\right)\right) \]
Calculating each coordinate:
\[ N\left(\frac{2}{3}\right) = \left(4 - 2, 5 - 8, 8 - 2\right) = (2, -3, 6) \]

### Step 7: Calculate the distance from point \( N(2, -3, 6) \) to the plane \( 2x - 2y + z + 5 = 0 \).

The formula for the distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
For our plane, \( A = 2, B = -2, C = 1, D = 5 \) and point \( N(2, -3, 6) \):
\[ d = \frac{|2(2) - 2(-3) + 1(6) + 5|}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{|4 + 6 + 6 + 5|}{\sqrt{4 + 4 + 1}} = \frac{|21|}{3} = 7 \]

### Final Answer:
The distance of point \( N \) from the plane is \( 7 \).

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