The length of perpendicular from the point `P(1,-1,2)` on the line `(x+1)/(2) = (y-2)/(-3) = (z+2)/(4)` is

Options

A. `sqrt6` units
B. `sqrt21` units
C. `sqrt29` units
D. `0` units

Answer

To find the length of the perpendicular from the point \( P(1, -1, 2) \) to the line given by the equations \( \frac{x+1}{2} = \frac{y-2}{-3} = \frac{z+2}{4} \), we can follow these steps:

### Step 1: Parametrize the Line
The given line can be expressed in parametric form. Let \( t \) be the parameter such that:
- \( x = -1 + 2t \)
- \( y = 2 - 3t \)
- \( z = -2 + 4t \)

### Step 2: Find the Coordinates of Point Q
Let \( Q \) be a point on the line. Thus, the coordinates of point \( Q \) can be expressed as:
- \( Q(-1 + 2t, 2 - 3t, -2 + 4t) \)

### Step 3: Find the Direction Ratios of Line PQ
The direction ratios of the line segment \( PQ \) from point \( P(1, -1, 2) \) to point \( Q \) are given by:
- \( PQ_x = (-1 + 2t) - 1 = 2t - 2 \)
- \( PQ_y = (2 - 3t) - (-1) = 3 - 3t \)
- \( PQ_z = (-2 + 4t) - 2 = 4t - 4 \)

So, the direction ratios of line \( PQ \) are:
- \( (2t - 2, 3 - 3t, 4t - 4) \)

### Step 4: Direction Ratios of the Given Line
From the line equation, the direction ratios (DRs) of the line are:
- \( (2, -3, 4) \)

### Step 5: Use the Condition for Perpendicularity
For the lines \( PQ \) and the given line to be perpendicular, the dot product of their direction ratios must equal zero:
\[ (2t - 2) \cdot 2 + (3 - 3t) \cdot (-3) + (4t - 4) \cdot 4 = 0 \]

Expanding this, we get:
\[ 4t - 4 - 9 + 9t + 16t - 16 = 0 \]
\[ (4t + 9t + 16t) - (4 + 9 + 16) = 0 \]
\[ 29t - 29 = 0 \]

### Step 6: Solve for t
From the equation \( 29t - 29 = 0 \), we find:
\[ t = 1 \]

### Step 7: Find the Coordinates of Point Q
Substituting \( t = 1 \) back into the parametric equations for \( Q \):
- \( Q = (-1 + 2 \cdot 1, 2 - 3 \cdot 1, -2 + 4 \cdot 1) \)
- \( Q = (1, -1, 2) \)

### Step 8: Calculate the Distance PQ
Now, we find the distance \( PQ \) using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the coordinates of \( P(1, -1, 2) \) and \( Q(1, -1, 2) \):
\[ d = \sqrt{(1 - 1)^2 + (-1 + 1)^2 + (2 - 2)^2} = \sqrt{0 + 0 + 0} = 0 \]

### Conclusion
The length of the perpendicular from the point \( P(1, -1, 2) \) to the line is \( 0 \) units.

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