Perpendicular distance of the point (1, 2, 3) from the line `(x-6)/3=(y-7)/2 = (z-7)/(-2)` is

Options

A. 5
B. 6
C. 7
D. 8

Answer

To find the perpendicular distance of the point \( P(1, 2, 3) \) from the line given by the equations \( \frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2} \), we can follow these steps:

### Step 1: Identify the direction ratios and a point on the line
The line can be expressed in parametric form. From the given equation, we can identify the direction ratios and a point on the line.

Let \( t \) be the parameter. Then we can write:
- \( x = 3t + 6 \)
- \( y = 2t + 7 \)
- \( z = -2t + 7 \)

From this, we can see that:
- The direction ratios of the line are \( \vec{b} = (3, 2, -2) \).
- A point on the line when \( t = 0 \) is \( Q(6, 7, 7) \).

### Step 2: Write the vector from point \( P \) to point \( Q \)
The vector \( \vec{PQ} \) from point \( P(1, 2, 3) \) to point \( Q(6, 7, 7) \) is given by:
\[ \vec{PQ} = Q - P = (6 - 1, 7 - 2, 7 - 3) = (5, 5, 4) \]

### Step 3: Set up the dot product condition for perpendicularity
For the vector \( \vec{PQ} \) to be perpendicular to the direction vector \( \vec{b} \), their dot product must be zero:
\[ \vec{b} \cdot \vec{PQ} = 0 \]
Calculating the dot product:
\[ (3, 2, -2) \cdot (5, 5, 4) = 3 \cdot 5 + 2 \cdot 5 + (-2) \cdot 4 \]
\[ = 15 + 10 - 8 = 17 \]
Since this is not zero, we need to find the point \( Q \) on the line that minimizes the distance.

### Step 4: Express point \( Q \) in terms of parameter \( t \)
We express the point \( Q \) on the line in terms of \( t \):
\[ Q(t) = (3t + 6, 2t + 7, -2t + 7) \]

### Step 5: Write the vector \( \vec{PQ} \) in terms of \( t \)
Now, we can express \( \vec{PQ} \) in terms of \( t \):
\[ \vec{PQ} = (3t + 6 - 1, 2t + 7 - 2, -2t + 7 - 3) = (3t + 5, 2t + 5, -2t + 4) \]

### Step 6: Set up the dot product condition again
We need \( \vec{b} \cdot \vec{PQ} = 0 \):
\[ (3, 2, -2) \cdot (3t + 5, 2t + 5, -2t + 4) = 0 \]
Calculating this:
\[ 3(3t + 5) + 2(2t + 5) - 2(-2t + 4) = 0 \]
\[ 9t + 15 + 4t + 10 + 4t - 8 = 0 \]
\[ (9t + 4t + 4t) + (15 + 10 - 8) = 0 \]
\[ 17t + 17 = 0 \]
\[ t = -1 \]

### Step 7: Find the coordinates of point \( Q \)
Substituting \( t = -1 \) into the parametric equations:
\[ Q(-1) = (3(-1) + 6, 2(-1) + 7, -2(-1) + 7) = (3, 5, 9) \]

### Step 8: Calculate the distance \( PQ \)
Now we calculate the distance \( PQ \):
\[ PQ = \sqrt{(3 - 1)^2 + (5 - 2)^2 + (9 - 3)^2} \]
\[ = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

### Final Answer
The perpendicular distance of the point \( (1, 2, 3) \) from the line is \( 7 \).

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