The foot of the perpendicular drawn from the origin to a plane is the point `(1, -3, 1)`. What is the intercept cut on the x-axis by the plane ?

Options

A. 1
B. 3
C. `sqrt11`
D. 11

Explanation

To find the intercept cut on the x-axis by the plane given that the foot of the perpendicular from the origin to the plane is the point (1, -3, 1), we can follow these steps:

### Step 1: Determine the normal vector of the plane
The foot of the perpendicular from the origin to the plane is given as the point (1, -3, 1). The vector from the origin (0, 0, 0) to this point is the normal vector of the plane. Thus, the normal vector \( \vec{n} \) can be represented as:
\[ \vec{n} = (1, -3, 1) \]

### Step 2: Write the equation of the plane
The general equation of a plane can be expressed as:
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]
where \( (x_0, y_0, z_0) \) is a point on the plane, and \( (a, b, c) \) are the direction ratios of the normal vector. Substituting \( (x_0, y_0, z_0) = (1, -3, 1) \) and \( (a, b, c) = (1, -3, 1) \), we get:
\[ 1(x - 1) - 3(y + 3) + 1(z - 1) = 0 \]

### Step 3: Simplify the equation
Expanding the equation:
\[ x - 1 - 3y - 9 + z - 1 = 0 \]
Combining like terms gives:
\[ x - 3y + z - 11 = 0 \]

### Step 4: Rearranging to find intercepts
Rearranging the equation to express it in intercept form:
\[ x - 3y + z = 11 \]
To convert this into intercept form, we can rewrite it as:
\[ \frac{x}{11} - \frac{y}{\frac{11}{3}} + \frac{z}{11} = 1 \]

### Step 5: Identify the x-intercept
In the intercept form \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \), the x-intercept is represented by \( a \). From our equation:
\[ \frac{x}{11} + \frac{y}{\frac{11}{3}} + \frac{z}{11} = 1 \]
we can see that the x-intercept \( a = 11 \).

### Final Answer
Thus, the intercept cut on the x-axis by the plane is:
\[ \text{X-intercept} = 11 \]