The equation of the plane passing through the point `(1, 1, 0)` and perpendicular to the line `overset(to)( r) = 2 hat(i) + 3 hat(j) + 4 hat(k) + lambda ( 3 hat(i) + 4 hat(j) + 5hat(k) )` is

Options

A. `3x -4y + 5z =7`
B. `3x + 4y + 5z =7`
C. `x +y - 4z =9`
D. `3x+4y-12=0`

Answer

To find the equation of the plane passing through the point \( (1, 1, 0) \) and perpendicular to the given line, we will follow these steps:

### Step 1: Identify the direction vector of the line
The equation of the line is given as:
\[ \vec{r} = 2\hat{i} + 3\hat{j} + 4\hat{k} + \lambda(3\hat{i} + 4\hat{j} + 5\hat{k}) \]
From this equation, we can identify the direction vector of the line, which is:
\[ \vec{d} = 3\hat{i} + 4\hat{j} + 5\hat{k} \]

### Step 2: Identify the normal vector of the plane
Since the plane is perpendicular to the line, the normal vector \( \vec{n} \) of the plane is the same as the direction vector of the line:
\[ \vec{n} = 3\hat{i} + 4\hat{j} + 5\hat{k} \]

### Step 3: Use the point-normal form of the plane equation
The equation of a plane in point-normal form is given by:
\[ \vec{n} \cdot (\vec{r} - \vec{r_0}) = 0 \]
where \( \vec{r_0} \) is a point on the plane. Here, \( \vec{r_0} = (1, 1, 0) \) and \( \vec{r} = (x, y, z) \).

### Step 4: Substitute the values into the equation
Substituting the normal vector and the point into the equation gives:
\[ (3\hat{i} + 4\hat{j} + 5\hat{k}) \cdot ((x - 1)\hat{i} + (y - 1)\hat{j} + z\hat{k}) = 0 \]

### Step 5: Calculate the dot product
Calculating the dot product:
\[ 3(x - 1) + 4(y - 1) + 5z = 0 \]
Expanding this, we get:
\[ 3x - 3 + 4y - 4 + 5z = 0 \]

### Step 6: Rearranging the equation
Rearranging the equation gives:
\[ 3x + 4y + 5z = 7 \]

### Final Equation of the Plane
Thus, the equation of the plane is:
\[ 3x + 4y + 5z = 7 \]