The equation of the plane which contains the origin and the line of intersectio of the plane `vecr.veca=d_(1)` and `vecr.vecb=d_(2)` is

Options

A. `vecr.(d_(1)veca-d_(2)vecb)=0`
B. `vecr.(d_(1)veca+d_(2)vecb)=0`
C. `vecr.(d_(2)veca+d_(1)vecb)=0`
D. `vecr.(d_(2)veca-d_(2)vecb)=0`

Explanation

To find the equation of the plane that contains the origin and the line of intersection of the planes given by the equations \( \vec{r} \cdot \vec{a} = d_1 \) and \( \vec{r} \cdot \vec{b} = d_2 \), we can follow these steps:

### Step 1: Understand the problem
We need to find a plane that contains the origin and the line of intersection of the two given planes. The general form of a plane that contains the line of intersection of two planes can be expressed as a linear combination of the two plane equations.

### Step 2: Write the general equation of the family of planes
The equation of the plane that contains the intersection of the two planes can be written as:
\[ P_1 + \lambda P_2 = 0 \]
where \( P_1: \vec{r} \cdot \vec{a} - d_1 = 0 \) and \( P_2: \vec{r} \cdot \vec{b} - d_2 = 0 \).

### Step 3: Substitute the plane equations
Substituting the equations of the planes into the family of planes equation gives:
\[ (\vec{r} \cdot \vec{a} - d_1) + \lambda (\vec{r} \cdot \vec{b} - d_2) = 0 \]

### Step 4: Set the condition for passing through the origin
Since the plane must pass through the origin, we substitute \( \vec{r} = 0 \) into the equation:
\[ (0 \cdot \vec{a} - d_1) + \lambda (0 \cdot \vec{b} - d_2) = 0 \]
This simplifies to:
\[ -d_1 - \lambda d_2 = 0 \]
From this, we can solve for \( \lambda \):
\[ \lambda = -\frac{d_1}{d_2} \]

### Step 5: Substitute \( \lambda \) back into the plane equation
Now we substitute \( \lambda \) back into the equation of the family of planes:
\[ \vec{r} \cdot \vec{a} - d_1 - \left(-\frac{d_1}{d_2}\right)(\vec{r} \cdot \vec{b} - d_2) = 0 \]
This expands to:
\[ \vec{r} \cdot \vec{a} - d_1 + \frac{d_1}{d_2} \vec{r} \cdot \vec{b} - d_1 = 0 \]
Combining like terms gives:
\[ \vec{r} \cdot \vec{a} + \frac{d_1}{d_2} \vec{r} \cdot \vec{b} - 2d_1 = 0 \]

### Step 6: Rearranging the equation
Rearranging the equation yields:
\[ \vec{r} \cdot \left(\vec{a} + \frac{d_1}{d_2} \vec{b}\right) = 2d_1 \]

### Step 7: Final form of the equation
Thus, the equation of the required plane is:
\[ \vec{r} \cdot \left(d_2 \vec{a} - d_1 \vec{b}\right) = 0 \]

### Conclusion
The final equation of the plane that contains the origin and the line of intersection of the two given planes is:
\[ \vec{r} \cdot (d_2 \vec{a} - d_1 \vec{b}) = 0 \]