A straight line `vecr=veca+lambda vecb` meets the plane `vecr. vec n=0 at p.` Then position vector of P is

Options

A. `veca+(veca.vecn)/(vecb.vecn)vecb`
B. `veca(vecb.vecn)/(veca.vecn)vecb`
C. `veca-(veca.vecn)/(vecb.vecn)`
D. none of these

Explanation

To solve the problem step by step, we need to find the position vector of the point \( P \) where the line intersects the plane. The line is given by the equation:

\[ \vec{r} = \vec{a} + \lambda \vec{b} \]

and the plane is defined by the equation:

\[ \vec{r} \cdot \vec{n} = 0 \]

### Step 1: Substitute the line equation into the plane equation

Since the point \( P \) lies on both the line and the plane, we can substitute the equation of the line into the equation of the plane:

\[ (\vec{a} + \lambda \vec{b}) \cdot \vec{n} = 0 \]

### Step 2: Expand the dot product

Expanding the dot product gives:

\[ \vec{a} \cdot \vec{n} + \lambda (\vec{b} \cdot \vec{n}) = 0 \]

### Step 3: Solve for \( \lambda \)

Rearranging the equation to solve for \( \lambda \):

\[ \lambda (\vec{b} \cdot \vec{n}) = -\vec{a} \cdot \vec{n} \]

Thus, we can express \( \lambda \) as:

\[ \lambda = -\frac{\vec{a} \cdot \vec{n}}{\vec{b} \cdot \vec{n}} \]

### Step 4: Substitute \( \lambda \) back into the line equation

Now that we have \( \lambda \), we can substitute it back into the line equation to find the position vector of point \( P \):

\[ \vec{r} = \vec{a} + \lambda \vec{b} = \vec{a} - \frac{\vec{a} \cdot \vec{n}}{\vec{b} \cdot \vec{n}} \vec{b} \]

### Step 5: Final expression for the position vector of \( P \)

Thus, the position vector of point \( P \) where the line meets the plane is:

\[ \vec{r} = \vec{a} - \frac{\vec{a} \cdot \vec{n}}{\vec{b} \cdot \vec{n}} \vec{b} \]

### Conclusion

The final position vector of \( P \) is:

\[ \vec{P} = \vec{a} - \frac{\vec{a} \cdot \vec{n}}{\vec{b} \cdot \vec{n}} \vec{b} \]

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