Question

The equation of the plane passing through the line of intersection of the planes \( \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1 \) and \( \vec{r} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) + 4 = 0 \) and parallel to the x-axis is :

• A. \(\vec{r} \cdot (\hat{j} - 3\hat{k}) + 6 = 0\)
• B. \(\vec{r} \cdot (\hat{j} + 3\hat{k}) + 6 = 0\)
• C. \(\vec{r} \cdot (\hat{j} - 3\hat{k}) - 6 = 0\)
• D. \(\vec{r} \cdot (\hat{j} - 3\hat{k}) + 6 = 0\)

Hint:

The equation of a plane passing through the intersection of \(P_1=0\) and \(P_2=0\) is \(P_1 + \lambda P_2 = 0\). A plane is parallel to a line if the plane's normal vector is perpendicular to the line's direction vector (i.e., their dot product is zero).

Answer 1

The Correct Option is A

Step 1: Write the equation of the family of planes.
The equation of any plane passing through the line of intersection of two planes P\(_1\) = 0 and P\(_2\) = 0 is given by P\(_1\) + \(\lambda\)P\(_2\) = 0. First, write the given plane equations in Cartesian form or standard vector form: P\(_1\): \(\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) - 1 = 0 \implies x + y + z - 1 = 0\) P\(_2\): \(\vec{r} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) + 4 = 0 \implies 2x + 3y - z + 4 = 0\) The equation of the required plane is: \[ (x + y + z - 1) + \lambda(2x + 3y - z + 4) = 0 \] \[ (1+2\lambda)x + (1+3\lambda)y + (1-\lambda)z + (-1+4\lambda) = 0 \] Step 2: Use the condition that the plane is parallel to the x-axis.
The normal vector to this plane is \(\vec{n} = (1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1-\lambda)\hat{k}\). The direction vector of the x-axis is \(\vec{d} = \hat{i} = (1, 0, 0)\). If the plane is parallel to the x-axis, its normal vector must be perpendicular to the direction vector of the x-axis. Therefore, their dot product is zero: \(\vec{n} \cdot \vec{d} = 0\). \[ ((1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1-\lambda)\hat{k}) \cdot (\hat{i}) = 0 \] \[ 1+2\lambda = 0 \implies \lambda = -1/2 \] Step 3: Substitute the value of \(\lambda\) back into the plane equation.
Substitute \(\lambda = -1/2\) into the equation of the family of planes: \[ (1 + 2(-1/2))x + (1 + 3(-1/2))y + (1 - (-1/2))z + (-1 + 4(-1/2)) = 0 \] \[ (1 - 1)x + (1 - 3/2)y + (1 + 1/2)z + (-1 - 2) = 0 \] \[ 0x - \frac{1}{2}y + \frac{3}{2}z - 3 = 0 \] Multiply the entire equation by -2 to clear the fractions and match the options: \[ y - 3z + 6 = 0 \] Step 4: Convert the equation to vector form.
The equation \(y - 3z + 6 = 0\) can be written in vector form as \((x\hat{i} + y\hat{j} + z\hat{k}) \cdot (0\hat{i} + 1\hat{j} - 3\hat{k}) + 6 = 0\). \[ \vec{r} \cdot (\hat{j} - 3\hat{k}) + 6 = 0 \] This matches option (A).