Find the equation of the plane passing through the intersection of r.(i + 3j - k) = 6 and r.(3i - j - 4k) = 0 at distance 1/sqrt(6) from origin.

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Step-by-Step Solution

Step 1: Formulate the equation of the plane

The equation of a plane passing through the intersection of two planes $\mathbf{r} \cdot \mathbf{n}_1 = d_1$ and $\mathbf{r} \cdot \mathbf{n}_2 = d_2$ is given by $\mathbf{r} \cdot (\mathbf{n}_1 + \lambda \mathbf{n}_2) = d_1 + \lambda d_2$. We substitute the given normal vectors and constants into this general form to get the equation of the required plane in terms of $\lambda.

Step 2: Convert to Cartesian form

To work with the distance formula, it's convenient to convert the vector equation of the plane into its Cartesian form. We replace $\mathbf{r}$ with $x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ and perform the dot product.

Step 3: Apply the distance from origin formula

The distance of a plane $Ax + By + Cz + D = 0$ from the origin $(0,0,0)$ is given by the formula $\frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$. We are given that this distance is $\frac{1}{\sqrt{6}}$. We substitute the coefficients from our plane equation and set it equal to the given distance.

Step 4: Solve for $\lambda$

To find the value of $\lambda$, we square both sides of the equation obtained in the previous step and simplify it into a quadratic equation. This will allow us to solve for $\lambda.

Step 5: Solve the quadratic equation for $\lambda$

We use the quadratic formula $\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to solve for $\lambda$. Here, $a = 26$, $b = 8$, and $c = -205$. We calculate the discriminant and then the two possible values for $\lambda.

Step 6: Calculate values of $\lambda$

We calculate the two possible values for $\lambda$ using the positive and negative roots from the quadratic formula. These values will lead to two possible plane equations.

Step 7: Substitute $\lambda$ values to find plane equations

Finally, we substitute each calculated value of $\lambda$ back into the Cartesian equation of the plane we derived in Step 2. This gives us the two possible equations for the plane that satisfy all the given conditions.