1. Key Concepts and Formulas

• Equation of a Plane Containing Two Intersecting Lines: If two lines $\vec{r} = \vec{a_1} + \lambda \vec{d_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{d_2}$ intersect at a point $\vec{a}$, the plane containing them has a normal vector $\vec{n} = \vec{d_1} \times \vec{d_2}$. The Cartesian equation of the plane is then $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$, where $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$ and $\vec{a} = x_0\hat{i} + y_0\hat{j} + z_0\hat{k}$.
• Perpendicular Distance from a Point to a Plane: The perpendicular distance $D$ from a point $P(x_1, y_1, z_1)$ to a plane $Ax+By+Cz+D'=0$ is given by the formula: $D = \frac{|Ax_1+By_1+Cz_1+D'|}{\sqrt{A^2+B^2+C^2}}$ Alternatively, if the plane equation is in the form $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$, the distance from $(x_1, y_1, z_1)$ is: $D = \frac{|A(x_1-x_0)+B(y_1-y_0)+C(z_1-z_0)|}{\sqrt{A^2+B^2+C^2}}$

2. Step-by-Step Solution

Step 1: Identify the Common Point and Direction Vectors of the Lines

The two given lines are: Line 1: $\vec{r}=(\hat{i}+\hat{j})+\lambda(\hat{i}+a \hat{j}-\hat{k})$ Line 2: $\vec{r}=(\hat{i}+\hat{j})+\mu(-\hat{i}+\hat{j}-a \hat{k})$

Explanation: We first extract the necessary information to define the plane. Both lines pass through the point $\hat{i}+\hat{j}$, which means they intersect at $(1,1,0)$. This point will be used as $\vec{a}$ (or $(x_0, y_0, z_0)$) for the plane equation. The direction vectors are $\vec{d_1}$ and $\vec{d_2}$.

From the equations, we have:

• Common point on the plane: $\vec{a} = \hat{i}+\hat{j}$, which corresponds to $(x_0, y_0, z_0) = (1,1,0)$.
• Direction vector of Line 1: $\vec{d_1} = \hat{i}+a \hat{j}-\hat{k}$.
• Direction vector of Line 2: $\vec{d_2} = -\hat{i}+\hat{j}-a \hat{k}$.

Step 2: Determine the Normal Vector to the Plane

Explanation: The normal vector to the plane is perpendicular to any two non-parallel vectors lying in the plane. Since $\vec{d_1}$ and $\vec{d_2}$ lie in the plane, their cross product will yield a vector normal to the plane.

We calculate $\vec{n} = \vec{d_1} \times \vec{d_2}$: $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & a & -1 \\ -1 & 1 & -a \end{vmatrix}$ Expanding the determinant: $\vec{n} = \hat{i}(a(-a) - (-1)(1)) - \hat{j}(1(-a) - (-1)(-1)) + \hat{k}(1(1) - a(-1))$ $\vec{n} = \hat{i}(-a^2 + 1) - \hat{j}(-a - 1) + \hat{k}(1 + a)$ $\vec{n} = (1-a^2)\hat{i} + (a+1)\hat{j} + (a+1)\hat{k}$ We can factor $(a+1)$ from each component: $\vec{n} = (1-a)(1+a)\hat{i} + (1+a)\hat{j} + (1+a)\hat{k}$ Important Note: If $a=-1$, then $\vec{d_1} = \hat{i}-\hat{j}-\hat{k}$ and $\vec{d_2} = -\hat{i}+\hat{j}+\hat{k}$. In this case, $\vec{d_1} = -\vec{d_2}$, meaning the lines are coincident (as they share a point). Coincident lines do not define a unique plane, and their cross product would be $\vec{0}$. Since the problem expects a valid plane and distance, we assume $a \neq -1$. Therefore, we can divide the normal vector by the scalar $(a+1)$ to simplify it, as any scalar multiple of a normal vector is also a normal vector: $\vec{n}' = \frac{\vec{n}}{a+1} = (1-a)\hat{i} + 1\hat{j} + 1\hat{k}$ This simplified normal vector makes subsequent calculations easier. So, the coefficients for the plane equation are $A = (1-a)$, $B=1$, and $C=1$.

Step 3: Formulate the Equation of the Plane

Explanation: Now that we have a normal vector $\vec{n}' = (1-a)\hat{i} + \hat{j} + \hat{k}$ and a point $(x_0, y_0, z_0) = (1,1,0)$ on the plane, we can write its Cartesian equation using the formula $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$.

Substituting the values: $(1-a)(x-1) + (1)(y-1) + (1)(z-0) = 0$ $(1-a)(x-1) + (y-1) + z = 0$ This is the Cartesian equation of the plane.

Step 4: Apply the Perpendicular Distance Formula

Explanation: We are given that the perpendicular distance from the point $(2,1,4)$ to this plane is $\sqrt{3}$. We will use the distance formula to set up an equation in terms of 'a'.

The given point is $(x_1, y_1, z_1) = (2,1,4)$. The plane equation is $(1-a)(x-1) + (y-1) + z = 0$. The distance formula is $D = \frac{|A(x_1-x_0)+B(y_1-y_0)+C(z_1-z_0)|}{\sqrt{A^2+B^2+C^2}}$.

Substituting the values: $D = \frac{|(1-a)(2-1) + (1-1) + (4-0)|}{\sqrt{(1-a)^2 + 1^2 + 1^2}}$ $D = \frac{|(1-a)(1) + 0 + 4|}{\sqrt{(1-a)^2 + 1 + 1}}$ $D = \frac{|1-a+4|}{\sqrt{(1-a)^2 + 2}}$ $D = \frac{|5-a|}{\sqrt{(1-a)^2 + 2}}$ We are given that $D = \sqrt{3}$: $\frac{|5-a|}{\sqrt{(1-a)^2 + 2}} = \sqrt{3}$

Step 5: Solve the Equation for 'a'

Explanation: To solve for 'a', we need to eliminate the square roots. We square both sides of the equation.

Squaring both sides: $\left(\frac{|5-a|}{\sqrt{(1-a)^2 + 2}}\right)^2 = (\sqrt{3})^2$ $\frac{(5-a)^2}{(1-a)^2 + 2} = 3$ Now, expand and simplify the equation: $(5-a)^2 = 3((1-a)^2 + 2)$ $25 - 10a + a^2 = 3(1 - 2a + a^2 + 2)$ $25 - 10a + a^2 = 3(a^2 - 2a + 3)$ $25 - 10a + a^2 = 3a^2 - 6a + 9$ Rearrange into a standard quadratic equation $Ax^2+Bx+C=0$: $0 = 3a^2 - a^2 - 6a + 10a + 9 - 25$ $0 = 2a^2 + 4a - 16$ Divide by 2 to simplify: $a^2 + 2a - 8 = 0$ Factor the quadratic equation: We need two numbers that multiply to -8 and add to 2 (these are 4 and -2). $(a+4)(a-2) = 0$ This gives two possible values for 'a': $a+4=0 \Rightarrow a = -4$ $a-2=0 \Rightarrow a = 2$

The problem asks for the largest value of $a$. Comparing the two values, $2$ is larger than $-4$. Both $a=2$ and $a=-4$ satisfy the condition $a \neq -1$ (from Step 2).

Thus, the largest value of $a$ is $2$.

3. Common Mistakes & Tips

• Cross Product Calculation: Be meticulous with the signs and arithmetic when calculating the cross product. Errors here will propagate throughout the solution.
• Absolute Value: Remember the absolute value in the numerator of the distance formula. Squaring both sides correctly handles this.
• Algebraic Simplification: Carefully expand squared terms and combine like terms to avoid errors in solving the quadratic equation.
• Conditional Validity: Always consider conditions like $a \neq -1$ where a unique plane might not be defined.

4. Summary

To determine the value of a parameter 'a' for a plane defined by two intersecting lines, given its perpendicular distance from a point, the process involves several key steps. First, find a common point on the lines and their direction vectors. Then, calculate the normal vector to the plane using the cross product of the direction vectors. Formulate the Cartesian equation of the plane using the normal vector and the common point. Finally, apply the perpendicular distance formula, set it equal to the given distance, and solve the resulting equation for 'a'. From the multiple solutions for 'a', select the one that meets the problem's specific criteria (e.g., largest value).

5. Final Answer

The largest value of $a$ is $\boxed{2}$.