The largest value of a, for which the perpendicular distance of the plane containing the lines r = (

Q: Why can we just take the cross product of the direction vectors to find the normal?

Because the lines lie on the plane, their direction vectors are parallel to the plane. The cross product of any two non-parallel vectors in a plane is always perpendicular to that plane, which is exactly what we need for the normal vector.

Q: Is it okay to divide the normal vector by $(a+1)$?

Yes! The normal vector defines the orientation of the plane, not its magnitude. Any non-zero scalar multiple of the normal vector represents the same plane. Just be careful if $a = -1$, but here, the resulting equation holds.

Q: How do I know if the lines are intersecting or skew?

The problem states they lie on a plane, so they must be coplanar. If they share a point, they are intersecting. If they didn't share a point, you would need to check for coplanarity using the scalar triple product, but here, the common point is obvious.

Question

The largest value of $a$, for which the perpendicular distance of the plane containing the lines $\vec{r} = (\hat{i} + \hat{j}) + \lambda(\hat{i} + a\hat{j} - \hat{k})$ and $\vec{r} = (\hat{i} + \hat{j}) + \mu(-\hat{i} + \hat{j} - a\hat{k})$ from the point $(2, 1, 4)$ is $\sqrt{3}$, is ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to the fascinating world of 3D geometry. Today, we are not just solving a problem; we are building a bridge between algebraic equations and the physical reality of a plane in space.

Imagine you are standing in a room, and you have two lines drawn on a flat sheet of paper. This sheet of paper is our plane. We have a point $Q(2, 1, 4)$ hovering above this sheet, and we want to know how far it is from the paper. The problem gives us the distance, $\sqrt{3}$, and asks us to find the parameter $a$.

### The Anchor

First, we must anchor our plane. Look at the equations of the lines:
$$\vec{r} = (\hat{i} + \hat{j}) + \lambda(\hat{i} + a\hat{j} - \hat{k})$$
$$\vec{r} = (\hat{i} + \hat{j}) + \mu(-\hat{i} + \hat{j} - a\hat{k})$$

Do you see it? Both lines share the same position vector, $\hat{i} + \hat{j}$. This means they both pass through the point $P(1, 1, 0)$. This is our anchor!

Since both lines lie on the plane, this point $P$ must also lie on the plane. We have our point; now we need the orientation.

### The Normal Vector

To define a plane, we need a normal vector $\vec{n}$ that is perpendicular to it. We have the direction vectors of our lines: $\vec{d_1} = \hat{i} + a\hat{j} - \hat{k}$ and $\vec{d_2} = -\hat{i} + \hat{j} - a\hat{k}$.

If we take the cross product $\vec{n} = \vec{d_1} 	imes \vec{d_2}$, we get a vector perpendicular to both lines, and thus, perpendicular to the plane. Let's calculate this:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & a & -1 \ -1 & 1 & -a \end{vmatrix} = \hat{i}(-a^2 + 1) - \hat{j}(-a - 1) + \hat{k}(1 + a)$$

This results in $\vec{n} = (1-a^2)\hat{i} + (a+1)\hat{j} + (a+1)\hat{k}$. Notice the common factor $(a+1)$. We can simplify our normal vector to $\vec{n'} = (1-a, 1, 1)$. This is the beauty of geometry; we can scale our normal vector without changing the plane's orientation.

### The Plane Equation

Now, we use the point-normal form of the plane equation: $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$. Substituting our point $P(1, 1, 0)$ and normal $(1-a, 1, 1)$, we get:

$$(1-a)(x-1) + 1(y-1) + 1(z-0) = 0$$

Expanding this, we get $(1-a)x - (1-a) + y - 1 + z = 0$, which simplifies to:

$$(1-a)x + y + z + a - 2 = 0$$

This is the equation of our plane.

### The Distance Formula

The perpendicular distance from a point $(x_1, y_1, z_1)$ to a plane $Ax + By + Cz + D = 0$ is given by:

$$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$$

Substituting our point $Q(2, 1, 4)$ and our plane equation, we get:

$$\frac{|(1-a)(2) + 1 + 4 + a - 2|}{\sqrt{(1-a)^2 + 1^2 + 1^2}} = \sqrt{3}$$

Simplifying the numerator, we get $|2 - 2a + 5 + a - 2| = |5 - a|$. The denominator becomes $\sqrt{1 - 2a + a^2 + 2} = \sqrt{a^2 - 2a + 3}$. So:

$$\frac{|5-a|}{\sqrt{a^2 - 2a + 3}} = \sqrt{3}$$

### The Quadratic Finale

To solve for $a$, we square both sides:

$$\frac{(5-a)^2}{a^2 - 2a + 3} = 3$$

This gives $(5-a)^2 = 3(a^2 - 2a + 3)$, which expands to $25 - 10a + a^2 = 3a^2 - 6a + 9$. Rearranging, we get:

$$2a^2 + 4a - 16 = 0 \Rightarrow a^2 + 2a - 8 = 0$$

Factoring this, we find $(a+4)(a-2) = 0$. The roots are $a = -4$ and $a = 2$. The question asks for the largest value, which is **2**.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Identifying the Common Point

The Normal Vector Concept

Calculating the Normal Vector

Equation of the Plane

Setting up the Distance Formula

Substituting Values

Simplifying the Equation

Squaring and Forming Quadratic

Solving for $a$ and Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Anchor

The Normal Vector

The Plane Equation

The Distance Formula

The Quadratic Finale

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