Let the lines L1 : r = λ(i + 2j + 3k), λ ∈ R and L2 : r = (i + 3j + k) + μ(i + j + 5k), μ ∈ R inters

Q: Wait, why do we need to check the third equation after solving for $\lambda$ and $\mu$ using the first two?

This is a fantastic question! In 3D geometry, two lines might not intersect; they could be 'skew' (neither parallel nor intersecting). By solving for $\lambda$ and $\mu$ using only the $x$ and $y$ components, you are forcing an intersection in the $xy$-plane. You must verify these values in the $z$-component equation to ensure the lines actually meet in 3D space. If they didn't satisfy the $z$-equation, the lines would be skew.

Q: Can I use any point on the lines to find the plane equation?

No, you cannot. The problem explicitly states that the plane must pass through the intersection point $S$. If you pick an arbitrary point on $L_1$ or $L_2$ that is not $S$, you will generate a plane that is parallel to the lines but does not contain them (or pass through their intersection), leading to an incorrect value for the constant $d$.

Q: What if the cross product of the direction vectors results in a zero vector?

If the cross product $\vec{b}_1 \times \vec{b}_2$ is the zero vector, it means the direction vectors are collinear (parallel). In that case, the lines are parallel, and you wouldn't have a unique intersection point $S$ to define the plane. However, in this problem, the direction vectors are clearly not proportional, so they are not parallel, ensuring a valid normal vector exists.

Question

Let the lines $L_1 : \vec{r} = \lambda(\hat{i} + 2\hat{j} + 3\hat{k}), \lambda \in \mathbb{R}$ and $L_2 : \vec{r} = (\hat{i} + 3\hat{j} + \hat{k}) + \mu(\hat{i} + \hat{j} + 5\hat{k}), \mu \in \mathbb{R}$ intersect at the point $S$. If a plane $ax + by - z + d = 0$ passes through $S$ and is parallel to both the lines $L_1$ and $L_2$, then the value of $a + b + d$ is equal to ____.

Sigma Percentile · Accepted Answer

### The Geometry of Intersection: A 3D Odyssey

Imagine you are standing in a vast, empty room. You have two thin, glowing laser beams, $L_1$ and $L_2$, cutting through the darkness.

They are not parallel, and they are not skew—they meet at a single, precise point $S$. Our mission today is to find this point and then construct a plane that perfectly aligns with both beams.

### Phase 1: The Algebraic Dance of Intersection

We start with the equations of our lines:

$$L_1 : \vec{r} = \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$$

$$L_2 : \vec{r} = (\hat{i} + 3\hat{j} + \hat{k}) + \mu(\hat{i} + \hat{j} + 5\hat{k})$$

To find the intersection point $S$, we equate the components. If the lines meet, there must exist a specific $\lambda$ and $\mu$ such that the $x, y,$ and $z$ coordinates are identical. This gives us a system:

1. $x: \lambda = 1 + \mu$
2. $y: 2\lambda = 3 + \mu$
3. $z: 3\lambda = 1 + 5\mu$

We solve the first two equations. Substituting $\lambda = 1 + \mu$ into the second equation, we get $2(1 + \mu) = 3 + \mu$, which simplifies to $2 + 2\mu = 3 + \mu$, yielding $\mu = 1$.

Plugging this back, we find $\lambda = 2$. Now, the crucial step: verification. We check these values in the third equation: $3(2) = 1 + 5(1) \Rightarrow 6 = 6$.

The consistency confirms that the lines do indeed intersect. Substituting $\lambda = 2$ into $L_1$, we find our point $S = (2, 4, 6)$.

### Phase 2: The Normal Vector—The Anchor of the Plane

Now, we need a plane passing through $S(2, 4, 6)$ that is parallel to both lines. A plane is defined by its normal vector $\vec{n}$.

If the plane is parallel to both lines, its normal must be perpendicular to the direction vectors $\vec{b}_1 = (1, 2, 3)$ and $\vec{b}_2 = (1, 1, 5)$. The cross product is our tool of choice here:

$$\vec{n} = \vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 1 & 1 & 5 \end{vmatrix}$$

Expanding this determinant, we get:

$$\vec{n} = \hat{i}(10 - 3) - \hat{j}(5 - 3) + \hat{k}(1 - 2) = 7\hat{i} - 2\hat{j} - \hat{k}$$

This vector $\vec{n} = (7, -2, -1)$ is the backbone of our plane equation.

### Phase 3: The Final Assembly

The equation of a plane passing through $(x_0, y_0, z_0)$ with normal $(a, b, c)$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$. Substituting our point $S(2, 4, 6)$ and normal $(7, -2, -1)$:

$$7(x - 2) - 2(y - 4) - 1(z - 6) = 0$$

$$7x - 14 - 2y + 8 - z + 6 = 0$$

$$7x - 2y - z = 0$$

Comparing this to the given form $ax + by - z + d = 0$, we identify $a = 7$, $b = -2$, and $d = 0$.

The sum **$a + b + d = 7 - 2 + 0 = 5$**. We have navigated the 3D space, verified the intersection, and constructed the plane with precision.

Enter Numerical Value:

Visualized Solution

Visualizing the Lines $L_{1}$ and $L_{2}$

Equating Components for Intersection

Solving for $λ$ and $μ$

Verifying the Intersection

Finding the Coordinates of $S$

Normal to the Plane

Setting up the Cross Product

Calculating the Normal Vector

Equation of the Plane

Simplifying the Equation

Finding $a, b,$ and $d$

Final Calculation: $a + b + d$

The Sigma Insight: Equation of a Plane

The Geometry of Intersection

Phase 1

Phase 2

Phase 3

Similar Questions

Let a plane $P$ contain two lines $r = \hat{i} + λ (\hat{i} + \hat{j}), λ \in R$ and $r = - \hat{j} + μ (\hat{j} - \hat{k}), μ \in R$ . If $Q (α, β, γ)$ is the foot of the perpendicular drawn from the point $M (1, 0, 1)$ to $P$ , then $3 (α + β + γ)$ equals

If the equation of the plane passing through the line of intersection of the planes $2 x - y + z = 3, 4 x - 3 y + 5 z + 9 = 0$ and parallel to the line $\frac{x + 1}{- 2} = \frac{y + 3}{4} = \frac{z - 2}{5}$ is $a x + b y + cz + 6 = 0$ , then $a + b + c$ is equal to

If the lines $r = (\hat{i} - \hat{j} + \hat{k}) + λ (3 \hat{j} - \hat{k})$ and $r = (α \hat{i} - \hat{j}) + μ (2 \hat{i} - 3 \hat{k})$ are co-planar, then distance of the plane containing these two lines from the point $(α, 0, 0)$ is :

Let the plane $a x + b y + cz = d$ pass through $(2, 3, - 5)$ and is perpendicular to the planes $2 x + y - 5 z = 10$ and $3 x + 5 y - 7 z = 12$ . If $a, b, c, d$ are integers $d > 0$ and $g cd (∣ a ∣, ∣ b ∣, ∣ c ∣, d) = 1$ , then the value of $a + 7 b + c + 20 d$ is equal to

Let the plane passing through the point $(- 1, 0, - 2)$ and perpendicular to each of the planes $2 x + y - z = 2$ and $x - y - z = 3$ be $a x + b y + cz + 8 = 0$ . Then the value of $a + b + c$ is equal to:

List-I

List-II

If the plane $2 x - y + 2 z + 3 = 0$ has the distances $1/3$ and $2/3$ units from the planes $4 x - 2 y + 4 z + λ = 0$ and $2 x - y + 2 z + μ = 0$ , respectively, then the maximum value of $λ + μ$ is equal to :

Let the equation of the plane passing through the line $x - 2 y - z - 5 = 0 = x + y + 3 z - 5$ and parallel to the line $x + y + 2 z - 7 = 0 = 2 x + 3 y + z - 2$ be $a x + b y + cz = 65$ . Then the distance of the point $(a, b, c)$ from the plane $2 x + 2 y - z + 16 = 0$ is ______.