Let L1 : r = (i - j + 2k) + λ(i - j + 2k), λ ∈ R, L2 : r = (j - k) + μ(3i + j + pk), μ ∈ R and L3 :

Q: Wait, why do we use the dot product for $L_1 \perp L_2$? Can't we use the angle formula?

You absolutely can, but remember that the angle formula $\cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{|a||b|}$ relies on the dot product! If the lines are perpendicular, $\theta = 90^\circ$, and $\cos(90^\circ) = 0$. Therefore, the numerator $\vec{a} \cdot \vec{b}$ must be zero. Using the dot product directly is the most efficient, 'JEE-smart' way to enforce orthogonality.

Q: Why does $L_3$ pass through the origin? The equation looks weird.

Look closely at the equation $L_3 : \vec{r} = \delta(l\hat{i} + m\hat{j} + n\hat{k})$. If you set the parameter $\delta = 0$, you get $\vec{r} = 0\hat{i} + 0\hat{j} + 0\hat{k}$, which is the origin $(0,0,0)$. This confirms that $L_3$ is a line passing through the origin with a direction vector $\vec{v} = l\hat{i} + m\hat{j} + n\hat{k}$.

Q: Is the order of the cross product important? What if I did $\vec{b_2} \times \vec{b_1}$?

Great question! The cross product is anti-commutative, meaning $\vec{b_2} \times \vec{b_1} = -(\vec{b_1} \times \vec{b_2})$. This would give you a vector pointing in the exact opposite direction. However, since $L_3$ is a line, any scalar multiple of the direction vector is valid. Your result would just be a negative version of the correct direction, which still defines the same line in space. You are safe either way!

Question

Let $L_1 : \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda(\hat{i} - \hat{j} + 2\hat{k}), \lambda \in \mathbb{R}$, $L_2 : \vec{r} = (\hat{j} - \hat{k}) + \mu(3\hat{i} + \hat{j} + p\hat{k}), \mu \in \mathbb{R}$ and $L_3 : \vec{r} = \delta(l\hat{i} + m\hat{j} + n\hat{k}), \delta \in \mathbb{R}$ be three lines such that $L_1$ is perpendicular to $L_2$ and $L_3$ is perpendicular to both $L_1$ and $L_2$. Then the point which lies on $L_3$ is

Options:

(A) $(-1, 7, 4)$

(B) $(-1, -7, 4)$

(C) $(1, 7, -4)$

(D) $(1, -7, 4)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow JEE aspirant. Today, we are not just solving a problem; we are navigating the elegant architecture of three-dimensional space. Many students fear 3D geometry, viewing it as a chaotic mess of vectors and parameters.

But I want you to see it differently. Imagine you are standing in a vast, empty room. You have two lines, $L_1$ and $L_2$, stretching out like beams of light. Our mission is to find a third line, $L_3$, that stands perfectly perpendicular to both. This is the essence of spatial reasoning.

### The Mystery of the Unknown

Let us begin by examining our tools. We are given $L_1$ and $L_2$. Every line in 3D space is defined by a point it passes through and a direction vector.

For $L_1$, the direction vector is clearly $\vec{b_1} = \hat{i} - \hat{j} + 2\hat{k}$. For $L_2$, we have $\vec{b_2} = 3\hat{i} + \hat{j} + p\hat{k}$.

But wait—there is a shadow over $L_2$. We have an unknown parameter $p$. How do we uncover it? The problem gives us the key: $L_1$ is perpendicular to $L_2$.

In the language of vectors, this is a golden rule. When two vectors are perpendicular, their dot product must vanish into nothingness. Mathematically, we write this as:

$$\vec{b_1} \cdot \vec{b_2} = 0$$

Substituting our components, we get:

$$(\hat{i} - \hat{j} + 2\hat{k}) \cdot (3\hat{i} + \hat{j} + p\hat{k}) = 0$$

Performing the dot product is like a dance of corresponding components: $(1)(3) + (-1)(1) + (2)(p) = 0$. This simplifies beautifully to $3 - 1 + 2p = 0$, or $2 + 2p = 0$.

Solving this, we find **$p = -1$**. Just like that, the mystery of $L_2$ is solved. Our direction vector $\vec{b_2}$ is now fully revealed as $3\hat{i} + \hat{j} - \hat{k}$.

### The Power of the Cross Product

Now, we face the main challenge. We need a line $L_3$ that is perpendicular to both $L_1$ and $L_2$. In 3D geometry, when you need a vector that is simultaneously perpendicular to two others, there is only one tool you should reach for: the cross product.

The cross product $\vec{b_1} 	imes \vec{b_2}$ creates a new vector that is orthogonal to the plane containing both $\vec{b_1}$ and $\vec{b_2}$. Let us set up the determinant, the engine of the cross product:

$$\vec{b_3} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 2 \ 3 & 1 & -1 \end{vmatrix}$$

Take a deep breath. Let us expand this carefully. For the $\hat{i}$ component, we cover the first column and row, leaving us with $(-1)(-1) - (2)(1) = 1 - 2 = -1$.

For the $\hat{j}$ component, remember the negative sign: $-((1)(-1) - (2)(3)) = -(-1 - 6) = -(-7) = 7$. Finally, for the $\hat{k}$ component, we have $(1)(1) - (-1)(3) = 1 + 3 = 4$.

Thus, our direction vector for $L_3$ is **$\vec{b_3} = -\hat{i} + 7\hat{j} + 4\hat{k}$**.

### Final Verification

We have arrived at the final stage. The equation of $L_3$ is given as $\vec{r} = \delta(-\hat{i} + 7\hat{j} + 4\hat{k})$. This means any point on this line is simply a scalar multiple of our direction vector.

If we choose a value for the parameter $\delta$, we generate a point on the line. Let us test $\delta = 1$. This gives us the point $(-1, 7, 4)$.

Look at your options. There it is, staring back at you. It is not just a coordinate; it is the result of your logical journey through the dot product and the cross product. You have successfully navigated the 3D space, conquered the unknown parameter, and identified the line.

Select Answer:

Visualized Solution

Direction of $L_{1}$

Direction of $L_{2}$

Perpendicularity Condition

Dot Product Setup

Solve for $p$

Direction of $L_{3}$

Cross Product Setup

Expand Determinant

Equation of $L_{3}$

Verify Options

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Mystery of the Unknown

The Power of the Cross Product

Final Verification

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