Let the position vectors of two points P and Q be 3i - j + 2k and i + 2j - 4k, respectively. Let R a

Q: Wait, why do we use $\lambda$ and $\mu$ instead of just solving for $x, y, z$ directly?

Great question! In 3D geometry, a line is not just a single point; it is an infinite set of points. By using parameters like $\lambda$ and $\mu$, we create a 'map' for every point on that line. This allows us to express any point on the line as a function of a single variable, making it much easier to set up equations for intersection.

Q: Why do we use the cross product to find the direction of $\vec{TA}$?

Think of the cross product as a geometric 'magic wand.' If you have two vectors, their cross product is guaranteed to be perpendicular to the plane containing both of them. Since $\vec{TA}$ must be perpendicular to both lines, it must be parallel to the cross product of their direction vectors.

Q: Does it matter if we take the positive or negative case for $\vec{TA}$?

Not for the final answer! The problem asks for the modulus of the position vector of $A$. Whether $\vec{TA}$ points in one direction or the opposite, the length of the resulting vector $\vec{OA}$ remains the same because the squared components in the modulus calculation will yield the same sum. It is a beautiful symmetry in the problem.

Question

Let the position vectors of two points $P$ and $Q$ be $3\hat{i} - \hat{j} + 2\hat{k}$ and $\hat{i} + 2\hat{j} - 4\hat{k}$, respectively. Let $R$ and $S$ be two points such that the direction ratios of lines $PR$ and $QS$ are $(4,-1,2)$ and $(-2, 1, -2)$, respectively. Let lines $PR$ and $QS$ intersect at $T$. If the vector $\vec{TA}$ is perpendicular to both $PR$ and $QS$ and the length of vector $\vec{TA}$ is $\sqrt{5}$ units, then the modulus of a position vector of $A$ is :

Options:

(A) $\sqrt{482}$

(B) $\sqrt{171}$

(C) $\sqrt{5}$

(D) $\sqrt{227}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional room. You have two lines, $PR$ and $QS$, stretching out into the void. They are not parallel; they are destined to meet at a single, precise point $T$.

This is the heart of our problem. We are not just solving equations; we are navigating the architecture of space.

### The Parametric Dance

To find where these lines meet, we must first understand how they move. We are given the starting points $P(3, -1, 2)$ and $Q(1, 2, -4)$ and their direction ratios.

Think of these ratios as the 'velocity' or the 'heading' of the lines. We introduce two parameters, $\lambda$ and $\mu$, to describe any point on these lines.

For line $PR$, any point can be written as $(3 + 4\lambda, -1 - \lambda, 2 + 2\lambda)$. Similarly, for line $QS$, any point is $(1 - 2\mu, 2 + \mu, -4 - 2\mu)$.

By equating the $x$ and $y$ coordinates, we create a system of linear equations:

$$3 + 4\lambda = 1 - 2\mu$$

$$-1 - \lambda = 2 + \mu$$

Solving this system is our first victory. Subtracting the equations reveals $\lambda = 2$, which leads us to $\mu = -5$. Checking these in the $z$-coordinate confirms our intersection point $T$ is $(11, -3, 6)$.

### The Perpendicularity Challenge

Now, we introduce a new vector, $\vec{TA}$. We are told it is perpendicular to both lines.

How do we find a direction that is orthogonal to two different lines simultaneously? We turn to the cross product. By taking the cross product of the direction vectors $(4, -1, 2)$ and $(-2, 1, -2)$, we find the normal vector $\vec{n} = (0, 4, 2)$.

This vector $\vec{n}$ is the 'anchor' for the direction of $\vec{TA}$. To make it useful, we normalize it into a unit vector:

$$\hat{n} = \frac{2\hat{j} + \hat{k}}{\sqrt{5}}$$

Since the length of $\vec{TA}$ is given as $\sqrt{5}$, we simply multiply this length by our unit vector to get:

$$\vec{TA} = \pm (2\hat{j} + \hat{k})$$

### The Final Leap

We are almost there. To find the position vector of $A$, we use the triangle law of vector addition: $\vec{OA} = \vec{OT} + \vec{TA}$.

Substituting our coordinates for $T$ and the two possible vectors for $\vec{TA}$, we get two potential positions for $A$: $(11, -1, 7)$ or $(11, -5, 5)$.

Finally, we calculate the modulus:

$$|\vec{OA}| = \sqrt{11^2 + (-1)^2 + 7^2} = \sqrt{121 + 1 + 49} = \sqrt{171}$$

Whether we chose the positive or negative case, the result is the same. The elegance of the math ensures that no matter which way we look, the distance remains constant.

You have successfully navigated the 3D landscape and arrived at the solution: **$\sqrt{171}$**.

Select Answer:

Visualized Solution

Visualizing Points $P$ and $Q$

Defining the Lines $P R$ and $QS$

Parametric Equations of the Lines

Setting up the Intersection Equations

Solving for $λ$ and $μ$

Finding the Coordinates of $T$

Direction of the Perpendicular Vector $T A$

Calculating the Cross Product

Finding the Unit Normal Vector

Determining the Vector $T A$

Finding the Position Vector of $A$

Calculating the Modulus $∣ O A ∣$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Parametric Dance

The Perpendicularity Challenge

The Final Leap

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