Key Concepts and Formulas

1. Vector Equation of a Line: A line passing through a point with position vector $\vec{a}$ and parallel to a direction vector $\vec{d}$ can be represented as $\vec{r} = \vec{a} + \lambda \vec{d}$, where $\lambda$ is a scalar parameter.
2. Intersection of Lines: If two lines intersect, their position vectors are identical at the point of intersection. Equating the corresponding components (x, y, z) allows us to form a system of linear equations to solve for the parameters.
3. Cross Product for Perpendicular Vectors: The cross product of two non-parallel vectors, $\vec{u} \times \vec{v}$, results in a vector that is perpendicular to both $\vec{u}$ and $\vec{v}$. This property is crucial for finding a vector perpendicular to a plane defined by two direction vectors.
4. Magnitude of a Vector: For a vector $\vec{v} = x\widehat i + y\widehat j + z\widehat k$, its magnitude (or length) is given by $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$.
5. Position Vector Modulus: The modulus of the position vector of a point A with coordinates $(x, y, z)$ is $|\overrightarrow{OA}| = \sqrt{x^2 + y^2 + z^2}$, where O is the origin.

---

Step-by-Step Solution

Step 1: Define Position Vectors and Direction Vectors

We begin by translating the given information into vector notation. The position vector of point P is $\vec{p} = 3\widehat i - \widehat j + 2\widehat k$. The position vector of point Q is $\vec{q} = \widehat i + 2\widehat j - 4\widehat k$.

The direction ratios of line PR are $(4, -1, 2)$. This implies that the direction vector for line PR can be taken as $\vec{d}_{PR} = 4\widehat i - \widehat j + 2\widehat k$. Similarly, the direction ratios of line QS are $(-2, 1, -2)$. Thus, the direction vector for line QS can be taken as $\vec{d}_{QS} = -2\widehat i + \widehat j - 2\widehat k$.

Step 2: Formulate the Equations of Lines PR and QS

Using the vector equation of a line $\vec{r} = \vec{a} + \lambda \vec{d}$:

• Equation of Line PR: This line passes through P with position vector $\vec{p}$ and is parallel to $\vec{d}_{PR}$. $L_{PR}: \vec{r}_1 = (3\widehat i - \widehat j + 2\widehat k) + \lambda (4\widehat i - \widehat j + 2\widehat k)$ where $\lambda$ is a scalar parameter.

• Equation of Line QS: This line passes through Q with position vector $\vec{q}$ and is parallel to $\vec{d}_{QS}$. $L_{QS}: \vec{r}_2 = (\widehat i + 2\widehat j - 4\widehat k) + \mu (-2\widehat i + \widehat j - 2\widehat k)$ where $\mu$ is a scalar parameter.

Step 3: Find the Intersection Point T

Lines PR and QS intersect at T. This means that at point T, their position vectors are equal. Let $\vec{t}$ be the position vector of T. Equating $\vec{r}_1$ and $\vec{r}_2$: $(3 + 4\lambda)\widehat i + (-1 - \lambda)\widehat j + (2 + 2\lambda)\widehat k = (1 - 2\mu)\widehat i + (2 + \mu)\widehat j + (-4 - 2\mu)\widehat k$ Equating the corresponding components:

1. $3 + 4\lambda = 1 - 2\mu \Rightarrow 4\lambda + 2\mu = -2 \Rightarrow 2\lambda + \mu = -1$
2. $-1 - \lambda = 2 + \mu \Rightarrow \lambda + \mu = -3$
3. $2 + 2\lambda = -4 - 2\mu \Rightarrow 2\lambda + 2\mu = -6 \Rightarrow \lambda + \mu = -3$ (This is consistent with equation 2)

Now we solve the system of equations for $\lambda$ and $\mu$: Subtract equation (2) from equation (1): $(2\lambda + \mu) - (\lambda + \mu) = -1 - (-3)$ $\lambda = 2$

Substitute $\lambda = 2$ into equation (2): $2 + \mu = -3$ $\mu = -5$

Now, substitute $\lambda = 2$ into the equation for $L_{PR}$ (or $\mu = -5$ into the equation for $L_{QS}$) to find the position vector of T: $\vec{t} = (3\widehat i - \widehat j + 2\widehat k) + 2(4\widehat i - \widehat j + 2\widehat k)$ $\vec{t} = (3 + 8)\widehat i + (-1 - 2)\widehat j + (2 + 4)\widehat k$ $\vec{t} = 11\widehat i - 3\widehat j + 6\widehat k$ So, the coordinates of point T are $(11, -3, 6)$.

Step 4: Determine the Direction of $\overrightarrow{TA}$

We are given that the vector $\overrightarrow{TA}$ is perpendicular to both $\overrightarrow{PR}$ and $\overrightarrow{QS}$. This means $\overrightarrow{TA}$ is parallel to the cross product of their direction vectors, $\vec{d}_{PR} \times \vec{d}_{QS}$. Let's calculate the cross product: $\vec{d}_{PR} \times \vec{d}_{QS} = (4\widehat i - \widehat j + 2\widehat k) \times (-2\widehat i + \widehat j - 2\widehat k)$ $= \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 4 & -1 & 2 \\ -2 & 1 & -2 \end{vmatrix}$ $= \widehat i ((-1)(-2) - (2)(1)) - \widehat j ((4)(-2) - (2)(-2)) + \widehat k ((4)(1) - (-1)(-2))$ $= \widehat i (2 - 2) - \widehat j (-8 + 4) + \widehat k (4 - 2)$ $= 0\widehat i - (-4)\widehat j + 2\widehat k$ $= 4\widehat j + 2\widehat k$ Let $\vec{d}_{TA} = 4\widehat j + 2\widehat k$ be a direction vector for $\overrightarrow{TA}$.

Step 5: Find the Vector $\overrightarrow{TA}$

The length of vector $\overrightarrow{TA}$ is given as $\sqrt{5}$ units. First, we find the unit vector in the direction of $\vec{d}_{TA}$: $|\vec{d}_{TA}| = \sqrt{0^2 + 4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$ The unit vector $\widehat u_{TA} = \frac{\vec{d}_{TA}}{|\vec{d}_{TA}|} = \frac{4\widehat j + 2\widehat k}{2\sqrt{5}} = \frac{2\widehat j + \widehat k}{\sqrt{5}}$.

Since the length of $\overrightarrow{TA}$ is $\sqrt{5}$, we can write $\overrightarrow{TA}$ as: $\overrightarrow{TA} = \pm (\text{length of } \overrightarrow{TA}) \times \widehat u_{TA}$ $\overrightarrow{TA} = \pm \sqrt{5} \times \frac{2\widehat j + \widehat k}{\sqrt{5}}$ $\overrightarrow{TA} = \pm (2\widehat j + \widehat k)$ We can choose either direction; the modulus of the position vector of A will be the same. Let's choose the positive direction: $\overrightarrow{TA} = 2\widehat j + \widehat k$

Step 6: Find the Position Vector of A

We know that $\overrightarrow{TA} = \vec{a} - \vec{t}$, where $\vec{a}$ is the position vector of A. Therefore, $\vec{a} = \vec{t} + \overrightarrow{TA}$. Substitute the values of $\vec{t}$ and $\overrightarrow{TA}$: $\vec{a} = (11\widehat i - 3\widehat j + 6\widehat k) + (2\widehat j + \widehat k)$ $\vec{a} = 11\widehat i + (-3 + 2)\widehat j + (6 + 1)\widehat k$ $\vec{a} = 11\widehat i - \widehat j + 7\widehat k$

Step 7: Calculate the Modulus of the Position Vector of A

Finally, we need to find the modulus (length) of the position vector of A, $|\vec{a}|$. $|\vec{a}| = \sqrt{11^2 + (-1)^2 + 7^2}$ $|\vec{a}| = \sqrt{121 + 1 + 49}$ $|\vec{a}| = \sqrt{171}$

---

Common Mistakes & Tips

• Cross Product Calculation: Be meticulous with the signs and calculations when computing the cross product. A common error is swapping the order of subtraction in the determinant expansion, leading to sign errors.
• Solving System of Equations: Carefully verify the values of the parameters ($\lambda$ and $\mu$) by substituting them back into all three component equations to ensure consistency.
• Direction of Perpendicular Vector: Remember that a vector perpendicular to two given vectors can point in two opposite directions. However, for quantities like magnitude or modulus of a position vector, this choice often does not affect the final numerical answer.
• Distinguishing Position Vectors from Direction Vectors: Clearly understand that $\vec{p}$ and $\vec{q}$ are position vectors of points, while $\vec{d}_{PR}$ and $\vec{d}_{QS}$ are direction vectors of lines.

---

Summary

This problem involved finding the position vector of a point A by first determining the intersection point T of two lines. We then used the property that $\overrightarrow{TA}$ is perpendicular to both lines to find its direction via the cross product of the line's direction vectors. Knowing the length of $\overrightarrow{TA}$, we constructed the vector $\overrightarrow{TA}$ and subsequently found the position vector of A. Finally, the modulus of A's position vector was calculated.

The final answer is $\boxed{\sqrt{171}}$ which corresponds to option (A).