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Step-by-Step Solution
Step 1: Identify the Given Vectors and Directions
• The position vector of point P is
$ \overrightarrow{P} = 3\hat{i} - \hat{j} + 2\hat{k} $.
• The position vector of point Q is
$ \overrightarrow{Q} = \hat{i} + 2\hat{j} - 4\hat{k} $.
• The direction ratios (D.R.s) of line PR are given by
$ (4, -1, 2). $
• The direction ratios (D.R.s) of line QS are given by
$ (-2, 1, -2). $
Step 2: Write the Parametric Equations of the Lines PR and QS
• Any point on line PR can be written as:
$$
\overrightarrow{r} = \overrightarrow{P} + \lambda \, (4\hat{i} - \hat{j} + 2\hat{k}).
$$
In coordinate form, a generic point $ (x,\, y,\, z) $ on PR is:
$$
(3 + 4\lambda,\, -1 - \lambda,\, 2 + 2\lambda).
$$
• Any point on line QS can be written as:
$$
\overrightarrow{r} = \overrightarrow{Q} + \mu \, (-2\hat{i} + \hat{j} - 2\hat{k}).
$$
In coordinate form, a generic point $ (x,\, y,\, z) $ on QS is:
$$
(1 - 2\mu,\, 2 + \mu,\, -4 - 2\mu).
$$
Step 3: Find the Intersection Point T
Point T lies on both lines, so its coordinates must satisfy both parametric forms simultaneously:
$$
3 + 4\lambda = 1 - 2\mu, \quad -1 - \lambda = 2 + \mu, \quad 2 + 2\lambda = -4 - 2\mu.
$$
From the first two equations (it is sufficient to solve two for $\lambda$ and $\mu$):
1) $3 + 4\lambda = 1 - 2\mu \; \Rightarrow \; \mu + 2\lambda = -1$
2) $-1 - \lambda = 2 + \mu \; \Rightarrow \; \mu + \lambda = -3$
Solving these simultaneously,
$$
\lambda = 2, \quad \mu = -5.
$$
Plug these back in to get T:
$$
T = (3 + 4\cdot 2,\, -1 - 2,\, 2 + 2\cdot 2) = (11,\, -3,\, 6).
$$
Step 4: Determine the Direction of Vector TA
It is given that $ \overrightarrow{TA} $ is perpendicular to both $ \overrightarrow{PR} $ and $ \overrightarrow{QS} $. A vector perpendicular to two given vectors can be found using their cross product:
$$
\overrightarrow{v}_{QS} \times \overrightarrow{v}_{PR}
=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k}\\
-2 & 1 & -2 \\
4 & -1 & 2
\end{vmatrix}.
$$
Compute this determinant:
1) Expand along the first row:
$$
= \hat{i} \bigl((1)(2) - (-2)(-1)\bigr)
- \hat{j} \bigl((-2)(2) - (-2)(4)\bigr)
+ \hat{k} \bigl((-2)(-1) - (1)(4)\bigr).
$$
2) Simplify each term:
$ \hat{i} (2 - 2) = 0 \hat{i}, $
$ -\hat{j} \bigl(-4 - (-8)\bigr) = -\hat{j} ( -4 + 8 ) = -\hat{j} (4) = -4\hat{j}, $
$ \hat{k} \bigl(2 - 4\bigr) = -2\hat{k}. $
Therefore,
$$
\overrightarrow{v}_{QS} \times \overrightarrow{v}_{PR} = 0\hat{i} - 4\hat{j} - 2\hat{k}.
$$
Hence, $ \overrightarrow{TA} $ is parallel to $ \left\langle 0, -4, -2 \right\rangle. $
Step 5: Write the Parametric Form for Line TA
Since T is $ (11, -3, 6) $, and direction ratios for TA are $ (0, -4, -2) $, any point A on TA can be expressed as:
$$
\overrightarrow{r} = (11, -3, 6) + \lambda\,(0, -4, -2).
$$
Thus, the coordinates of A are:
$$
A = (11,\, -3 - 4\lambda,\, 6 - 2\lambda).
$$
Step 6: Use the Given Magnitude Condition $|\overrightarrow{TA}| = \sqrt{5}$
The vector $ \overrightarrow{TA} $ from $T$ to $A$ is
$$
(0,\, -4\lambda,\, -2\lambda).
$$
Its magnitude is:
$$
|\overrightarrow{TA}|
= \sqrt{(0)^2 + (-4\lambda)^2 + (-2\lambda)^2}
= \sqrt{16\lambda^2 + 4\lambda^2}
= \sqrt{20\lambda^2}
= \sqrt{5} \quad \text{(given)}.
$$
So we have:
$$
\sqrt{20\lambda^2} = \sqrt{5},
$$
which gives
$$
\sqrt{20}\,|\lambda| = \sqrt{5} \quad \Rightarrow \quad \sqrt{20} \,|\lambda| = \sqrt{5}.
$$
Hence,
$$
|\lambda| = \frac{\sqrt{5}}{\sqrt{20}} = \frac{1}{2}.
$$
Therefore,
$$
\lambda = \pm \frac{1}{2}.
$$
Step 7: Find the Coordinates of A for Each Value of $\lambda$
Case 1: $ \lambda = \frac{1}{2} $
$
A = \Bigl(11,\; -3 - 4\Bigl(\tfrac{1}{2}\Bigr),\; 6 - 2\Bigl(\tfrac{1}{2}\Bigr)\Bigr)
= (11,\; -3 - 2,\; 6 - 1)
= (11,\; -5,\; 5).
$
Case 2: $ \lambda = -\frac{1}{2} $
$
A = \Bigl(11,\; -3 - 4\Bigl(-\tfrac{1}{2}\Bigr),\; 6 - 2\Bigl(-\tfrac{1}{2}\Bigr)\Bigr)
= (11,\; -3 + 2,\; 6 + 1)
= (11,\; -1,\; 7).
$
In either case, we will see that the modulus $|A|$ (distance from origin) will turn out to be the same if the result is $ \sqrt{171} $.
Step 8: Compute the Modulus of A
Let's compute $|A|$ using the first set of coordinates $ (11, -5, 5) $:
$$
|A| = \sqrt{(11)^2 + (-5)^2 + (5)^2}
= \sqrt{121 + 25 + 25}
= \sqrt{171}.
$$
Similarly, for the second possible A (11, -1, 7), the square of its distance from origin is:
$$
11^2 + (-1)^2 + 7^2 = 121 + 1 + 49 = 171.
$$
So in both cases,
$$
|A| = \sqrt{171}.
$$
Final Answer
The modulus of the position vector of point A is
$$
\sqrt{171}.
$$
