Question

Opposite sides of a square are along the lines :
$\vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$
$\vec{r} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(2\hat{i} + 3\hat{j} + 6\hat{k})$
Find the area of the square if direction ratios of other pair of opposite sides of the square are given by $\langle -3, 6, p \rangle$. Also, find the value of $p$.}

Hint:

For distance between parallel lines, you can also use the projection formula or pick a point on one line and find its perpendicular distance to the other. The orthogonality condition ($\vec{b_1} \cdot \vec{b_2} = 0$) is the fastest way to find missing parameters in geometry problems.

Answer 1

Concept:
The two given lines are parallel, so the distance between them represents the side length of the square.
The area of a square is: \[ \text{Area}=s^2 \] Also, adjacent sides of a square are perpendicular, so their direction vectors have zero dot product. 
Step 1: Find the side length of the square
For two parallel lines, the distance between them is: \[ d=\frac{|(\vec{a_2}-\vec{a_1})\times\vec{b}|}{|\vec{b}|} \] Here: \[ \vec{a_2}-\vec{a_1} =(3-1)\hat{i}+(3-2)\hat{j}+(-5+4)\hat{k} \] \[ \vec{a_2}-\vec{a_1}=2\hat{i}+\hat{j}-\hat{k} \] The direction vector is: \[ \vec{b}=2\hat{i}+3\hat{j}+6\hat{k} \] Now: \[ (\vec{a_2}-\vec{a_1})\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix} \] \[ =9\hat{i}-14\hat{j}+4\hat{k} \] Also: \[ |\vec{b}|=\sqrt{2^2+3^2+6^2}=7 \] Therefore: \[ d= \frac{\sqrt{9^2+(-14)^2+4^2}}{7} \] \[ d=\frac{\sqrt{293}}{7} \] Hence, the side length is: \[ s=\frac{\sqrt{293}}{7} \] 
Step 2: Calculate the area of the square
Using: \[ \text{Area}=s^2 \] \[ \text{Area} = \left(\frac{\sqrt{293}}{7}\right)^2 \] \[ \text{Area}=\frac{293}{49} \] 
Step 3: Find the value of \(p\)
The direction vectors of adjacent sides are: \[ \langle 2,3,6\rangle \] and \[ \langle -3,6,p\rangle \] Since adjacent sides are perpendicular: \[ (2)(-3)+(3)(6)+(6)(p)=0 \] \[ -6+18+6p=0 \] \[ 12+6p=0 \] \[ p=-2 \] 
Final Answer:
Therefore: \[ \boxed{\text{Area}=\frac{293}{49}\text{ sq. units}} \] and \[ \boxed{p=-2} \]