Question

Opposite sides of a square are along the lines : \[ \vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) \quad \text{and} \quad \vec{r} = 3\hat{i} + 3\hat{j} - 5\hat{k} + \mu(2\hat{i} + 3\hat{j} + 6\hat{k}) \] 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:

Always double check vector components during cross-product evaluations. Finding the distance between parallel lines gives you the exact side length of the square, and squaring it gives the total area directly.

Answer 1

Concept:
• The distance between two parallel lines $\vec{r} = \vec{a}_1 + \lambda \vec{b}$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}$ represents the side length ($d$) of the square: \[ d = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|} \]
• Since a square has perpendicular adjacent sides, the direction vector of the first pair ($\vec{b}$) must be perpendicular to the direction vector of the second pair ($\vec{c}$), meaning their dot product is zero: $\vec{b} \cdot \vec{c} = 0$.

Step 1: Calculate the value of $p$ using orthogonality.
The direction vector of the given parallel lines is $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$. The direction ratios of the other pair of sides are given as $\langle -3, 6, p \rangle$, which translates to a direction vector $\vec{c} = -3\hat{i} + 6\hat{j} + p\hat{k}$. Since adjacent sides of a square are perpendicular: \[ \vec{b} \cdot \vec{c} = 0 \implies (2)(-3) + (3)(6) + (6)(p) = 0 \] \[ -6 + 18 + 6p = 0 \implies 12 + 6p = 0 \implies 6p = -12 \implies p = -2 \]

Step 2: Set up parameters to find the distance between the parallel lines.
From the line equations, extract the position vectors: \[ \vec{a}_1 = \hat{i} + 2\hat{j} - 4\hat{k}, \quad \vec{a}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k} \] Calculate the difference vector $(\vec{a}_2 - \vec{a}_1)$: \[ \vec{a}_2 - \vec{a}_1 = (3 - 1)\hat{i} + (3 - 2)\hat{j} + (-5 - (-4))\hat{k} = 2\hat{i} + \hat{j} - \hat{k} \]

Step 3: Compute the cross product $(\vec{a}_2 - \vec{a}_1) \times \vec{b$.}
\[ (\vec{a}_2 - \vec{a}_1) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 2 & 1 & -1 2 & 3 & 6 \end{vmatrix} \] Expanding along the first row: \[ = \hat{i}((1)(6) - (-1)(3)) - \hat{j}((2)(6) - (-1)(2)) + \hat{k}((2)(3) - (1)(2)) \] \[ = \hat{i}(6 + 3) - \hat{j}(12 + 2) + \hat{k}(6 - 2) = 9\hat{i} - 14\hat{j} + 4\hat{k} \]

Step 4: Determine the magnitude of the cross product and the direction vector.
\[ |(\vec{a}_2 - \vec{a}_1) \times \vec{b}| = \sqrt{9^2 + (-14)^2 + 4^2} = \sqrt{81 + 196 + 16} = \sqrt{293} \] \[ |\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \] The perpendicular distance (side length $d$) is: \[ d = \frac{\sqrt{293}}{7} \]

Step 5: Calculate the area of the square.
The area of a square is equal to the square of its side length ($d^2$): \[ \text{Area} = d^2 = \left(\frac{\sqrt{293}}{7}\right)^2 = \frac{293}{49} \approx 5.98 \text{ square units} \]