Q1825: Opposite sides of a square are along the lines : $\vec{r}=\h...

Class CBSE Class 12 Mathematics Three Dimensional Geometry Q #1825

COMPETENCY BASED

APPLY

5 Marks 2026 AISSCE(Board Exam) LA

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

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AI Tutor Explanation

Detailed Solution

Step 1: Identify the distance between parallel lines

The lines are given by $\vec{r} = \vec{a_1} + \lambda\vec{b}$ and $\vec{r} = \vec{a_2} + \mu\vec{b}$, where $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$. The distance $d$ between these parallel lines is given by: $$d = \frac{|\vec{b} \times (\vec{a_2} - \vec{a_1})|}{|\vec{b}|}$$ Here, $\vec{a_2} - \vec{a_1} = (3-1)\hat{i} + (3-2)\hat{j} + (-5 - (-4))\hat{k} = 2\hat{i} + \hat{j} - \hat{k}$.

Step 2: Calculate the cross product and magnitude

$\vec{b} \times (\vec{a_2} - \vec{a_1}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 2 & 1 & -1 \end{vmatrix} = \hat{i}(-3-6) - \hat{j}(-2-12) + \hat{k}(2-6) = -9\hat{i} + 14\hat{j} - 4\hat{k}$. The magnitude is $\sqrt{(-9)^2 + 14^2 + (-4)^2} = \sqrt{81 + 196 + 16} = \sqrt{293}$. The magnitude of $\vec{b}$ is $\sqrt{2^2 + 3^2 + 6^2} = \sqrt{4+9+36} = 7$. Thus, $d = \frac{\sqrt{293}}{7}$.

Step 3: Calculate the area of the square

Since $d$ represents the side length of the square, the area is $d^2 = \frac{293}{49}$ square units.

Final Answer: 293/49

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Pedagogical Audit

Bloom's Analysis: This is an APPLY question because it requires the student to translate geometric properties of a square into vector algebraic operations.

Knowledge Dimension: PROCEDURAL

Justification: The student must follow a specific sequence of vector operations (cross product, magnitude, distance formula) to reach the solution.

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the application of 3D Geometry concepts beyond standard textbook problems.

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