Class CBSE Class 12 Mathematics Three Dimensional Geometry Q #1395
KNOWLEDGE BASED
REMEMBER
5 Marks 2025 AISSCE(Board Exam) LA
Find the shortest distance between the lines: $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$ and $\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$.
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Detailed Solution

Step 1: Identify the direction vectors and a point on each line

The given lines are: Line 1: $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$ Line 2: $\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$ For Line 1, the direction vector $\vec{b_1} = 2\hat{i} + \hat{j} - 3\hat{k}$ and a point on the line is $A(-1, 1, 9)$. For Line 2, the direction vector $\vec{b_2} = 2\hat{i} - 7\hat{j} + 5\hat{k}$ and a point on the line is $B(3, -15, 9)$.

Step 2: Find the vector connecting the two points

The vector connecting the points $A$ and $B$ is: $\vec{AB} = (3 - (-1))\hat{i} + (-15 - 1)\hat{j} + (9 - 9)\hat{k} = 4\hat{i} - 16\hat{j} + 0\hat{k}$

Step 3: Calculate the cross product of the direction vectors

The cross product of the direction vectors $\vec{b_1}$ and $\vec{b_2}$ is: $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 2 & -7 & 5 \end{vmatrix} = (5 - 21)\hat{i} - (10 - (-6))\hat{j} + (-14 - 2)\hat{k} = -16\hat{i} - 16\hat{j} - 16\hat{k}$

Step 4: Find the magnitude of the cross product

The magnitude of the cross product is: $|\vec{b_1} \times \vec{b_2}| = \sqrt{(-16)^2 + (-16)^2 + (-16)^2} = \sqrt{3 \cdot 16^2} = 16\sqrt{3}$

Step 5: Calculate the scalar triple product

The scalar triple product is the dot product of $\vec{AB}$ and $\vec{b_1} \times \vec{b_2}$: $\vec{AB} \cdot (\vec{b_1} \times \vec{b_2}) = (4\hat{i} - 16\hat{j} + 0\hat{k}) \cdot (-16\hat{i} - 16\hat{j} - 16\hat{k}) = (4)(-16) + (-16)(-16) + (0)(-16) = -64 + 256 + 0 = 192$

Step 6: Calculate the shortest distance

The shortest distance $d$ between the lines is given by: $d = \frac{|\vec{AB} \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} = \frac{|192|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = \frac{12\sqrt{3}}{3} = 4\sqrt{3}$

Final Answer: $4\sqrt{3}$

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Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because it requires recalling the formula for the shortest distance between two skew lines and applying it directly.
Knowledge Dimension: PROCEDURAL
Justification: The question involves applying a specific algorithm or method (formula) to calculate the shortest distance between two lines in 3D space. This requires knowing the steps and order of operations.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It is a standard textbook problem. The question directly tests the student's knowledge of the formula and its application.

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