Class CBSE Class 12 Mathematics Three Dimensional Geometry Q #1442
KNOWLEDGE BASED
REMEMBER
5 Marks 2025 AISSCE(Board Exam) LA
Find the point on the line $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-4}{3}$ at a distance of $2\sqrt{2}$ units from the point (-1, -1, 2).
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Detailed Solution

Step 1: Express the general point on the line

Let the given line be $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-4}{3} = \lambda$. Then, any point on the line can be expressed as $(3\lambda + 1, 2\lambda - 1, 3\lambda + 4)$.

Step 2: Use the distance formula

Let the point on the line be $P(3\lambda + 1, 2\lambda - 1, 3\lambda + 4)$ and the given point be $A(-1, -1, 2)$. The distance between $P$ and $A$ is given as $2\sqrt{2}$. Using the distance formula: $$PA = \sqrt{((3\lambda + 1) - (-1))^2 + ((2\lambda - 1) - (-1))^2 + ((3\lambda + 4) - 2)^2} = 2\sqrt{2}$$ $$PA = \sqrt{(3\lambda + 2)^2 + (2\lambda)^2 + (3\lambda + 2)^2} = 2\sqrt{2}$$

Step 3: Simplify and solve for λ

Squaring both sides: $$(3\lambda + 2)^2 + (2\lambda)^2 + (3\lambda + 2)^2 = (2\sqrt{2})^2$$ $$9\lambda^2 + 12\lambda + 4 + 4\lambda^2 + 9\lambda^2 + 12\lambda + 4 = 8$$ $$22\lambda^2 + 24\lambda + 8 = 8$$ $$22\lambda^2 + 24\lambda = 0$$ $$2\lambda(11\lambda + 12) = 0$$ So, $\lambda = 0$ or $\lambda = -\frac{12}{11}$

Step 4: Find the points

Case 1: $\lambda = 0$ The point is $(3(0) + 1, 2(0) - 1, 3(0) + 4) = (1, -1, 4)$ Case 2: $\lambda = -\frac{12}{11}$ The point is $(3(-\frac{12}{11}) + 1, 2(-\frac{12}{11}) - 1, 3(-\frac{12}{11}) + 4) = (-\frac{36}{11} + \frac{11}{11}, -\frac{24}{11} - \frac{11}{11}, -\frac{36}{11} + \frac{44}{11}) = (-\frac{25}{11}, -\frac{35}{11}, \frac{8}{11})$

Final Answer: (1, -1, 4) and (-25/11, -35/11, 8/11)

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Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because it requires recalling the distance formula and the parametric form of a line.
Knowledge Dimension: PROCEDURAL
Justification: The question requires applying a specific procedure (distance formula and parametric representation of a line) to find the solution.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly tests the student's understanding of 3D geometry concepts from the textbook.

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