Q1827: Find the foot of the perpendicular from the point (0, 2, 3) ...

Class CBSE Class 12 Mathematics Three Dimensional Geometry Q #1827

COMPETENCY BASED

APPLY

5 Marks 2026 AISSCE(Board Exam) LA

Find the foot of the perpendicular from the point (0, 2, 3) on the line $\frac{-x-3}{-5}=\frac{1-y}{-2}=\frac{3z+12}{9}$ and hence find the length of the perpendicular.

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AI Tutor Explanation

Detailed Solution

Step 1: Standardize the line equation

The given line is $\frac{-(x+3)}{-5} = \frac{-(y-1)}{-2} = \frac{3(z+4)}{9}$. Simplifying this, we get the standard form: $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3} = \lambda$.

Step 2: Express general point on the line

Any point $P$ on the line is given by $P(5\lambda - 3, 2\lambda + 1, 3\lambda - 4)$.

Step 3: Apply perpendicularity condition

Let $A = (0, 2, 3)$. The direction ratios of $AP$ are $(5\lambda - 3 - 0, 2\lambda + 1 - 2, 3\lambda - 4 - 3) = (5\lambda - 3, 2\lambda - 1, 3\lambda - 7)$. Since $AP$ is perpendicular to the line with direction ratios $(5, 2, 3)$, their dot product is zero: $5(5\lambda - 3) + 2(2\lambda - 1) + 3(3\lambda - 7) = 0$. Solving gives $25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0$, which simplifies to $38\lambda = 38$, so $\lambda = 1$.

Step 4: Calculate foot and length

Substituting $\lambda = 1$ into $P$, the foot is $(2, 3, -1)$. The length $AP = \sqrt{(2-0)^2 + (3-2)^2 + (-1-3)^2} = \sqrt{4 + 1 + 16} = \sqrt{21}$.

Final Answer: Foot: (2, 3, -1), Length: \sqrt{21} units

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Pedagogical Audit

Bloom's Analysis: This is an APPLY question because the student must utilize the concept of direction ratios and the dot product of perpendicular vectors to solve a geometric problem.

Knowledge Dimension: PROCEDURAL

Justification: The problem requires a multi-step algorithmic approach involving vector algebra and coordinate geometry.

Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the student's ability to manipulate line equations and apply spatial reasoning beyond rote memorization.

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MCQ_SINGLE

The lines $\frac{1-x}{2}=\frac{y-1}{3}=\frac{z}{1}$ and $\frac{2x-3}{2p}=\frac{y}{-1}=\frac{z-4}{7}$ are perpendicular to each other for p equal to: