Class CBSE Class 12 Mathematics Three Dimensional Geometry Q #1853
COMPETENCY BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
The length of perpendicular drawn from point $(2, 5, 7)$ on line $\frac{x}{1}=\frac{y}{0}=\frac{z}{0}$ is
(A) 2
(B) 5
(C) $\sqrt{74}$
(D) $\sqrt{78}$
Prev Finish
Correct Answer: C

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Detailed Solution

Step 1: Identify the line and point

The given line is x/1 = y/0 = z/0. This represents the x-axis, where any point on the line is of the form (k, 0, 0). The given point is P(2, 5, 7).

Step 2: Find the foot of the perpendicular

Let the foot of the perpendicular from P(2, 5, 7) to the line be M(k, 0, 0). The direction vector of the line is d = (1, 0, 0). The vector PM is (k-2, -5, -7). Since PM is perpendicular to the line, their dot product is zero:

$$(k-2)(1) + (-5)(0) + (-7)(0) = 0$$

This gives k - 2 = 0, so k = 2. Thus, the foot of the perpendicular is M(2, 0, 0).

Step 3: Calculate the distance

The length of the perpendicular is the distance between P(2, 5, 7) and M(2, 0, 0):

$$d = \sqrt{(2-2)^2 + (5-0)^2 + (7-0)^2}$$ $$d = \sqrt{0^2 + 5^2 + 7^2} = \sqrt{25 + 49} = \sqrt{74}$$

Final Answer: \sqrt{74}

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must utilize the concept of vector projection and perpendicularity conditions to solve a geometric problem.
Knowledge Dimension: PROCEDURAL
Justification: The question requires a step-by-step algorithmic approach involving coordinate geometry and vector dot products.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the student's ability to apply 3D geometry concepts to find distances between points and lines, a core competency in the Vector Algebra and 3D Geometry unit.

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