Class CBSE Class 12 Mathematics Three Dimensional Geometry Q #1352
COMPETENCY BASED
UNDERSTAND
5 Marks 2024 AISSCE(Board Exam) LA
Find the co-ordinates of the foot of the perpendicular drawn from the point (2, 3, -8) to the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$ Also, find the perpendicular distance of the given point from the line.
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Detailed Solution

Step 1: Rewrite the equation of the line in standard form

The given equation of the line is $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$. We rewrite it in the standard form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$. So, we have $\frac{x-4}{-2}=\frac{y-0}{6}=\frac{z-1}{-3}$.

Step 2: Express the coordinates of a general point on the line

Let the coordinates of any point on the line be $P$. Then, we can express the coordinates of $P$ in terms of a parameter $\lambda$ as follows: $\frac{x-4}{-2}=\frac{y-0}{6}=\frac{z-1}{-3} = \lambda$ $x = 4 - 2\lambda$ $y = 6\lambda$ $z = 1 - 3\lambda$ So, the coordinates of $P$ are $(4 - 2\lambda, 6\lambda, 1 - 3\lambda)$.

Step 3: Find the direction ratios of the line joining the given point and the foot of the perpendicular

Let $A = (2, 3, -8)$ be the given point. Let $P$ be the foot of the perpendicular from $A$ to the line. The direction ratios of the line $AP$ are given by: $a_1 = (4 - 2\lambda) - 2 = 2 - 2\lambda$ $b_1 = 6\lambda - 3$ $c_1 = (1 - 3\lambda) - (-8) = 9 - 3\lambda$

Step 4: Use the perpendicularity condition

Since $AP$ is perpendicular to the given line, the dot product of their direction ratios is zero. The direction ratios of the given line are $-2, 6, -3$. So, we have: $(-2)(2 - 2\lambda) + (6)(6\lambda - 3) + (-3)(9 - 3\lambda) = 0$ $-4 + 4\lambda + 36\lambda - 18 - 27 + 9\lambda = 0$ $49\lambda - 49 = 0$ $\lambda = 1$

Step 5: Find the coordinates of the foot of the perpendicular

Substitute $\lambda = 1$ in the coordinates of $P$: $x = 4 - 2(1) = 2$ $y = 6(1) = 6$ $z = 1 - 3(1) = -2$ So, the coordinates of the foot of the perpendicular are $(2, 6, -2)$.

Step 6: Calculate the perpendicular distance

The perpendicular distance is the distance between $A(2, 3, -8)$ and $P(2, 6, -2)$. $AP = \sqrt{(2-2)^2 + (6-3)^2 + (-2-(-8))^2}$ $AP = \sqrt{0^2 + 3^2 + 6^2}$ $AP = \sqrt{0 + 9 + 36}$ $AP = \sqrt{45} = 3\sqrt{5}$

Final Answer: The coordinates of the foot of the perpendicular are (2, 6, -2) and the perpendicular distance is $3\sqrt{5}$ units.

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply their understanding of 3D geometry, direction ratios, and perpendicularity conditions to solve the problem.
Knowledge Dimension: PROCEDURAL
Justification: The question requires a series of steps to be performed in a specific order, such as finding the general point on a line, using the perpendicularity condition, and calculating the distance.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question requires the student to apply concepts of 3D geometry to find the foot of the perpendicular and the perpendicular distance, demonstrating a deeper understanding and application of the concepts.

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