Class CBSE Class 12 Mathematics Three Dimensional Geometry Q #878
KNOWLEDGE BASED
APPLY
3 Marks 2023 SA
(a) Find the coordinates of the foot of the perpendicular drawn from the point P(0,2,3) to the line (x+3)/5 = (y-1)/2 = (z+4)/3. OR (b) Three vectors →a, →b and →c satisfy the condition →a + →b + →c = →0. Evaluate the quantity μ = →a·→b + →b·→c + →c·→a, if |→a|=3, |→b|=4 and |→c|=2.
Prev Next

AI Tutor Explanation

Powered by Gemini

Step-by-Step Solution

Part (a): Finding the foot of the perpendicular

  1. Express the general point on the line: Let (x+3)/5 = (y-1)/2 = (z+4)/3 = λ. Then, x = 5λ - 3, y = 2λ + 1, z = 3λ - 4. So, any point on the line can be represented as Q(5λ - 3, 2λ + 1, 3λ - 4).

  2. Find the direction ratios of PQ: The direction ratios of the line PQ are (5λ - 3 - 0, 2λ + 1 - 2, 3λ - 4 - 3) = (5λ - 3, 2λ - 1, 3λ - 7).

  3. Use the perpendicularity condition: Since PQ is perpendicular to the given line, the dot product of their direction ratios is zero. 5(5λ - 3) + 2(2λ - 1) + 3(3λ - 7) = 0. 25λ - 15 + 4λ - 2 + 9λ - 21 = 0. 38λ - 38 = 0. λ = 1.

  4. Find the coordinates of the foot of the perpendicular: Substitute λ = 1 into the coordinates of Q. Q(5(1) - 3, 2(1) + 1, 3(1) - 4) = Q(2, 3, -1).

Part (b): Evaluating μ

  1. Use the given condition: →a + →b + →c = →0

  2. Square both sides: (→a + →b + →c)·(→a + →b + →c) = 0. |→a|^2 + |→b|^2 + |→c|^2 + 2(→a·→b + →b·→c + →c·→a) = 0.

  3. Substitute the given magnitudes: 3^2 + 4^2 + 2^2 + 2μ = 0. 9 + 16 + 4 + 2μ = 0. 29 + 2μ = 0.

  4. Solve for μ: 2μ = -29. μ = -29/2.

Correct Answer: (a) (2, 3, -1) OR (b) -29/2

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires students to apply their knowledge of 3D geometry (finding the foot of the perpendicular) or vector algebra (dot product and magnitude relationships) to solve a specific problem.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to execute a series of steps or algorithms to arrive at the solution. For part (a), this involves finding the general point on the line, using the dot product to ensure perpendicularity, and solving for the parameter. For part (b) it involves using the given condition to expand the square of the magnitude and then solving for the required quantity.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding and application of standard formulas and procedures related to lines in 3D space and vector algebra, as covered in the textbook.

More from this Chapter

MCQ_SINGLE
The value of $\lambda$ for which the angle between the lines $\vec{r}=\hat{i}+\hat{j}+\hat{k}+p(2\hat{i}+\hat{j}+2\hat{k})$ and $\vec{r}=(1+q)\hat{i}+(1+q\lambda)\hat{j}+(1+q)\hat{k}$ is $\frac{\pi}{2}$ :
MCQ_SINGLE
14. Distance of the point $(p, q, r)$ from y-axis is :
MCQ_SINGLE
The equation of a line parallel to the vector \(3\hat{i}+\hat{j}+2\hat{k}\) and passing through the point \((4, -3, 7)\) is:
LA
Find the value of $b$ so that the lines $\frac{x-1}{2}=\frac{y-b}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=z$ are intersecting lines. Also, find the point of intersection of these given lines.
MCQ_SINGLE
The Cartesian equation of the line passing through the point (1, -3, 2) and parallel to the line: \(\vec{r}=(2+\lambda)\hat{i}+\lambda\hat{j}+(2\lambda-1)\hat{k}\) is:
View All Questions