Class CBSE Class 12 Mathematics Three Dimensional Geometry Q #881
KNOWLEDGE BASED
APPLY
2 Marks 2023 VSA
25. (a) Find the vector equation of the line passing through the point $(2, 1, 3)$ and perpendicular to both the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3} ; \frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$
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Step-by-Step Solution

  1. Let the direction ratios of the given lines be $\vec{b_1} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b_2} = -3\hat{i} + 2\hat{j} + 5\hat{k}$.

  2. The line perpendicular to both lines will have direction ratios proportional to the cross product of $\vec{b_1}$ and $\vec{b_2}$.

    $\vec{b} = \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & 2 & 5 \end{vmatrix} = (10-6)\hat{i} - (5+9)\hat{j} + (2+6)\hat{k} = 4\hat{i} - 14\hat{j} + 8\hat{k}$

  3. The position vector of the point $(2, 1, 3)$ is $\vec{a} = 2\hat{i} + \hat{j} + 3\hat{k}$.

  4. The vector equation of the line is given by $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\lambda$ is a scalar.

    $\vec{r} = (2\hat{i} + \hat{j} + 3\hat{k}) + \lambda (4\hat{i} - 14\hat{j} + 8\hat{k})$

Correct Answer: $\vec{r} = (2\hat{i} + \hat{j} + 3\hat{k}) + \lambda (4\hat{i} - 14\hat{j} + 8\hat{k})$

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires students to apply their knowledge of vector equations of lines and cross products to find the equation of a line satisfying given conditions.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a specific procedure to find the vector equation of the line, involving finding the direction vector using cross product and then forming the equation.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding of vector algebra and 3D geometry concepts as covered in the textbook.

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