Class CBSE Class 12 Mathematics Vector Algebra Q #1279
COMPETENCY BASED
UNDERSTAND
3 Marks 2024 AISSCE(Board Exam) SA
The position vectors of vertices of $\Delta$ ABC are $A(2\hat{i}-\hat{j}+\hat{k}),$ $B(\hat{i}-3\hat{j}-5\hat{k})$ and $C(3\hat{i}-4\hat{j}-4\hat{k})$ Find all the angles of $\Delta$ AВС.
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Detailed Solution

Step 1: Find the vectors representing the sides of the triangle.

We need to find the vectors $\overrightarrow{AB}$, $\overrightarrow{BC}$, and $\overrightarrow{CA}$. $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (\hat{i}-3\hat{j}-5\hat{k}) - (2\hat{i}-\hat{j}+\hat{k}) = -\hat{i} - 2\hat{j} - 6\hat{k}$ $\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (3\hat{i}-4\hat{j}-4\hat{k}) - (\hat{i}-3\hat{j}-5\hat{k}) = 2\hat{i} - \hat{j} + \hat{k}$ $\overrightarrow{CA} = \overrightarrow{OA} - \overrightarrow{OC} = (2\hat{i}-\hat{j}+\hat{k}) - (3\hat{i}-4\hat{j}-4\hat{k}) = -\hat{i} + 3\hat{j} + 5\hat{k}$

Step 2: Calculate the magnitudes of the vectors.

We need to find the magnitudes of $\overrightarrow{AB}$, $\overrightarrow{BC}$, and $\overrightarrow{CA}$. $|\overrightarrow{AB}| = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}$ $|\overrightarrow{BC}| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ $|\overrightarrow{CA}| = \sqrt{(-1)^2 + (3)^2 + (5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$

Step 3: Find the angles using the dot product formula.

We use the formula $\cos{\theta} = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| |\overrightarrow{b}|}$ to find the angles. Angle A: $\cos{A} = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|} = \frac{(-\hat{i} - 2\hat{j} - 6\hat{k}) \cdot (\hat{i} - 3\hat{j} - 5\hat{k})}{\sqrt{41} \sqrt{35}} = \frac{-1 + 6 + 30}{\sqrt{41} \sqrt{35}} = \frac{35}{\sqrt{41} \sqrt{35}} = \frac{\sqrt{35}}{\sqrt{41}}$ $A = \cos^{-1}(\frac{\sqrt{35}}{\sqrt{41}})$ Angle B: $\cos{B} = \frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BA}| |\overrightarrow{BC}|} = \frac{(\hat{i} + 2\hat{j} + 6\hat{k}) \cdot (2\hat{i} - \hat{j} + \hat{k})}{\sqrt{41} \sqrt{6}} = \frac{2 - 2 + 6}{\sqrt{41} \sqrt{6}} = \frac{6}{\sqrt{41} \sqrt{6}} = \frac{\sqrt{6}}{\sqrt{41}}$ $B = \cos^{-1}(\frac{\sqrt{6}}{\sqrt{41}})$ Angle C: $\cos{C} = \frac{\overrightarrow{CB} \cdot \overrightarrow{CA}}{|\overrightarrow{CB}| |\overrightarrow{CA}|} = \frac{(-2\hat{i} + \hat{j} - \hat{k}) \cdot (-\hat{i} + 3\hat{j} + 5\hat{k})}{\sqrt{6} \sqrt{35}} = \frac{2 + 3 - 5}{\sqrt{6} \sqrt{35}} = \frac{0}{\sqrt{6} \sqrt{35}} = 0$ $C = \cos^{-1}(0) = \frac{\pi}{2}$

Step 4: State the final answer.

The angles of the triangle are: $A = \cos^{-1}(\frac{\sqrt{35}}{\sqrt{41}})$ $B = \cos^{-1}(\frac{\sqrt{6}}{\sqrt{41}})$ $C = \frac{\pi}{2}$

Final Answer: $A = \cos^{-1}(\frac{\sqrt{35}}{\sqrt{41}}), B = \cos^{-1}(\frac{\sqrt{6}}{\sqrt{41}}), C = \frac{\pi}{2}$

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Pedagogical Audit
Bloom's Analysis: This is an UNDERSTAND question because it requires the student to understand the concepts of vectors, dot product, and how to use them to find angles in a triangle. The student needs to comprehend the relationship between position vectors and the sides of a triangle.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply a specific procedure (using vector subtraction and the dot product formula) to calculate the angles of the triangle. It involves knowing the steps and algorithms to solve the problem.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It assesses the student's ability to apply vector algebra concepts to solve a geometrical problem, requiring them to demonstrate understanding and application rather than rote memorization.

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