Class CBSE Class 12 Mathematics Vector Algebra Q #1724
COMPETENCY BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
Three points $A(0,1,1)$, $B(2,0,-1)$ and $C(1,0,3)$ form $\Delta ABC$. The ar ($\Delta ABC$) is:
(A) $\frac{\sqrt{53}}{2}$ sq. units
(B) $\sqrt{53}$ sq. units
(C) $\frac{\sqrt{11}}{2}$ sq. units
(D) $\sqrt{11}$ sq. units
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Correct Answer: A

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Detailed Solution

Step 1: Define vectors for two sides of the triangle

To find the area of triangle ABC, we first determine two vectors originating from the same vertex, say A. Let $\vec{AB}$ and $\vec{AC}$ be these vectors. $$ \vec{AB} = (2-0)\hat{i} + (0-1)\hat{j} + (-1-1)\hat{k} = 2\hat{i} - \hat{j} - 2\hat{k} $$ $$ \vec{AC} = (1-0)\hat{i} + (0-1)\hat{j} + (3-1)\hat{k} = \hat{i} - \hat{j} + 2\hat{k} $$

Step 2: Calculate the cross product

The area of the triangle is given by $\frac{1}{2} |\vec{AB} \times \vec{AC}|$. First, compute the cross product: $$ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & -2 \\ 1 & -1 & 2 \end{vmatrix} $$ $$ = \hat{i}(-2 - 2) - \hat{j}(4 - (-2)) + \hat{k}(-2 - (-1)) $$ $$ = -4\hat{i} - 6\hat{j} - \hat{k} $$

Step 3: Calculate the magnitude and final area

Find the magnitude of the resulting vector: $$ |\vec{AB} \times \vec{AC}| = \sqrt{(-4)^2 + (-6)^2 + (-1)^2} = \sqrt{16 + 36 + 1} = \sqrt{53} $$ The area is $\frac{1}{2} \times \sqrt{53}$.

Final Answer: \frac{\sqrt{53}}{2} sq. units

AI Suggestion: Option A

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must utilize the vector cross-product formula to solve a geometric problem.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the execution of a specific sequence of mathematical steps (vector subtraction, determinant calculation, and magnitude computation).
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the application of Vector Algebra in 3D geometry, a core component of the Class 12 Mathematics curriculum.

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