Area (in sq. units) of the triangle with vertices A (1,1,1) B (1,2,3) and C (2,3,1) is: Option 1: 1/2sqrt(21) Option 2: sqrt(21) Option 3: 3 Option 4: 3/2

The answer is '1/2sqrt(21)'. View full solution:

CUET Mathematics - Area (in sq. units) of the triangle with vertices A (1,1,1) B (1,2,3) and C (2,3,1) is: | PYQs + Solutions

CUET Mathematics 2026 26th May Shift 2

Algebra

Medium

core

$\dfrac{1}{2}\sqrt{21}$

$\sqrt{21}$

$\dfrac{3}{2}$

✅ Correct Option: 1

Related questions:

25 May Shift 1

Match List I with List II

LIST I	LIST II
A. The area of parallelogram determined by vectors $2\hat{i}$ and $3\hat{j}$	I. 2
B. The value of $(\hat{i} \times \hat{j}) \cdot \hat{k} + (\hat{j} \times \hat{k}) \cdot \hat{i}$	II. 4
C. The value of a for which the vectors $2\hat{i} - 3\hat{j} + 4\hat{k}$ and $a\hat{i} - 6\hat{j} + 8\hat{k}$ are collinear.	III. 0
D. The value of $\lambda$ for which the vectors $2\hat{i} + \hat{j} + \hat{k}$ and $2\hat{i} - 4\hat{j} + \lambda\hat{k}$ are perpendicular	IV. 6

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16 July Shift 2

If $|\vec{a}| = |\vec{b}| = |\vec{a} + \vec{b}| = 1$ , then $|\vec{a} - \vec{b}|$ is equal to:

15th May Shift 1

Match List-I with List-II

LIST-I	LIST-II
A. $\vec{a}.\vec{c} + \vec{b}.\vec{c}$	I. $-25$
B. $\vec{a}.\vec{b} + \vec{a}.\vec{c}$	II. $-338$
C. $\vec{a}.\vec{b} + \vec{b}.\vec{c}$	III. $-169$
D. $2\left(\vec{a}.\vec{b} + \vec{b}.\vec{c} + \vec{c}.\vec{a}\right)$	IV. $-144$

Choose the correct answer from the options given below: