NIMCET 2015 — Mathematics PYQ

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Question

NIMCET 2015 — Mathematics PYQ

NIMCET | Mathematics | 2015

If $\vec{AC} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{BD} = -\hat{i} + 3\hat{j} + 2\hat{k}$ , then area of the quadrilateral $ABCD$ is:

Choose the correct answer:

Explanation

1. Identify the Diagonal Vectors:

Here, the diagonals are $\vec{AC}$ and $\vec{BD}$ :

\vec{d_1} = \vec{AC} = 2\hat{i} + \hat{j} + \hat{k}

\vec{d_2} = \vec{BD} = -\hat{i} + 3\hat{j} + 2\hat{k}

2. Calculate the Cross Product ( $\vec{d_1} \times \vec{d_2}$ ):

\vec{AC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ -1 & 3 & 2 \end{vmatrix}

Expanding the determinant:

= \hat{i}(1 \cdot 2 - 3 \cdot 1) - \hat{j}(2 \cdot 2 - (-1) \cdot 1) + \hat{k}(2 \cdot 3 - (-1) \cdot 1)

= \hat{i}(2 - 3) - \hat{j}(4 + 1) + \hat{k}(6 + 1)

= -\hat{i} - 5\hat{j} + 7\hat{k}

3. Find the Magnitude of the Cross Product:

| \vec{AC} \times \vec{BD} | = \sqrt{(-1)^2 + (-5)^2 + (7)^2}

= \sqrt{1 + 25 + 49}

= \sqrt{75}

= \sqrt{25 \times 3} = 5\sqrt{3}

4. Calculate the Area:

\text{Area} = \frac{1}{2} | \vec{AC} \times \vec{BD} |

\text{Area} = \frac{1}{2} (5\sqrt{3}) = \frac{5}{2}\sqrt{3} \text{ sq. units}

Correct Option:

(a) $\frac{5}{2}\sqrt{3}$