Let A(2, 3, 5) and C(–3, 4, –2) be opposite vertices of a parallelogram ABCD. If the diagonal , then the area of the parallelogram is equal to [2024]

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Q.
Let A(2, 3, 5) and C(–3, 4, –2) be opposite vertices of a parallelogram ABCD. If the diagonal $\vec{B D} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , then the area of the parallelogram is equal to [2024]

1 $\frac{1}{2} \sqrt{410}$

2 $\frac{1}{2} \sqrt{474}$

3 $\frac{1}{2} \sqrt{586}$

4 $\frac{1}{2} \sqrt{306}$

Ans.
(2)

Area of parallelogram ABCD

$= \frac{1}{2}$ $| \vec{A C} \times \vec{B D} |$

$= \frac{1}{2}$ $| (- 5 \hat{i} + \hat{j} - 7 \hat{j}) \times (\hat{i} + 2 \hat{j} + 3 \hat{k}) |$

$= \frac{1}{2}$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 5 & 1 & - 7 \\ 1 & 2 & 3 \end{matrix} | |$

$= \frac{1}{2}$ $| (3 + 14) \hat{i} - (- 15 + 7) \hat{j} + (- 10 - 1) \hat{k} |$

$= \frac{1}{2}$ $| (17 \hat{i} + 8 \hat{j} - 11 \hat{k}) |$

$= \frac{1}{2} \sqrt{289 + 64 + 121} = \frac{1}{2} \sqrt{474}$ sq. units.

Solution

Q.
Let A(2, 3, 5) and C(–3, 4, –2) be opposite vertices of a parallelogram ABCD. If the diagonal $\vec{B D} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , then the area of the parallelogram is equal to [2024]

1 $\frac{1}{2} \sqrt{410}$

2 $\frac{1}{2} \sqrt{474}$

3 $\frac{1}{2} \sqrt{586}$

4 $\frac{1}{2} \sqrt{306}$

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Solution

Q. Let A(2, 3, 5) and C(–3, 4, –2) be opposite vertices of a parallelogram ABCD. If the diagonal BD→=i^+2j^+3k^, then the area of the parallelogram is equal to [2024]

1 12410

2 12474

3 12586

4 12306

Ans. (2) Area of parallelogram ABCD =12|AC→×BD→| =12|(–5i^+j^–7j^)×(i^+2j^+3k^)| =12||i^j^k^–51–7123|| =12|(3+14)i^–(-15+7)j^+(–10–1)k^| =12|(17i^+8j^–11k^)| =12289+64+121=12474 sq. units.

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Q.
Let A(2, 3, 5) and C(–3, 4, –2) be opposite vertices of a parallelogram ABCD. If the diagonal $\vec{B D} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , then the area of the parallelogram is equal to [2024]

1 $\frac{1}{2} \sqrt{410}$

2 $\frac{1}{2} \sqrt{474}$

3 $\frac{1}{2} \sqrt{586}$

4 $\frac{1}{2} \sqrt{306}$