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JEE Mathematics 2024 Question with Solution

Let A(2,3,5)A(2,3,5) and C(−3,4,−2)C(-3,4,-2) be opposite vertices of a parallelogram ABCDABCD. If the diagonal BD→=i^+2j^+3k^\overrightarrow{BD} = \hat{i} + 2\hat{j} + 3\hat{k}, then the area of the parallelogram is equal to:

  • A

    12410\frac{1}{2}\sqrt{410}

  • B

    12474\frac{1}{2}\sqrt{474}

  • C

    12586\frac{1}{2}\sqrt{586}

  • D

    12306\frac{1}{2}\sqrt{306}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Opposite vertices are A(2,3,5)A(2,3,5) and C(−3,4,−2)C(-3,4,-2), and diagonal BD→=i^+2j^+3k^\overrightarrow{BD} = \hat{i} + 2\hat{j} + 3\hat{k}.

Find: The area of parallelogram ABCDABCD.

First find the other diagonal vector:

AC→=(−3−2)i^+(4−3)j^+(−2−5)k^=−5i^+j^−7k^\overrightarrow{AC} = (-3-2)\hat{i} + (4-3)\hat{j} + (-2-5)\hat{k} = -5\hat{i} + \hat{j} - 7\hat{k}

For a parallelogram, area is half the magnitude of the cross product of its diagonals:

Area=12∣AC→×BD→∣\text{Area} = \frac{1}{2}\left|\overrightarrow{AC} \times \overrightarrow{BD}\right|

Now compute the cross product:

AC→×BD→=∣i^j^k^−51−7123∣\overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -7 \\ 1 & 2 & 3 \end{vmatrix} =i^(1⋅3−(−7)⋅2)−j^((−5)⋅3−(−7)⋅1)+k^((−5)⋅2−1⋅1)= \hat{i}(1\cdot 3 - (-7)\cdot 2) - \hat{j}((-5)\cdot 3 - (-7)\cdot 1) + \hat{k}((-5)\cdot 2 - 1\cdot 1) =17i^+8j^−11k^= 17\hat{i} + 8\hat{j} - 11\hat{k}

Its magnitude is:

∣AC→×BD→∣=172+82+(−11)2=289+64+121=474\left|\overrightarrow{AC} \times \overrightarrow{BD}\right| = \sqrt{17^2 + 8^2 + (-11)^2} = \sqrt{289+64+121} = \sqrt{474}

Therefore,

Area=12474\text{Area} = \frac{1}{2}\sqrt{474}

So, the correct option is B.

Diagonal Formula

Given: Diagonals AC→\overrightarrow{AC} and BD→\overrightarrow{BD} of parallelogram ABCDABCD.

Find: Area of the parallelogram.

Use the direct result:

Area of parallelogram=12∣AC→×BD→∣\text{Area of parallelogram} = \frac{1}{2}\left|\overrightarrow{AC} \times \overrightarrow{BD}\right|

Here,

AC→=−5i^+j^−7k^,BD→=i^+2j^+3k^\overrightarrow{AC} = -5\hat{i} + \hat{j} - 7\hat{k}, \qquad \overrightarrow{BD} = \hat{i} + 2\hat{j} + 3\hat{k}

Then,

AC→×BD→=17i^+8j^−11k^\overrightarrow{AC} \times \overrightarrow{BD} = 17\hat{i} + 8\hat{j} - 11\hat{k}

so

∣AC→×BD→∣=474\left|\overrightarrow{AC} \times \overrightarrow{BD}\right| = \sqrt{474}

Hence,

Area=12474\text{Area} = \frac{1}{2}\sqrt{474}

Therefore, the correct option is B.

Common mistakes

  • Using ∣AC→×BD→∣|\overrightarrow{AC} \times \overrightarrow{BD}| directly as the area is incorrect because diagonals of a parallelogram give twice the area through the cross product. Use 12∣AC→×BD→∣\frac{1}{2}|\overrightarrow{AC} \times \overrightarrow{BD}| instead.

  • Computing AC→\overrightarrow{AC} incorrectly is a common error. The vector must be terminal point minus initial point, so AC→=C−A=(−5,1,−7)\overrightarrow{AC} = C-A = (-5,1,-7), not A−CA-C.

  • Making a sign mistake in the determinant expansion changes the magnitude. While expanding the cross product, handle the j^\hat{j} term carefully because it carries a negative sign in cofactor expansion.

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