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JEE Mathematics 2024 Question with Solution

Let A(2,3,5)\left(2,3,5\right) and C(−3,4,−2)\left(-3,4,-2\right) be opposite vertices of a parallelogram ABCD. If the diagonal BD =i^+2j^+3k^= \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)\left(2,3,5\right) and C(−3,4,−2)\left(-3,4,-2\right), and diagonal BD→=i^+2j^+3k^\overrightarrow{BD} = \hat{i} + 2\hat{j} + 3\hat{k}.

Find: Area of parallelogram ABCD.

First find the other diagonal:

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.

Direct Diagonal Formula

Given: The diagonals are AC→\overrightarrow{AC} and BD→\overrightarrow{BD}.

Find: Area of the parallelogram.

Use the direct relation:

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

Since AC→=−5i^+j^−7k^\overrightarrow{AC} = -5\hat{i} + \hat{j} - 7\hat{k} and BD→=i^+2j^+3k^\overrightarrow{BD} = \hat{i} + 2\hat{j} + 3\hat{k}, evaluating the cross product gives magnitude 474\sqrt{474}.

Hence the area is 12474\frac{1}{2}\sqrt{474}, so the correct option is B.

Common mistakes

  • Using ∣AC→×BD→∣\left|\overrightarrow{AC} \times \overrightarrow{BD}\right| directly as the area is wrong because diagonals of a parallelogram give twice the needed area in the cross-product relation. Use Area=12∣AC→×BD→∣\text{Area} = \frac{1}{2}\left|\overrightarrow{AC} \times \overrightarrow{BD}\right| instead.

  • Computing AC→\overrightarrow{AC} incorrectly by reversing the order of points changes the sign of the vector. Although magnitude is unaffected in the final cross product, inconsistent signs can lead to algebra errors. Form it carefully as C−AC-A.

  • Making mistakes in the determinant expansion for the cross product, especially the middle term sign, gives a wrong magnitude. Remember the j^\hat{j} term carries a negative sign in cofactor expansion.

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