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JEE Mathematics 2024 Question with Solution

The area of a quadrilateral ABCD with vertices A(3,1,−1)A(3, 1, -1), B(53,73,13)B\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), C(2,2,1)C(2, 2, 1), D(103,23,−13)D\left(\frac{10}{3}, \frac{2}{3}, -\frac{1}{3}\right) is:

  • A

    423\frac{4\sqrt{2}}{3}

  • B

    523\frac{5\sqrt{2}}{3}

  • C

    222\sqrt{2}

  • D

    223\frac{2\sqrt{2}}{3}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The vertices are A(3,1,−1)A(3,1,-1), B(53,73,13)B\left(\frac{5}{3},\frac{7}{3},\frac{1}{3}\right), C(2,2,1)C(2,2,1) and D(103,23,−13)D\left(\frac{10}{3},\frac{2}{3},-\frac{1}{3}\right).

Find: The area of quadrilateral ABCDABCD.

The solution states that the correct option is A. However, the working shown there uses a completely different set of points, so that algebra does not belong to this question. Therefore, we use the diagonals formula for the given quadrilateral:

Area of quadrilateral=12∣AC⃗×BD⃗∣\text{Area of quadrilateral} = \frac{1}{2}\left|\vec{AC} \times \vec{BD}\right|

Now,

AC⃗=C−A=(2−3, 2−1, 1−(−1))=(−1,1,2)\vec{AC} = C-A = (2-3,\,2-1,\,1-(-1)) = (-1,1,2) BD⃗=D−B=(103−53, 23−73, −13−13)=(53,−53,−23)\vec{BD} = D-B = \left(\frac{10}{3}-\frac{5}{3},\,\frac{2}{3}-\frac{7}{3},\,-\frac{1}{3}-\frac{1}{3}\right) = \left(\frac{5}{3},-\frac{5}{3},-\frac{2}{3}\right)

Their cross product is

AC⃗×BD⃗=∣i^j^k^−11253−53−23∣=(83,83,0)\vec{AC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 2 \\ \frac{5}{3} & -\frac{5}{3} & -\frac{2}{3} \end{vmatrix} = \left(\frac{8}{3},\frac{8}{3},0\right)

So,

∣AC⃗×BD⃗∣=(83)2+(83)2=823\left|\vec{AC} \times \vec{BD}\right| = \sqrt{\left(\frac{8}{3}\right)^2+\left(\frac{8}{3}\right)^2} = \frac{8\sqrt{2}}{3}

Hence,

Area=12⋅823=423\text{Area} = \frac{1}{2} \cdot \frac{8\sqrt{2}}{3} = \frac{4\sqrt{2}}{3}

Therefore, the area is 423\frac{4\sqrt{2}}{3}, so the correct option is A.

Common mistakes

  • Using the side vectors of only one triangle instead of the diagonals of the quadrilateral is incorrect here. For this figure, use 12∣AC⃗×BD⃗∣\frac{1}{2}\left|\vec{AC}\times\vec{BD}\right| for the area of the quadrilateral.

  • Subtracting coordinates in the wrong order while forming AC⃗\vec{AC} or BD⃗\vec{BD} changes the sign pattern of components. Write each vector carefully as terminal point minus initial point.

  • Computing the cross product determinant incorrectly is a common conceptual error. Keep the expansion systematic and remember that the middle component carries a minus sign in the determinant expansion.

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