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JEE Mathematics 2023 Question with Solution

Let the position vectors of the points A, B, C and D be 5i+5j+2λk5\mathbf{i} + 5\mathbf{j} + 2\lambda \mathbf{k}, i+2j+3k\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}, −2i+xj+4k-2\mathbf{i} + x\mathbf{j} + 4\mathbf{k} and −i+5j+6k-\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}. Let the set S={λ∈R:S = \{\lambda \in \mathbb{R}: the points A, B, C and D are coplanar}\}. Then ∑λ∈S(λ+2)2\sum_{\lambda \in S}(\lambda + 2)^2 is equal to:

  • A

    99

  • B

    1313

  • C

    2525

  • D

    4141

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The position vectors of points A, B, C and D are given. We need the values of λ\lambda for which these four points are coplanar.

Find: The value of ∑λ∈S(λ+2)2\sum_{\lambda \in S}(\lambda + 2)^2.

For four points to be coplanar, the scalar triple product of AB→\overrightarrow{AB}, AC→\overrightarrow{AC} and AD→\overrightarrow{AD} must be zero.

The vectors are:

AB→=i+2j+3k−(5i+5j+2λk)=−4i−3j+(3−2λ)k\overrightarrow{AB} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} - (5\mathbf{i} + 5\mathbf{j} + 2\lambda \mathbf{k}) = -4\mathbf{i} - 3\mathbf{j} + (3 - 2\lambda)\mathbf{k} AC→=−2i+λj+4k−(5i+5j+2λk)=−7i+(λ−5)j+(4−2λ)k\overrightarrow{AC} = -2\mathbf{i} + \lambda \mathbf{j} + 4\mathbf{k} - (5\mathbf{i} + 5\mathbf{j} + 2\lambda \mathbf{k}) = -7\mathbf{i} + (\lambda - 5)\mathbf{j} + (4 - 2\lambda)\mathbf{k} AD→=−i+5j+6k−(5i+5j+2λk)=−6i+0j+(6−2λ)k\overrightarrow{AD} = -\mathbf{i} + 5\mathbf{j} + 6\mathbf{k} - (5\mathbf{i} + 5\mathbf{j} + 2\lambda \mathbf{k}) = -6\mathbf{i} + 0\mathbf{j} + (6 - 2\lambda)\mathbf{k}

So the coplanarity condition is:

∣−4−33−2λ−7λ−54−2λ−606−2λ∣=0\begin{vmatrix} -4 & -3 & 3 - 2\lambda \\ -7 & \lambda - 5 & 4 - 2\lambda \\ -6 & 0 & 6 - 2\lambda \end{vmatrix} = 0

Expanding the determinant gives:

−4λ2+20λ−24=0-4\lambda^2 + 20\lambda - 24 = 0

Dividing by −4-4:

λ2−5λ+6=0\lambda^2 - 5\lambda + 6 = 0

Factoring:

(λ−2)(λ−3)=0(\lambda - 2)(\lambda - 3) = 0

Hence, the possible values are λ=2\lambda = 2 and λ=3\lambda = 3. Therefore,

S={2,3}S = \{2, 3\}

Now compute:

∑λ∈S(λ+2)2=(2+2)2+(3+2)2=16+25=41\sum_{\lambda \in S}(\lambda + 2)^2 = (2 + 2)^2 + (3 + 2)^2 = 16 + 25 = 41

Therefore, the value of ∑λ∈S(λ+2)2\sum_{\lambda \in S}(\lambda + 2)^2 is 4141. The correct option is D.

Common mistakes

  • Using the coplanarity condition incorrectly. Four points are coplanar only when the scalar triple product of three displacement vectors from the same point is zero. Do not test pairwise collinearity; instead form AB→\overrightarrow{AB}, AC→\overrightarrow{AC} and AD→\overrightarrow{AD} and set their determinant equal to zero.

  • Subtracting coordinates in the wrong order while forming vectors. If AB→=B⃗−A⃗\overrightarrow{AB} = \vec{B} - \vec{A} is written incorrectly as A⃗−B⃗\vec{A} - \vec{B} for only one vector, the determinant changes inconsistently. Form all three vectors from the same initial point in a consistent order.

  • Mishandling the parameter λ\lambda during determinant expansion. Sign errors in terms like 3−2λ3 - 2\lambda, 4−2λ4 - 2\lambda, and 6−2λ6 - 2\lambda can lead to the wrong quadratic equation. Expand carefully and simplify step by step before factoring.

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