MCQMediumJEE 2023Cross Product

JEE Mathematics 2023 Question with Solution

Let for a triangle ABC, AB=2i^+j^+3k^,CB=i^+j^+k^,CA=4i^+3j^+4k^\overrightarrow{AB} = -2\hat{i} + \hat{j} + 3\hat{k}, \quad \overrightarrow{CB} = \hat{i} + \hat{j} + \hat{k}, \quad \overrightarrow{CA} = 4\hat{i} + 3\hat{j} + 4\hat{k} If λ\lambda = 00 and the area of triangle ABC is 565\sqrt{6}, then CBCB is equal to:

  • A

    108108

  • B

    6060

  • C

    5454

  • D

    120120

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: AB=2i^+j^+3k^\overrightarrow{AB} = -2\hat{i} + \hat{j} + 3\hat{k}, CA=4i^+3j^+4k^\overrightarrow{CA} = 4\hat{i} + 3\hat{j} + 4\hat{k}, and area of triangle ABC is 565\sqrt{6}.

Find: The required value asked in the options.

From the solution working, let the third component be δ\delta and write

CB=AB+CA=(2+4)i^+(1+3)j^+(3+δ)k^=2i^+4j^+(3+δ)k^\overrightarrow{CB} = \overrightarrow{AB} + \overrightarrow{CA} = (-2+4)\hat{i} + (1+3)\hat{j} + (3+\delta)\hat{k} = 2\hat{i} + 4\hat{j} + (3+\delta)\hat{k}

Then

CB×CA=(δ9)i^+(12+2δ)j^10k^\overrightarrow{CB} \times \overrightarrow{CA} = (\delta-9)\hat{i} + (12+2\delta)\hat{j} - 10\hat{k}

So,

CB×CA=(δ9)2+(12+2δ)2+100\left|\overrightarrow{CB} \times \overrightarrow{CA}\right| = \sqrt{(\delta-9)^2 + (12+2\delta)^2 + 100}

Using area of triangle,

12CB×CA=56\frac{1}{2}\left|\overrightarrow{CB} \times \overrightarrow{CA}\right| = 5\sqrt{6}

Hence,

CB×CA=106\left|\overrightarrow{CB} \times \overrightarrow{CA}\right| = 10\sqrt{6}

Therefore,

(δ9)2+(12+2δ)2+100=106\sqrt{(\delta-9)^2 + (12+2\delta)^2 + 100} = 10\sqrt{6}

Squaring both sides,

(δ9)2+(12+2δ)2+100=600(\delta-9)^2 + (12+2\delta)^2 + 100 = 600 δ218δ+81+4δ2+48δ+144+100=600\delta^2 - 18\delta + 81 + 4\delta^2 + 48\delta + 144 + 100 = 600 5δ2+30δ+325=6005\delta^2 + 30\delta + 325 = 600 5δ2+30δ275=05\delta^2 + 30\delta - 275 = 0 δ2+6δ55=0\delta^2 + 6\delta - 55 = 0 (δ+11)(δ5)=0(\delta+11)(\delta-5) = 0

So δ=5\delta = 5.

Then

CB=2i^+4j^+8k^\overrightarrow{CB} = 2\hat{i} + 4\hat{j} + 8\hat{k}

Now,

CBCA=(2)(4)+(4)(3)+(8)(5)=8+12+40=60\overrightarrow{CB} \cdot \overrightarrow{CA} = (2)(4) + (4)(3) + (8)(5) = 8 + 12 + 40 = 60

Thus the working gives the value 6060. However, the solution explicitly marks Option A as correct, while the answer key marks (2). Since the solution is the primary source, the correct option is taken as A.

Answer Discrepancy Note

The extracted question text appears corrupted: it states CB=i^+j^+k^\overrightarrow{CB} = \hat{i} + \hat{j} + \hat{k} in the question, but the solution recomputes CB\overrightarrow{CB} using an unknown third component δ\delta and finally evaluates a dot product equal to 6060.

So there is a mismatch between the displayed question, the option values, and the conclusion in the solution. The numerical result obtained from the worked solution is 6060, which corresponds to Option B, but the solution says The Correct Option is A. Following the instruction that the solution is the primary source, the recorded answer is A, while preserving the discrepancy here.

Common mistakes

  • Using the area formula with a dot product instead of a cross product is incorrect. The area of a triangle formed by two vectors is 12a×b\frac{1}{2}\left|\vec{a} \times \vec{b}\right|, not 12(ab)\frac{1}{2}(\vec{a}\cdot\vec{b}). First compute the cross product magnitude, then use the area condition.

  • Forgetting the factor 12\frac{1}{2} for the triangle area leads to a wrong equation. The magnitude CB×CA\left|\overrightarrow{CB} \times \overrightarrow{CA}\right| gives the area of the parallelogram, so the triangle area is half of that.

  • Making determinant sign errors while computing CB×CA\overrightarrow{CB} \times \overrightarrow{CA} changes the quadratic in δ\delta. Expand the determinant carefully and preserve the alternating signs of components.

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