JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let for a triangle ABC, $\vec{AB} = -2\hat{i} + \hat{j} + 3\hat{k}$ $\vec{CB} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$ $\vec{CA} = 4\hat{i} + 3\hat{j} + \delta\hat{k}$ If $\delta > 0$ and the area of the triangle ABC is $5\sqrt{6}$ , then $\vec{CB} \cdot \vec{CA}$ is equal to:

Choose the correct answer:

Explanation

Solution

Step 1: Triangle Law of Vectors ka use karke $\delta$ nikalna

Ek triangle $ABC$ mein:

$\vec{AB} + \vec{BC} + \vec{CA} = 0$

Humein diya gaya hai:

$\vec{AB} = -2\hat{i} + \hat{j} + 3\hat{k}$
$\vec{CA} = 4\hat{i} + 3\hat{j} + \delta\hat{k}$
$\vec{CB} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k} \implies \vec{BC} = -(\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k})$

Lekin hum seedha $\vec{AB} = \vec{CB} - \vec{CA}$ bhi likh sakte hain:

$\vec{AB} = \vec{CB} - \vec{CA} \implies \vec{CB} = \vec{AB} + \vec{CA}$

$\vec{CB} = (-2+4)\hat{i} + (1+3)\hat{j} + (3+\delta)\hat{k}$

$\vec{CB} = 2\hat{i} + 4\hat{j} + (3+\delta)\hat{k}$

Step 2: Area of Triangle ka use karke $\delta$ ki value nikalna

Area of $\Delta ABC = \frac{1}{2} |\vec{AB} \times \vec{CA}| = 5\sqrt{6}$

Pehle $\vec{AB} \times \vec{CA}$ nikaalte hain:

$\vec{AB} \times \vec{CA} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 3 \\ 4 & 3 & \delta \end{vmatrix} = \hat{i}(\delta - 9) - \hat{j}(-2\delta - 12) + \hat{k}(-6 - 4)$

$= (\delta - 9)\hat{i} + (2\delta + 12)\hat{j} - 10\hat{k}$

Ab iska magnitude:

$|\vec{AB} \times \vec{CA}| = \sqrt{(\delta - 9)^2 + (2\delta + 12)^2 + (-10)^2} = 2 \times 5\sqrt{6} = 10\sqrt{6}$

Square karne par:

$(\delta^2 - 18\delta + 81) + (4\delta^2 + 48\delta + 144) + 100 = 600$

$5\delta^2 + 30\delta + 325 = 600$

$5\delta^2 + 30\delta - 275 = 0$

$\delta^2 + 6\delta - 55 = 0$

$(\delta + 11)(\delta - 5) = 0$

Kyunki $\delta > 0$ , isliye $\delta = 5$ .

Step 3: $\vec{CB} \cdot \vec{CA}$ ki value nikalna

$\delta = 5$ rakhne par:

$\vec{CA} = 4\hat{i} + 3\hat{j} + 5\hat{k}$
$\vec{CB} = 2\hat{i} + 4\hat{j} + (3+5)\hat{k} = 2\hat{i} + 4\hat{j} + 8\hat{k}$

Ab dot product:

$\vec{CB} \cdot \vec{CA} = (2)(4) + (4)(3) + (8)(5)$

$= 8 + 12 + 40 = 60$

Correct Option: (2) 60

Choose the correct answer:

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