JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

If $\vec{a} = \hat{i} + 2\hat{k}, \vec{b} = \hat{i} + \hat{j} + \hat{k}, \vec{c} = 7\hat{i} - 3\hat{j} + 4\hat{k}, \vec{r} \times \vec{b} + \vec{b} \times \vec{c} = \vec{0}$ and $\vec{r} \cdot \vec{a} = 0$ . Then $\vec{r} \cdot \vec{c}$ is equal to:

Choose the correct answer:

Explanation

Solution

Pehli equation se:

\vec{r} \times \vec{b} - \vec{c} \times \vec{b} = \vec{0} \implies (\vec{r} - \vec{c}) \times \vec{b} = \vec{0}

Iska matlab $(\vec{r} - \vec{c})$ aur $\vec{b}$ collinear hain:

\vec{r} - \vec{c} = m\vec{b} \implies \vec{r} = \vec{c} + m\vec{b}

Ab $\vec{r} \cdot \vec{a} = 0$ ka use karte hain:

(\vec{c} + m\vec{b}) \cdot \vec{a} = 0 \implies \vec{c} \cdot \vec{a} + m(\vec{b} \cdot \vec{a}) = 0

$\vec{c} \cdot \vec{a} = (7)(1) + (-3)(0) + (4)(2) = 7 + 8 = 15$
$\vec{b} \cdot \vec{a} = (1)(1) + (1)(0) + (1)(2) = 1 + 2 = 3$

Toh, $15 + m(3) = 0 \implies m = -5$ .

Ab $\vec{r} = \vec{c} - 5\vec{b}$ .

Humein $\vec{r} \cdot \vec{c}$ nikalna hai:

\vec{r} \cdot \vec{c} = (\vec{c} - 5\vec{b}) \cdot \vec{c} = |\vec{c}|^2 - 5(\vec{b} \cdot \vec{c})

$|\vec{c}|^2 = 7^2 + (-3)^2 + 4^2 = 49 + 9 + 16 = 74$
$\vec{b} \cdot \vec{c} = (1)(7) + (1)(-3) + (1)(4) = 7 - 3 + 4 = 8$

\vec{r} \cdot \vec{c} = 74 - 5(8) = 74 - 40 = 34

Sahi Option: (4)

Choose the correct answer:

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