Let a ? = 3 i ^ + 2 j ^ + k ^ , b ? = 2 i ^ – j ^ + 3 k ^ and c ? be a vector such that ( a ? + b ? ) × c ? = 2 ( a ? × b ? ) + 24 j ^ – 6 k ^ and ( a ? – b ? + i ^ ) · c ? = – 3 . Then | c ? | 2 is equal to [2024]

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Solution

Q.
Let $\vec{a} = 3 \hat{i} + 2 \hat{j} + \hat{k}$ , $\vec{b} = 2 \hat{i} - \hat{j} + 3 \hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a} + \vec{b}) \times \vec{c} = 2 (\vec{a} \times \vec{b}) + 24 \hat{j} - 6 \hat{k}$ and $(\vec{a} - \vec{b} + \hat{i}) \cdot \vec{c} = - 3$ . Then $| \vec{c} |^{2}$ is equal to __________. [2024]

Ans.
(38)

We have, $\vec{a} = 3 \hat{i} + 2 \hat{j} + \hat{k}$ , $\vec{b} = 2 \hat{i} - \hat{j} + 3 \hat{k}$

Let $\vec{c} = x \hat{i} + y \hat{j} + z \hat{k}$

$(\vec{a} + \vec{b}) \times \vec{c} = (3 \hat{i} + 2 \hat{j} + \hat{k} + 2 \hat{i} - \hat{j} + 3 \hat{k}) \times (x \hat{i} + y \hat{j} + z \hat{k})$

$= (5 \hat{i} + \hat{j} + 4 \hat{k}) \times (x \hat{i} + y \hat{j} + z \hat{k})$

$= (z - 4 y) \hat{i} - (5 z - 4 x) \hat{j} + (5 y - x) \hat{k}$

$\vec{a} \times \vec{b} = 7 \hat{i} - 7 \hat{j} - 7 \hat{k}$

$∴ (z - 4 y) \hat{i} - (5 z - 4 x) \hat{j} + (5 y - x) \hat{k} = 2 (7 \hat{i} - 7 \hat{j} - 7 \hat{k}) + 24 \hat{j} - 6 \hat{k}$

On comparing, we get

z – 4y = 14 ... (i), 5z – 4x = –10 ... (ii), 5y – x = –20 ... (iii)

From (i), z = 14 +4y

Now, $(\vec{a} - \vec{b} + \hat{i}) \cdot \vec{c} = - 3$

$\Rightarrow (3 \hat{i} + 2 \hat{j} + \hat{k} - 2 \hat{i} + \hat{j} - 3 \hat{k} + \hat{i}) \cdot (x \hat{i} + y \hat{j} + z \hat{k}) = - 3$

$\Rightarrow (2 \hat{i} + 3 \hat{j} - 2 \hat{k}) \cdot (x \hat{i} + y \hat{j} + z \hat{k}) = - 3$

2x + 3y – 2z = –3

$\Rightarrow 2 x + 3 (\frac{x - 20}{5}) - 2 (\frac{4 x - 10}{5}) = - 3$

$\Rightarrow 10 x + 3 x - 60 - 8 x + 20 = - 15$

$\Rightarrow 5 x - 25 = 0 \Rightarrow x = 5 \Rightarrow y = \frac{5 - 20}{5} = - 3$

z = 14 + 4y = 2

$∴ \vec{c} = 5 \hat{i} - 3 \hat{j} + 2 \hat{k}; | \vec{c} |^{2} = 25 + 9 + 4 = 38$ .

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