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Let $\overrightarrow{\mathrm{a}}=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{\mathrm{c}}=2 \hat{i}-3 \hat{j}+2 \hat{k}$ . Then the number of vectors $\overrightarrow{\mathrm{b}}$ such that $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}}$ and $|\overrightarrow{\mathrm{b}}| \in\{1,2, \ldots ., 10\}$ is :
Option: 1
$0$

Option: 2
$1$

Option: 3
$2$

Option: 4
$3$

Answers (1)

best_answer

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

$\vec{b} \times \vec{c}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{x} \\ \hat{x} & y & z \\ 2 & -3 & 2 \end{array}\right|=\vec{a}$

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

$\mathrm{2y+3z= 1}$ ....(1) $\mathrm{2y-3z= 1}$ ......(2)

$\mathrm{3x-2y= -1}$ $\Rightarrow 3x+2y=1$ .......(3)

This system of equation is inconsistent So no such $\vec{b}$ exist.

Hence the correct answer is option 1.

Posted by

himanshu.meshram

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