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Step-by-Step Solution
Step 1: Understand the Given Information
We are given three vectors:
$\overrightarrow{a} = \hat{i} + \hat{j} + \hat{k}$
$\overrightarrow{c} = \hat{j} - \hat{k}$
$\overrightarrow{b}$ is a vector such that:
$\overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{c}$
$\overrightarrow{a} \cdot \overrightarrow{b} = 3$
We need to find the magnitude of $\overrightarrow{b}$, denoted by $\left|\overrightarrow{b}\right|$.
Step 2: Compute the Magnitude of $\overrightarrow{a}$ and $\overrightarrow{c}$
1. The vector $\overrightarrow{a} = \hat{i} + \hat{j} + \hat{k}$ has components $(1,1,1)$. Hence:
$\left|\overrightarrow{a}\right|
= \sqrt{1^2 + 1^2 + 1^2}
= \sqrt{3}.$
2. The vector $\overrightarrow{c} = \hat{j} - \hat{k}$ has components $(0, 1, -1)$. Hence:
$\left|\overrightarrow{c}\right|
= \sqrt{0^2 + 1^2 + (-1)^2}
= \sqrt{1 + 1}
= \sqrt{2}.$
Step 3: Express the Cross Product and Dot Product Relationships
From the property of the cross product, we know:
$\left|\overrightarrow{a} \times \overrightarrow{b}\right|
= \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\sin\theta,$
where $\theta$ is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$. Since $\overrightarrow{a} \times \overrightarrow{b} = \overrightarrow{c}$, it follows that
$\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\sin\theta = \left|\overrightarrow{c}\right|.$
Substituting the magnitudes found:
$\sqrt{3}\,\left|\overrightarrow{b}\right| \,\sin\theta = \sqrt{2}.\quad (1)$
From the property of the dot product, we know:
$\overrightarrow{a}\cdot \overrightarrow{b}
= \left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos\theta.$
Given $\overrightarrow{a}\cdot \overrightarrow{b} = 3$, we have
$\sqrt{3}\,\left|\overrightarrow{b}\right|\cos\theta = 3.\quad (2)$
Step 4: Use the Ratio to Find $\tan\theta$
Divide equation (1) by equation (2) to eliminate $\left|\overrightarrow{b}\right|$ and find $\tan\theta$:
$\displaystyle
\frac{\sqrt{3}\,\left|\overrightarrow{b}\right|\sin\theta}{\sqrt{3}\,\left|\overrightarrow{b}\right|\cos\theta}
= \frac{\sqrt{2}}{3}
\;\;\Longrightarrow\;\;
\tan\theta = \frac{\sqrt{2}}{3}.
$
Step 5: Determine $\sin\theta$
Using $\tan\theta = \frac{\sqrt{2}}{3}$, we can find $\sin\theta$ by recalling:
$\sin\theta = \frac{\tan\theta}{\sqrt{1 + \tan^2\theta}}.
$
Therefore,
$\displaystyle
\sin\theta
= \frac{\frac{\sqrt{2}}{3}}{\sqrt{1 + \left(\frac{\sqrt{2}}{3}\right)^2}}
= \frac{\sqrt{2}/3}{\sqrt{1 + \frac{2}{9}}}
= \frac{\sqrt{2}/3}{\sqrt{\frac{9}{9} + \frac{2}{9}}}
= \frac{\sqrt{2}/3}{\sqrt{\frac{11}{9}}}
= \frac{\sqrt{2}}{3} \cdot \frac{3}{\sqrt{11}}
= \frac{\sqrt{2}}{\sqrt{11}}.
$
Step 6: Substitute $\sin\theta$ Back to Find $\left|\overrightarrow{b}\right|$
Returning to equation (1):
$\sqrt{3}\,\left|\overrightarrow{b}\right|\sin\theta = \sqrt{2}.
$
Substitute $\sin\theta = \frac{\sqrt{2}}{\sqrt{11}}$:
$\sqrt{3}\,\left|\overrightarrow{b}\right|\left(\frac{\sqrt{2}}{\sqrt{11}}\right) = \sqrt{2}.
$
Isolate $\left|\overrightarrow{b}\right|$:
$\left|\overrightarrow{b}\right|
= \frac{\sqrt{2}}{\sqrt{3}} \cdot \frac{\sqrt{11}}{\sqrt{2}}
= \frac{\sqrt{11}}{\sqrt{3}}
= \sqrt{\frac{11}{3}}.
$
Step 7: Conclude the Result
Therefore, the magnitude of $\,\overrightarrow{b}\,$ is
$\boxed{\sqrt{\frac{11}{3}}}.
