22. Show that $$ |\boldsymbol{u} imes \boldsymbol{v}|=\sqrt{|\boldsymbol{u}|^{2}|\boldsymbol{v}|^{2}-(\boldsymbol{u} \cdot \boldsymbol{v})^{2}} $$

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22. Show that $$ |\boldsymbol{u} 	imes \boldsymbol{v}|=\sqrt{|\boldsymbol{u}|^{2}|\boldsymbol{v}|^{2}-(\boldsymbol{u} \cdot \boldsymbol{v})^{2}} $$

UpStudy AI · Accepted Answer

To show that $$ |\boldsymbol{u} \times \boldsymbol{v}| = \sqrt{|\boldsymbol{u}|^{2}|\boldsymbol{v}|^{2} - (\boldsymbol{u} \cdot \boldsymbol{v})^{2}} $$, we can use the properties of vectors and the definitions of the cross product and dot product.

### Step 1: Understand the Magnitude of the Cross Product
The magnitude of the cross product of two vectors $$ \boldsymbol{u} $$ and $$ \boldsymbol{v} $$ is given by:
$$
|\boldsymbol{u} \times \boldsymbol{v}| = |\boldsymbol{u}||\boldsymbol{v}|\sin(\theta)
$$
where $$ \theta $$ is the angle between the two vectors.

### Step 2: Use the Dot Product
The dot product of the two vectors is defined as:
$$
\boldsymbol{u} \cdot \boldsymbol{v} = |\boldsymbol{u}||\boldsymbol{v}|\cos(\theta)
$$

### Step 3: Relate Sine and Cosine
Using the identity $$ \sin^2(\theta) + \cos^2(\theta) = 1 $$, we can express $$ \sin(\theta) $$ in terms of the dot product:
$$
\sin^2(\theta) = 1 - \cos^2(\theta) = 1 - \left(\frac{\boldsymbol{u} \cdot \boldsymbol{v}}{|\boldsymbol{u}||\boldsymbol{v}|}\right)^2
$$

### Step 4: Substitute into the Magnitude of the Cross Product
Substituting $$ \sin(\theta) $$ into the expression for the magnitude of the cross product:
$$
|\boldsymbol{u} \times \boldsymbol{v}| = |\boldsymbol{u}||\boldsymbol{v}|\sqrt{1 - \left(\frac{\boldsymbol{u} \cdot \boldsymbol{v}}{|\boldsymbol{u}||\boldsymbol{v}|}\right)^2}
$$

### Step 5: Simplify the Expression
Now, we can simplify this expression:
$$
|\boldsymbol{u} \times \boldsymbol{v}| = |\boldsymbol{u}||\boldsymbol{v}|\sqrt{\frac{|\boldsymbol{u}|^2|\boldsymbol{v}|^2 - (\boldsymbol{u} \cdot \boldsymbol{v})^2}{|\boldsymbol{u}|^2|\boldsymbol{v}|^2}}
$$
$$
= \sqrt{|\boldsymbol{u}|^2|\boldsymbol{v}|^2 - (\boldsymbol{u} \cdot \boldsymbol{v})^2}
$$

### Conclusion
Thus, we have shown that:
$$
|\boldsymbol{u} \times \boldsymbol{v}| = \sqrt{|\boldsymbol{u}|^{2}|\boldsymbol{v}|^{2} - (\boldsymbol{u} \cdot \boldsymbol{v})^{2}}
$$
$$ |\boldsymbol{u} \times \boldsymbol{v}| = \sqrt{|\boldsymbol{u}|^{2}|\boldsymbol{v}|^{2} - (\boldsymbol{u} \cdot \boldsymbol{v})^{2}} $$

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