Question
Define $\vec{\text{a}}\times\vec{\text{b}}$ and prove that $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\big(\vec{\text{a}}.\vec{\text{b}}\big)\tan\theta,$ where $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
✓
Answer
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two non-parallel vectors, then the vectors product denoted by $\vec{\text{a}}\times\vec{\text{b}}$ is defined as $\vec{\text{a}}\times\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\hat{\text{ n}}.$
Here, $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ and $\hat{\text{n}}$ is the unit vector perpendicular to the plane of $\vec{\text{a}}$ and $\vec{\text{b}}$ such that $\vec{\text{a}},\vec{\text{b}}$ and $\hat{\text{n}}$ form a right
$\text{LHS}=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\times\frac{\cos\theta}{\cos\theta}$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\frac{\sin\theta}{\cos\theta}$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\tan\theta$
$=\big(\vec{\text{a}}.\vec{\text{b}}\big)\tan\theta$
$=\text{RHS}$
Hence proved
Here, $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ and $\hat{\text{n}}$ is the unit vector perpendicular to the plane of $\vec{\text{a}}$ and $\vec{\text{b}}$ such that $\vec{\text{a}},\vec{\text{b}}$ and $\hat{\text{n}}$ form a right
$\text{LHS}=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\times\frac{\cos\theta}{\cos\theta}$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\frac{\sin\theta}{\cos\theta}$
$=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\tan\theta$
$=\big(\vec{\text{a}}.\vec{\text{b}}\big)\tan\theta$
$=\text{RHS}$
Hence proved
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