Question

The magnitude of dot and cross product of two vector 6√6 and 6 respectively find the angle between them

Answer 1

Answer :

 $\boxed{\bf \theta=tan^{-1}(\frac{1}{\sqrt{6}})}}$

Explanation :

Let  $\vec{a} \ and \ \vec{b}$  are two vectors and θ is the angle between the two vectors.

 $\bigstar \ \underline{\underline{\bf Dot \ product:}}$

•  It is the scalar product of two vectors.
•  It says that two vectors can be multiplied to get the scalar quantity.

                  $\boxed{\vec{a} . \vec{b}=|\vec{a}||\vec{b}|cos\theta}$

 $\bigstar \ \underline{\underline{\bf Cross \ product:}}$

•   It is the vector product.
•   It says that two vectors can be multiplied to get the vector quantity.

                  $\boxed{\vec{a} \times \vec{b}=|\vec{a}||\vec{b}|sin\theta}$

Given,

  The magnitude of dot and cross product of two vector 6√6 and 6 respectively.

•        Dot product = 6√6

                 

        $\longrightarrow \ \ \ \ \vec{a}.\vec{b}=|\vec{a}||\vec{b}|cos\theta} \\\\ \longrightarrow \ \ \ \ 6\sqrt{6}=|\vec{a}||\vec{b}|cos\theta} \\\\ \longrightarrow \ \ \ \ |\vec{a}||\vec{b}|cos\theta} =6\sqrt{6} \rightarrow [1]$

•         Cross product = 6

         $\longrightarrow \ \ \ \ \vec{a} \times \vec{b}=|\vec{a}||\vec{b}|sin\theta \\\\ \longrightarrow \ \ \ \ 6=|\vec{a}||\vec{b}|sin\theta \\\\ \longrightarrow \ \ \ \ |\vec{a}||\vec{b}|sin\theta=6 \rightarrow [2]$

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Dividing equation [2] by equation [1]

         

             $\longrightarrow \ \ \bf \frac{|\vec{a}|\vec{b}|sin\theta}{|\vec{a}||\vec{b}|cos\theta} =\frac{6}{6\sqrt{6}} \\\\\\ \longrightarrow \ \ \frac{sin\theta}{cos\theta} =\frac{1}{\sqrt{6}} \\\\\\ \longrightarrow \ \ tan\theta=\frac{1}{\sqrt{6}} \\\\ \\ \longrightarrow \ \ \underline{\boxed{\theta=tan^{-1}(\frac{1}{\sqrt{6}})}}$