#### please solve RD sharma class 12 chapter 24 vector or cross product exercise Fill in the blanks question 1 maths textbook solution

$\therefore |\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}=|A|^{2}|B|^{2}$

Hint: Just use the formula of cross product & dot product

Given:

\begin{aligned} &|\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}= \\ \end{aligned}______________

Now,

\begin{aligned} &\vec{a} \times \vec{b}=|A||B| \sin \theta \\ &\therefore|\vec{a} \times \vec{b}|^{2}=|A|^{2}|B|^{2} \sin ^{2} \theta \ldots \ldots . .(1) \\ \end{aligned}

Similarly,

\begin{aligned} &\vec{a} \cdot \vec{b}=|A||B| \cos \theta \\ &|\vec{a} \cdot \vec{b}|^{2}=|A|^{2}|B|^{2} \cos ^{2} \theta \ldots \ldots . .(2) \\ \end{aligned}

\begin{aligned} &|\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}=|A|^{2}|B|^{2} \sin ^{2} \theta+|A|^{2}|B|^{2} \\ \end{aligned}

\begin{aligned} &=|A|^{2}|B|^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=|A|^{2}|B|^{2} \\ \end{aligned}

\begin{aligned} &\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right) \\ \end{aligned}

\begin{aligned} &\therefore|\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}=|A|^{2}|B|^{2} \end{aligned}