#### Provide solution for RD sharma maths class 12 chapter 24 Vector or cross product exercise Fill in the blanks question 4

\begin{aligned} &=\vec{b} \times \vec{a}\\ \end{aligned}

Hint: Just multiply & use

\begin{aligned} &\stackrel{\Lambda}{i \times} \stackrel{\Lambda}{i}=0\\ \end{aligned}

Given:

\begin{aligned} &(2 \vec{a}+3 \vec{b}) \times(5 \vec{a}+7 \vec{b})\\ &=2 \vec{a}(5 \vec{a}+7 \vec{b})+3 \vec{b}(5 \vec{a}+7 \vec{b})\\ &=10(\vec{a} \times \vec{a})+14(\vec{a} \times \vec{b})+15(\vec{b} \times \vec{a})+21(\vec{b} \times \vec{b})\\ \end{aligned}

We know that cross product of same vector is zero

i.e

\begin{aligned} &\stackrel{\Lambda}{i} \times \hat{i}=0\\ &\hat{j} \times\stackrel{\Lambda}{i}=0\\ \end{aligned}

\begin{aligned} &=10(0)+14(\vec{a} \times \vec{b})+15(\vec{b} \times \vec{a})+21(0)\\ &=14(\vec{a} \times \vec{b})-15(\vec{a} \times \vec{b})\\ \end{aligned}

\begin{aligned} &(\because(\vec{a} \times \vec{b})=-(\vec{b} \times \vec{a}))\\ &=-(\vec{a} \times \vec{b})\\ &=\vec{b} \times \vec{a} \end{aligned}