Get Answers to all your Questions

header-bg qa

Explain solution RD Sharma class 12 chapter Vector or Cross Product exercise 24.1 question 12 maths

Answers (1)


Answer         : These are unit vectors as well as perpendicular

Hint               : To solve this , we do magnitude of one by one

Given             :

                      \begin{aligned} &\vec{a}=\frac{1}{7}(2 \hat{\imath}+3 \hat{\jmath}+6 \hat{k}) \\\\ &\vec{b}=\frac{1}{7}(3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k}) \\\\ &\vec{c}=\frac{1}{7}(6 \hat{\imath}+2 \hat{\jmath}-3 \hat{k}) \end{aligned}  


Solution         :  |\vec{a}|=\frac{1}{7} \sqrt{2^{2}+3^{2}+6^{2}}


\begin{aligned} &=\frac{1}{7} \sqrt{49}=>\frac{7}{7}=1 \\\\ &|\vec{b}|=\frac{1}{7} \sqrt{49}=1 \\\\ &|\vec{c}|=\frac{1}{7} \sqrt{36+4+9} \end{aligned}

\begin{aligned} &=\frac{1}{7} \sqrt{49}=\frac{7}{7}=1 \\\\ &\vec{a} \cdot \vec{b}=0 \\\\ &\vec{b} \cdot \vec{c}=0 \end{aligned}

\begin{aligned} &\vec{a} \cdot \vec{b}=\frac{1}{49}[6-18+12] \\\\ &\vec{b} \cdot \vec{c}=\frac{1}{49}[18-12-6] \end{aligned}


Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support