Get Answers to all your Questions

header-bg qa

Need solution for RD sharma maths class 12 chapter 23 Scalar or dot product exercise Fill in the blanks question 11

Answers (1)

\frac{\pi}{3}

Hint:

               Angle between \vec{a} and \vec{b}

Given:

                \left |\vec{a} \right |=1\left |\vec{b} \right |=3 and \left |a-b \right |=\sqrt{7}. Find Angle between a and b.

Solution:

 \left |\vec{a} \right |=1\left |\vec{b} \right |=3 and \left |a-b \right |=\sqrt{7}                                                                           … (i)

\left |a-b \right |=\sqrt{7}

          Squaring on both side we get

\begin{aligned} &|a-b|^{2}=(\sqrt{7})^{2} \\ &|\vec{a}|^{2}+|\vec{b}|^{2}-2|\vec{a}| \cdot|\vec{b}| \cos \theta=(\sqrt{7})^{2} \end{aligned}

From (i)

\begin{aligned} &(1)^{2}+(3)^{2}-2(1)(3) \cos \theta=7 \\ &1+9-6 \cos \theta=7 \\ &10-7=6 \cos \theta \\ \end{aligned}

\begin{aligned} &\frac{3}{6}=\cos \theta \\ &\cos \theta=\frac{1}{2} \\ &\theta=\frac{\pi}{3} \end{aligned}

Posted by

Info Expert 29

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads