Get Answers to all your Questions

header-bg qa

Explain solution for RD Sharma maths class 12 chapter 26 Direction Cosines and Direction Ratios exercise 26.1 question 7 maths textbook solution

Answers (1)

Answer: Angle between two vectors is \cos ^{-1}\left(\frac{-18 \sqrt{2}}{35}\right)

Hint: Direction cosine is proportional to direction ratios

Given: \left(1, m_{1}, n_{1}\right)=(2,3,-6) \text { and }\left(I_{2}, m_{2}, n_{2}\right)=(3,-4,5) . Find Angle between two vectors.

Solution:

Let \theta be the angle between two vectors with direction ratios \left(a_{1}, b_{1}, c_{1}\right) \&\left(a_{2}, b_{2}, c_{2}\right)

As direction ratios are proportional to direction cosines

              \begin{aligned} &\left(a_{1}, b_{1}, c_{1}\right)=\left(I_{1}, m_{1}, n_{1}\right)=(2,3,-6) \\ &\left(a_{2}, b_{2}, c_{2}\right)=\left(I_{2}, m_{2}, n_{2}\right)=(3,-4,5) \end{aligned}

Now,

              \begin{aligned} &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &\cos \theta=\frac{(2)(3)+(3)(-4)+(-6)(5)}{\sqrt{2^{2}+(3)^{2}+(-6)^{2}} \sqrt{3^{2}+(-4)^{2}+5^{2}}} \\ &\cos \theta=\frac{6-12-30}{\sqrt{49} \sqrt{50}} \\ &\cos \theta=\frac{-36 \times \sqrt{2}}{7 \times 5 \times \sqrt{2} \times \sqrt{2}} \\ &\cos \theta=\frac{-18 \sqrt{2}}{35} \\ \end{aligned}

                     \theta=\cos ^{-1}\left(\frac{-18 \sqrt{2}}{35}\right)

     Therefore, the angle between two vectors is \cos ^{-1}\left(\frac{-18 \sqrt{2}}{35}\right)

Posted by

Infoexpert

View full answer

Crack CUET with india's "Best Teachers"

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