Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide Solution for RD Sharma Class 12 Chapter 27 Straight Line in Space Exercise Very Short Answer Question 20

Provide Solution for RD Sharma Class 12 Chapter 27 Straight Line in Space Exercise Very Short Answer Question 20

Answers (1)

Answer:

Required answer is  \cos ^{-1} \frac{19}{21}

Hint:

 Use the equation of a line in space.

Given:

\begin{aligned} &\vec{r}=(2 \hat{i}-5 \hat{j}+\hat{k})+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k}) \\ & \end{aligned}

\vec{r}=(7 \hat{i}-6 \hat{k})+\mu(\hat{i}+2 \hat{j}+2 k) 

Solution:

Let  \theta  be the angle between the given lines

The given lines are parallel to the vectors  \vec{b}_{1}=3 \hat{i}+2 \hat{j}+6 \hat{k} \text { and } \vec{b}_{2}=\hat{i}+2 \hat{j}+2 \hat{k}    respectively

So, the angle \theta  between the given lines are given by

\begin{aligned} &\cos \theta=\frac{\vec{b}_{1} \cdot \vec{b}_{2}}{\left|\vec{b}_{1}\right|\left|\vec{b}_{2}\right|} \\ & \end{aligned}

          =\frac{(3 \hat{i}+2 \hat{j}+6 \hat{k}) \cdot(\hat{i}+2 \hat{j}+2 \hat{k})}{\sqrt{3^{2}+2^{2}+6^{2}} \sqrt{1^{2}+2^{2}+2^{2}}}

          \begin{aligned} &=\frac{19}{\sqrt{49} \sqrt{9}} \\ & \end{aligned}

         =\frac{19}{21} \\

\theta=\cos ^{-1} \frac{19}{21}

Posted by

infoexpert27

View full answer

Crack CUET with india's "Best Teachers"

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

Similar Questions

Trending Articles/News

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question