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  • 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}

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