Three Dimensional Geometry Question 381
Question: The angle between the straight lines $ \vec{r}=(2-3t)\vec{i}+(1+2t)\vec{j}+(2+6t)\vec{k} $ and $ \vec{r}=(1+4s)\vec{i}+(2-s)\vec{j}+(8s-1)\vec{k} $ is
Options:
A) $ {{\cos }^{-1}}( \frac{\sqrt{41}}{34} ) $
B) $ {{\cos }^{-1}}( \frac{21}{34} ) $
C) $ {{\cos }^{-1}}( \frac{43}{63} ) $
D) $ {{\cos }^{-1}}( \frac{34}{63} ) $
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Answer:
Correct Answer: D
Solution:
[d] $ L_1={{\vec{r}}_1}=2\vec{i}+\vec{j}+2\vec{k}+t(-3\vec{i}+2\vec{j}+6\vec{k}) $ $ L_2={{\vec{r}}_2}=(\vec{i}+2\vec{j}-\vec{k})+s(4\vec{i}-\vec{j}+8\vec{k}) $
$ \therefore $ Angle between $ L_1 $ and $ L_2 $ is given by $ \cos \theta =\frac{(-3\vec{i}+2\vec{j}+6\vec{k}).(4\vec{i}-\vec{j}+8\vec{k})}{\sqrt{9+4+36}\sqrt{16+1+64}} $ $ =\frac{-12-2+48}{\sqrt{49}\sqrt{81}}=\frac{34}{7\times 9}=\frac{34}{63} $
$ \Rightarrow \theta ={{\cos }^{-1}}( \frac{34}{63} ) $
BETA
