A) \[{{\cos }^{-1}}\left( \frac{3}{\sqrt{17}} \right)\]
B) \[{{\cos }^{-1}}\left( \sqrt{\frac{3}{17}} \right)\]
C) \[{{\sin }^{-1}}\left( \frac{3}{\sqrt{17}} \right)\]
D) \[{{\sin }^{-1}}\left( \sqrt{\frac{3}{17}} \right)\]
Correct Answer: B
Solution :
\[3x+4y+z=1\times 2\] \[5x+8y+2z=-14\] \[6x+8y+2z=2\] \[x=16;4y+z=1-48\] \[4y+z=-47\] \[x,4y+z=-47\] \[(15,-12,1)\]and \[(15,-11,-3)\] \[x=15;\frac{y+12}{1}=\frac{z-1}{-4}\overrightarrow{r}=(15,12,1)+\lambda (0-1,-4)\] \[x-15=0;\frac{y+12}{1}=z\] \[\cos \theta =\frac{\overrightarrow{b}\cdot \overrightarrow{x}}{|b||x|}\] \[=\frac{(0\widehat{i}+1\widehat{j}-4k)(\widehat{j}+\widehat{j}k}{\sqrt{17\times 3}}\overrightarrow{x}=(1,1,1)\] \[=\frac{1-4}{\sqrt{51}}=\frac{3}{\sqrt{17\times 3}}=\sqrt{\frac{3}{17}}\] \[\theta ={{\cos }^{-1}}\left( \sqrt{\frac{3}{17}} \right)\]. Normal to \[3x+4y+z=1\] is \[3\widehat{i}4\widehat{j}+\widehat{k}\] Normal to \[5x+8y+2z=-14\] is \[5\widehat{i}+8\widehat{j}+2\widehat{k}\] The line at which these planes intersect is perpendicular to both normal, hence its direction ratios are directly proportional to the cross product vector of the normal So, the direction ratios of the line can be chosen as \[-\widehat{j}+4\widehat{k}\] So, the angle between the plane \[x+y+z+5=0\] and the line obtained is given by \[{{\sin }^{-1}}\frac{-1+4}{\sqrt{17}\sqrt{3}}={{\sin }^{-1}}\sqrt{\frac{3}{17}}\] So, option D is the correct answer.
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