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Find the angle betweenthe line `(x-2)/-1=(y+3)/2=(z+4)/3` and the plane `2x-3y+z=5`. |
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Answer» The given line is `(x-x_(1))/(a_(1))=(y-y_(1))/(b_(1))=(z-z_(1))/(c_(1))` where `a_(1)=-1, b_(1)=2 c_(1)=3` `The given plane is `a_(2)x+b_(2)y+c_(2)y+d=0` where `a_(2)=2, b_(2)=-3 c_(2)=1,d=-5` Let the required angle be `phi` Then `sin phi =(|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))}` `=|(-1)xx2+2xx(-3)+3xx1)|/({sqrt(-1)^(2)+2^(2)+3^(2))}.{sqrt(2^(2)+(-3)^(2)+1^(2))}` =|(-2-6+3)|/{sqrt(14xxsqrt(14))=|(-5)|/(14)=(5)/(14)` `rarr phi sin^(-1) (5/14)` Hence the anlge between the given line and the given plane is `sin^(-1)(5/14)` |
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