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The perpendicular distance from the origin to the plane containing the two lines, `(x+2)/3=(y-2)/5=(z+5)/7` and `(x-1)/1=(y-4)/4=(z+4)/7` is: (a) `11sqrt6` (b)`11/(sqrt6)` (c) 11 (d) `6sqrt(11)`A. `11sqrt(6)`B. `(11)/(sqrt(6))`C. 11D. `6sqrt(11)` |
Answer» Correct Answer - B Let the equation of plane, containing the two lines `(x+2)/(3)=(y-2)/(5)=(z+5)/(7)" and "(x-1)/(1)=(y-4)/(4)=(z+4)/(7)` is `a(x+2)+b(y-2)+c(z+5)=0" "...(i)` `because` Plane (i) contain lines, so `3a+5b+7c=0" "...(ii)` and`" "a+4b+7c=0" "...(iii)` From Eqs. (ii) and (iii) , we get `(a)/(35-28)=(b)/(7-21)=(c)/(12-5)` `implies" "(a)/(7)=(b)/(-14)=(c)/(7)implies(a)/(1)=(b)/(-2)=(c)/(1)` So, equation of plane will be `1(x+2)-2(y-2)+1(z+5)=0` `implies" "x-2y+z+11=0" "...(iv)` Now, perpendicular distance from origin to the plane is `=(11)/(sqrt(1+4+1))=(11)/(sqrt(6))` [`because` perpendicular distance from origin to the plane `ax+by+cz+d=0," is "(|d|)/(sqrt(a^(2)+b^(2)+c^(2)))`] |
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