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Let `P_(1)=x+y+z+1=0, P_(2)=x-y+2z+1=0,P_(3)=3x+y+4z+7=0` be three planes. Find the distance of line of intersection of planes `P_(1)=0` and `P_(2) =0` from the plane `P_(3) =0.`A. `(2)/sqrt(26)`B. `(4)/sqrt(26)`C. `sqrt((1)/(26))`D. `(7)/sqrt(26)` |
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Answer» Correct Answer - B Let direction ratios of line are a,b,c Now `a+b+c=0` `a-b+2c=0` `(a)/(3)=(b)/(-1)=(c )/(-2)` So direction ratios of line are `(3,-1,-2)` `x+y+z+1+0….(1)` `x-y+2z+1=0….(2)` Put z=0, we get `x=1, y=0` Equation of line of intersection of `P_(1)=0 P_(2)=0` is parallel to the plane `P_(3)=0`. Sodistance of point `(-1,0,0)` from `P_(3)=0` is `|(-3+7)/sqrt(9+1+16)| = (4)/sqrt(26)` |
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