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If `L_1` is the line of intersection of the planes `2x-2y+3x-2=0``x-y+z+1=0` and `L_2` is the line of the intersection of the planes `x+2y-z-3=0` `3x-y+2z-1=0`then the distance of the origin from the plane containing the lines `L_1` and `L_2` isA. `(1)/(4sqrt(2))`B. `(1)/(3sqrt(2))`C. `(1)/(2sqrt(2))`D. `(1)/(sqrt(2))` |
Answer» Correct Answer - B `L_(1)` is the line of intersection of the plane `2x-2y+3z-2=0` and `x-y+z+1=0` and `L_(2)` is the line of intersection of the plane `x+2y-z-3=0` and `3x-y+2z-1=0` Since `L_(1)` is parallel to `|{:(hati,hatj,hatk),(2,-2,3),(1,-1,1):}|=hati+hatj` `L_(2)` is parallel to `|{:(hati,hatj,hatk),(1,2,-1),(3,-1,2):}|=3hati-5hatj-7hatk` Also, `L_(2)` passes through `((5)/(7),(8)/(7),0).` [put z=0 in last two planes] So, equation of plane is `|{:(x-(5)/(7),y-(8)/(7),z),(1,1,0),(3,-5,-7):}|=0implies7x-7y+8z+3=0` Now, perpendicular distance from origin is `|(3)/(sqrt(7^(2)+7^(2)+8^(2)))|=(3)/(sqrt(162))=(1)/(3sqrt(2))` |
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