The distance of the point (1,3,–7) from the plane passing through the point (1,–1,–1), having normal perpendicular to both the lines x−11=y+2−2=z−43 and x−22=y+1−1=z+7−1, is:
A
20√74
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B
10√83
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C
5√83
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D
10√74
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Solution
The correct option is B10√83 As the plane is parallel to both the lines. So the cross product of direction ratios of the lines will give the normal vector of the plane (1,−2,3)×(2,−1,−1)=(5,7,3)
The equation of plane with normal and a point is →r.→n=→a.→n(→r−→a).→n=0.
So the equation of plane passing through the point (1,−1,−1) and direction ratios (5,7,3) is
5(x–1)+7(y+1)+3(z+1)=0
Hence, 5x+7y+3z+5=0 is the equation of the plane. So, the distance of the point P(1,3,–7) from the plane is d=∣∣∣5+21−21+5√25+49+9∣∣∣=∣∣∣10√83∣∣∣units