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The equation of the plane passing through the point (-1, 2, 1) and perpendicular to the line joining the points (-3, 1, 2) and (2, 3, 4) is · · · · · ·A. `rdot(5hati+2hatj+2hatk)=1`B. `rdot(5hati+2hatj+2hatk)=-1`C. `rdot(5hati-2hatj+2hatk)=-5`D. `rdot(5hati-2hatj-2hatk)=1` |
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Answer» Correct Answer - A Direction ratios of line joining points (-3,1,2) and (2,3,4) are (2+3,3-1,4-2) i.e.,(5,2,2) Since plane is perpendicular to the line joining the given points therefore direction ratio of normal to the plane are (5, 2, 2). `therefore` Required equation of plane passing through the point (-1, 2, 1) and having normal with direction ratio (5, 2, 2) is `a(X-x_(1)+b(y-y_(1))+c(z-z_(1))=0` `!5(x+1)+2(y-2)+2(z-1)=0` `!5x+5+2y-4+2z-2=0` `!5x+2y+2z-1=0` `!rdot(5ati+2hatj+2hatk)-1=0` |
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