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Parametric form of the equation of the planeis bar r =(2hati + hatk ) + lambda hati + mu (hati + 2hati-3 hatk) lambda and mu are parameters. Find normal to the plane and hence equation of the plane in normal form. Write its Cartesion form. |
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Answer» Solution :The EQUATION `barr = BARA + lambdabarb + mubarc` represents a plane passing through a point having position vector a and parallel to vectors `barb` and `BARC.` Here, `bara = 2hati + hatk, barb = hati, barc = hati + 2hatj - 3hatk` `therefore` The normal vector `barn` to the plane is give by `barnxxbarbxxbarc=|{:(hati,,hatj,,hatk),(1,,0,,0),(1,,2,,"-3"):}|` ` = (0-0) hati -(-3-0) hatj + (2-0)hatk` i.e., `barn = 3hatj + 2hatk` The equation of the plane is scalar product form is ` barr * (barb xx barc) = bara * (barb xx barc)` i.e., `barr * barn = bara * barn` where `bara * barn = (2hati + hatk) * (3hati + 2hatk)` ` = 2(0) + 0(3) + 1(2)` ` = 0+ 0 + 2 = 2` `therefore` the equation of the plane is scalar product form is `barr*(3hatj + 2hatk) =2` i.e., `barr * barn = 2`, where `barn = 3hatj + 2hatk` `therefore |barn| = sqrt(3^(2) + 2^(2)) = sqrt(13)` The equation `barr * barn = 2`can be written as `barr*(barn)/(|barn|)=(2)/(|barn|)` i.e., `barr*((3hatj+2hatk)/(sqrt(13)))=(2)/(sqrt(13))` i.e., `barr*((3)/(sqrt(13))hatj+(2)/(sqrt(13))hatk)=(2)/(sqrt(13))` This is the normal form of the equation of plane. If `barr = XHATI + yhatj + zhatk`, then equation (1) BECOMES `(xhati+yhatj+zhatk)*(3hatj+2hatk)=2` `therefore x(0)+y(3)+z(2)=2` ` therefore 3y + 2z = 2` This is the cartasian equation of the plane. |
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