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Parametric form of the equation of the plane is `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» Equation of the plane is `bar r = ( 2hati + hatk)+ lambda hati + mu ( hati + 2hatj-3hatk)" …(i)"` Comparing equation (i) With `bar r = bara + lambda barb+ mu barc` `veca=2(hati +hatk), vecb=hati, vec c= hati + 2hatj - 3hatk` `therefore ` Normal to the plane is `vecb xx vecc = |(hati, hatj, hatk),(1,0,0),(1,2,-3)|` `=hati (0-0)- hatj(-3-0)+hatk(2-0)` `3hatj+ 2hatk` Now, `bara. (barb xx barc) = (2hati + hatk). (3 hatj+ 2hatk)` `=0+0+2=2` `therefore` Equation of the plane in normal form is `vec r . (vecb xx vecc)= veca .(vecb xx vecc)` `vecr. (3hatj+2hatk)=2" ...(ii)"` If `vec r = xhati + yhatj+ zhatk`, then the equation becomes, `(x hati+y hatj+ zhatk) . (3hatj+2hatk)=2` `rArr" "x(0)+y(3)+z(2)=2` `rArr" "3y+2x=2` This is the Cartesian form of the equation of given plane. |
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