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Find the vector equation of the plane passing through the point `hati + hatj -2hatk` and `hati + 2hatj + hatk, 2hati - hatj + hatk.` |
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Answer» `veca, vecb, vecc` are the position vectors i.e., `veca=hati+hatj-2hatk, vecb=hati+2hatj+hatk, vecc=2hati-hatj+hatkand vecb-veca=0hati+hatj+3hatk` `vecc-veca=hati-2hatj+3hatk` Now, `(vecb-veca)xx(vecc-veca)=|{:(hati, hatj, hatk),(0,1,3),(1,-2,3):}|` `=hati(3+6)-hatj(0-3)+hatk(0-1)=9hati+3hatj-hatk` Let `vecn =(vecb-veca)xx(vecc-veca)=9hati+3hatj-hatk` Then, the vector equation of the required plane is `vecr*vecn=veca*vecn` `vecr(9hati+3hatj-hatk)=(hati+hatj-2hatk)(9hati+3hatj-hatk)` `vecr(9hati+3hatj-hatk)=9+3+2` `vecr(9hati+3hatj-hatk)=14` The cartensian equation of the plane is given by `(x hati+yhatj+zhatk)(9hati+3hatj-hatk)=14` `because"["vecr=x hati+yhatj+z hatj"]"` `9x+3y-x=14` |
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