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Show that distance of the point vec c f...

Show that distance of the point ` vec c` from the line joining ` vec aa n d vec b` is `(| vec bxx vec c+ vec cxx vec a+ vec axx vec b)/(| vec b- vec a|)`

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Given, Area of `ΔABC = 1/2 abs( vec(AB)) xx P=1/2 abs( vec(AB) xx vec(AC)) `

`P=abs(vecAB xx vec(AC) /vec(AB) )`

`P= abs((vecb−veca) xx (vecc−veca)) / abs(vecb−veca) `

`p=abs((vecb−veca) xx vecc−(vecb−veca)×veca)/ abs(vecb−veca) `

`p=abs (vecb xx vecc−veca xx vecc−vecb xx veca+veca xx veca )/ abs(vecb−veca) `

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[vec a + vec b, vec b + vec c, vec c + vec a] = 2 [vec a, vec b, vec c]

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[vec a, vec b + vec c, vec d] = [vec a, vec b, vec d] + [vec a, vec c, vec d]

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The length of the perpendicular form the origin to the plane passing through the point a and containing the line vec r= vec b+lambda vec c is a. ([ vec a vec b vec c])/(| vec axx vec b+ vec bxx vec c+ vec cxx vec a|) b. ([ vec a vec b vec c])/(| vec axx vec b+ vec bxx vec c|) c. ([ vec a vec b vec c])/(| vec bxx vec c+ vec cxx vec a|) d. ([ vec a vec b vec c])/(| vec cxx vec a+ vec axx vec b|)

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If the vectors vec a , vec b ,a n d vec c form the sides B C ,C Aa n dA B , respectively, of triangle A B C ,t h e n vec adot vec b+ vec bdot vec c+ vec cdot vec a=0 b. vec axx vec b= vec bxx vec c= vec cxx vec a c. vec adot vec b= vec bdot vec c= vec cdot vec a d. vec axx vec b+ vec bxx vec c+ vec cxx vec a=0

If vec a_|_ vec b , then vector vec v in terms of vec aa n d vec b satisfying the equation s vec vdot vec a=0a n d vec vdot vec b=1a n d[ vec v vec a vec b]=1 is vec b/(| vec b|^2)+( vec axx vec b)/(| vec axx vec b|^2) b. vec b/(| vec b|^)+( vec axx vec b)/(| vec axx vec b|^2) c. vec b/(| vec b|^2)+( vec axx vec b)/(| vec axx vec b|^) d. none of these

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