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If veca, vecb and vecc are the position vectors of the vertices A,B and C. respectively , of triangleABC. Prove that the perpendicualar distance of the vertex A from the base BC of the triangle ABC is (|vecaxxvecb+vecbxxvecc+veccxxveca|)/(|vecc-vecb|) |
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Answer» SOLUTION :`|VEC(BC)xxvec(BA)|=|vecaxxvecb+vecb+vecbxxvecc+veccxxveca|` `|vec(BC)||vec(BA)|sin B=|vecaxxvecbxxvecbxxvecc+veccxxveca|` `|vecc-vecb| (ABsinB)=|vecaxxvecb+vecbxxvecc+veccxxveca|` Therefore, the length of PERPENDICUALR from A on BC is `AL=ABsinB=(|vecaxxvecb+vecbxxvecc+veccxxveca|)/(|vecb-vecc|)`
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