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|)`

If veca , vecb, vecc are the position vectors of the vertices. A,B,C of a triangle ABC. Then the area of triangle ABC is

veca,vecb,vecc are the position vectors of vertices A, B, C respectively of a paralleloram, ABCD, ifnd the position vector of D.

Let the vectors veca,vecb,vecc be the position vectors of the vertices P,Q,R respectively of a triangle. Which of the following represents the area of the triangle? (A) 1/2|vecaxxvecb| (B) 1/2|vecbxxvecc| (C) 1/2 |veccxxveca| (D) 1/2|vecaxxvecb+vecbxxvecc+veccxxveca|

If veca, vecb and vecc are position vectors of A,B, and C respectively of DeltaABC and if|veca-vecb|=4,|vecb-vec(c)|=2, |vecc-veca|=3 , then the distance between the centroid and incenter of triangleABC is

If veca,vecb,vecc are the position vectors of A,B,C respectively prove that vecaxxvecb+vecbxxvecc+veccxxveca is a vector perpendicular to the plane ABC.

If veca, vecb, vecc are three vectors, then [(vecaxxvecb, vecbxxvecc, veccxxveca)]=

If veca+2vecb+3vecc=0 , then vecaxxvecb+vecbxxvecc+veccxxveca=

If veca , vecb , vecc are position vectors of the vertices A, B, C of a triangle ABC, show that the area of the triangle ABC is 1/2 ​ [ veca × vecb + vecb × vecc + vecc × veca ] . Also deduce the condition for collinearity of the points A, B and C.

For three vectors veca+vecb+vecc=0 , check if (vecaxxvecb)=(vecbxxvecc)=(veccxxveca)

Prove that (vecbxxvecc)xx(veccxxveca)=[veca vecb vecc]vecc