Question

Let the position vectors of the points A, Band C be `veca, vecb` and `vecc` respectively. Let Q be the point of intersection of the medians of the triangle ΔABC. Then `vec(QA) + vec(QB) + vec(QC)` =

Options

• `(veca + vecb + vecc)/2`

• `2veca + vecb + vecc`

• `veca + vecb + vecc`

• `vec(0)`

Answer 1

`vec(0)`

Explanation:

Let the position vectors of the point A, B and C be `veca, vecb` and `vecc` respectively.

Since Q is the centroid therefore

Position vector of Q = `(veca + vecb + vecc)/3`

∴ `vec(QA) = veca - (veca + vecb + vecc)/3 = (2veca - vecb - vecc)/3`

Similarly, `vec(QB) = (2vecb - veca - vecc)/3` and `vec(QC) = (2vecc - veca - vecb)/3`

Consider, `vec(QA) + vec(QB) + vec(QC) = (2veca + 2vecb + 2vecc)/3 - ((2veca + 2vecb + 2vecc)/3) = vec(0)`

∴ `vec(QA) + vec(QB) + vec(QC) = vec(0)`