Let the position vectors of the points A, B and C be a , b and c, respectively. Let Q be the point of intersection of the medians of the Δ A B C . Then, Q A + Q B + Q C is equal to

Question

Let the position vectors of the points A, B and C be a , b and c, respectively. Let Q be the point of intersection of the medians of the Δ A B C . Then, Q A + Q B + Q C is equal to (A) (a + b + c/2) (B) 2 a+b+c (C)  a+b+c (D) (a + b + c/3)

Tardigrade Admin · Accepted Answer

Since, O is the intersecting point of all the medians of triangle A B C. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/46f65b210b40c376323ae205435e4f64-.png" /> Hence, Q =( a + b + c /3) Now, d Q A = a -( a + b + c /3) =(2 a - b - c /3) Similarly, Q B =(2 b - a - c /3) and Q C =(2 c - a - b /3) Hence, Q A + Q B + Q C =(2 a - b - c /3) +(2 b - a - c /3)+(2 c - a - b /3)=0

Q. Let the position vectors of the points $A, B$ and $C$ be $a, b$ and $c$ , respectively. Let $Q$ be the point of intersection of the medians of the $Δ A BC .$ Then, $Q A + QB + QC$ is equal to

$\frac{a + b + c}{2}$

$2 a + b + c$

$a + b + c$

$\frac{a + b + c}{3}$

$0$

Q. Let the position vectors of the points A,B and C be a,b and c, respectively. Let Q be the point of intersection of the medians of the ΔABC. Then, QA+QB+QC is equal to

2a+b+c​

2a+b+c

a+b+c

3a+b+c​

0

Since, O is the intersecting point of all the medians of △ABC. Hence, Q=3a+b+c​ Now, dQA=a−3a+b+c​ =32a−b−c​ { Similarly, QB=32b−a−c​ and QC=32c−a−b​ Hence, QA+QB+QC=32a−b−c​ +32b−a−c​+32c−a−b​=0

Q. Let the position vectors of the points $A, B$ and $C$ be $a, b$ and $c$ , respectively. Let $Q$ be the point of intersection of the medians of the $Δ A BC .$ Then, $Q A + QB + QC$ is equal to

$\frac{a + b + c}{2}$

$2 a + b + c$

$a + b + c$

$\frac{a + b + c}{3}$

$0$