Let G be the centroid of a-triangle ABC. If AB=a,AC=b , then the bisector AB in terms of vectors a and b is

Question

Let G be the centroid of a-triangle ABC. If AB=a,AC=b , then the bisector AB in terms of vectors a and b is (A) (2/3)(a+b) (B) (1/6)(a+b) (C) (1/3)(a+b) (D) (1/2)(a+b)

Tardigrade Admin · Accepted Answer

Take A as origin. Then position vectors of A, B, C are vec0, veca, vecb respectively. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/b48fb9230c3e7dc38cf7d12ca6b201a0-.jpeg" /> ∴ position vector of centroid is (0+ veca+ vecb/3) i.e., ( veca+ vecb/3) ∴ AG=( veca+ vecb/3)

Q. Let G be the centroid of a-triangle ABC. If $A B = a, A C = b$ , then the bisector $A B$ in terms of vectors $a$ and $b$ is

$\frac{2}{3} (a + b)$

$\frac{1}{6} (a + b)$

$\frac{1}{3} (a + b)$

$\frac{1}{2} (a + b)$

Take $A$ as origin. Then position vectors of $A$ , $B$ , $C$ are $0$ , $a$ , $b$ respectively.

$∴$ position vector of centroid is
$\frac{0 + a + b}{3}$ i.e., $\frac{a + b}{3}$
$∴ A G = \frac{a + b}{3}$

Q. Let G be the centroid of a-triangle ABC. If AB=a,AC=b , then the bisector AB in terms of vectors a and b is

32​(a+b)

61​(a+b)

31​(a+b)

21​(a+b)