Let A(1,-1,2) and B(2,3,-1) be two points. If a point P divides AB internally in the ratio 2:3, then the position vector of P is

Question

Let A(1,-1,2) and B(2,3,-1) be two points. If a point P divides AB internally in the ratio 2:3, then the position vector of P is (A)  (1/√5)( hati+ hatj+ hatk)  (B)  (1/√3)( hati+6 hatj+ hatk)  (C)  (1/√3)( hati+ hatj+ hatk)  (D)  (1/5)(7 hati+3 hatj+4 hatk)

Tardigrade Admin · Accepted Answer

The position vector of P =(3A+2B/3+2) =(3(1,-1,2)+2(2,3,-1)/5) =((7,3,4)/5) =(1/5)(7 hati+3 hatj+4 hatk)

Q. Let $A (1, - 1, 2)$ and $B (2, 3, - 1)$ be two points. If a point $P$ divides $A B$ internally in the ratio $2 : 3,$ then the position vector of $P$ is

$\frac{1}{5} (\hat{i} + \hat{j} + \hat{k})$

$\frac{1}{3} (\hat{i} + 6 \hat{j} + \hat{k})$

$\frac{1}{3} (\hat{i} + \hat{j} + \hat{k})$

$\frac{1}{5} (7 \hat{i} + 3 \hat{j} + 4 \hat{k})$

$\frac{1}{5} (\hat{i} + \hat{j} + 9 \hat{k})$

The position vector of P
$= \frac{3 A + 2 B}{3 + 2}$
$= \frac{3 ( 1 , - 1 , 2 ) + 2 ( 2 , 3 , - 1 )}{5}$
$= \frac{( 7 , 3 , 4 )}{5}$
$= \frac{1}{5} (7 \hat{i} + 3 \hat{j} + 4 \hat{k})$

Q. Let A(1,−1,2) and B(2,3,−1) be two points. If a point P divides AB internally in the ratio 2:3, then the position vector of P is

5​1​(i^+j^​+k^)

3​1​(i^+6j^​+k^)

3​1​(i^+j^​+k^)

51​(7i^+3j^​+4k^)

5​1​(i^+j^​+9k^)

The position vector of P =3+23A+2B​ =53(1,−1,2)+2(2,3,−1)​ =5(7,3,4)​ =51​(7i^+3j^​+4k^)