Question

$IfA=(1,2,3)$ , $B=(2,3,4)$ and $C$ is a point of trisection of$AB$ such that $C_x+C_y=\frac{13}{3}$ then $C_z=$

• A. $\frac{10}{3}$
• B. $\frac{11}{3}$
• C. $\frac{11}{2}$
• D. $11$

Answer 1

The correct option is B $\frac{11}{3}$

Given : $A=(1,2,3),B=(2,3,4)$

$C$ is a point of trisection of $AB$

$⟹C=\frac{A+2B}{3}$.............($\because$ this cuts the line in the ratio of $1:2$)

$⟹C=\frac{(1,2,3)+2(2,3,4)}{3}$

$⟹C=\frac{(1+4,2+6,3+8)}{3}$

$⟹C=(\frac{5}{3},\frac{8}{3},\frac{11}{3})$

$\therefore C_x=\frac{5}{3},C_y=\frac{8}{3},C_z=\frac{11}{3}$

Now, $C_x+C_y=\frac{5}{3}+\frac{8}{3}=\frac{13}{3}$

Thus, $C_z=\frac{11}{3}$

Hence, option B is correct.