Question

The direction cosines of a line segment AB are $-\frac{2}{\sqrt{17}},\frac{3}{\sqrt{17}},-\frac{2}{\sqrt{17}}$. If $AB=\sqrt{17}$ and the coordinates of A are $(3,-6,10)$, then the coordinates of B are

• A. $(1,-2,4)$
• B. $(2,5,8)$
• C. $(-1,3,-8)$
• D. $(1,-3,8)$

Answer 1

The correct option is D $(1,-3,8)$

DC of the line $AB<\frac{-2}{\sqrt{17}},\frac{3}{\sqrt{17}},\frac{-2}{\sqrt{17}}>$

Since $AB=\sqrt{17}$

$\therefore$ DR of the line AB $=<-12,3,-2>$

DR of the line $AB$ $=$ Co-ordinates of $B$ - co-ordinates $A$

Let co-ordinates of $B$ be $(x,y,z)$

$\Rightarrow \therefore <-2,3,-2>=<x-3,y+6,z-10>$

Co-ordinates of $A$ :

$x-3=-2$

$\therefore x=1$

$y+6=3$

$\therefore y=-3$

$z-10=-2$

$\therefore z=8$

$\therefore <x,y,z>=<1,-3,8>$

option D is the correct option