# Thread: Vector problem

1. ## Vector problem

Givan a=3x+4y and b =(h+5)x-(2k-3)y,find in term of h,k,x and y,vector a-2b.If the vector a-2b is a zero vector and x and y are non-zero vectors,find the values of h and k.

2. Hello, mastermin346!

This is quite straight-forward.
Exactly where is your difficulty?

Given: . $\begin{array}{ccc}\vec a &= & 3\vec x+4\vec y \\ \vec b &=& (h+5)\vec x -(2k-3)\vec y\end{array}$

. . find $\vec a-2\vec b$ in terms of $h,k, \vec x,\vec y.$

$\vec a - 2\vec b \;=\;(3\vec x-4\vec y) - 2\bigg[(h+5)\vec x - (2k-3)\vec y\bigg]$

. . . . $=\;3\vec x+ 4\vec y - 2(h+5)\vec x + 2(2k-3)\vec y$

. . . . $=\; 3\vec x + 4\vec y - 2h\vec x - 10\vec x + 4k\vec y - 6\vec y$

. . . . $=\;-2h\vec x - 7\vec x + 4k\vec y - 2\vec y$

. . . . $=\;-(2h+7)\vec x + 2(2k-1)\vec y$

If the vector $a -2b$ is the zero vector and $x$ and $y$ are non-zero vectors,
. . find the values of $h$ and $k$.

If $\vec a - 2\vec b \:=\:\vec0$, then: . $\begin{array}{ccccccccc}\text{-}(2h+7) \;=\;0 & \Rightarrow & h \;=\;\text{-}\frac{7}{2} \\ \\[-3mm] 2(2k-1) \;=\;0 & \Rightarrow & k \;=\; \frac{1}{2} \end{array}$