Find $h$ for which $\frac1{15}\mid 2\vec x-\vec y(h)\mid =1.$

Question

I'm given two vectors which are $\vec x=3\vec i-2\vec j$ and $\vec y=\vec i-h\vec j$

Find the positive value of $h$ such that the unit vector $2\vec x-\vec y$ is $\frac{1}{15}(2\vec x-\vec y)$

How should I solve this question? Thanks in advance.

Answer 1

Our vector as a function of $h$:

$$2\vec x-\vec y=5\vec i-(4-h)\vec j$$

From the length of the vector above, the objective is to find an $h$ so that:

$$\sqrt{41-8h+h^2}=15.$$

Thus, we have to solve the following equation

$$h^2-8h-184=0.$$

The solutions are

$$h_{1,2}=4\pm\sqrt{200}.$$

For $h=4\pm\sqrt{200}$ the length of $2\vec x-\vec y$ is $15.$

So for these $h$'s

$$\frac1{15}(2\vec x-\vec y)$$

is a unit vector. Indeed,

$$\sqrt{5^2+(4-(4+\sqrt{200}))^2}=\sqrt{25+200}=15$$

and $$\sqrt{5^2+(4-(4-\sqrt{200}))^2}=\sqrt{25+(-\sqrt{200})^2}=15.$$