# Operations with Algebraic Vectors in R2

I was given a thinking question as my homework.

Here is the question

So basically what I've tried to do is: Since it's given that both the vectors start from the exact same point, and have the same magnitude, I tried to find the magnitude of PB by calculating the distance between vector P and B (a,0) + (0,2) and got sqrt((-a)^2 + 2

Following the same procedure for vector PA, I got sqrt((5-a)^2 - 0)

Now, I tried making both magnitudes equal to each other and solve for a since I know both vectors are to have the same magnitude, however, my result is different than the one in the textbook.

So my question is: am I missing anything? Does my logic even work here?

• "$|\vec{PB}|$" is the distance from $P$ to $B$. What is the formula for distance between two points? May 7, 2019 at 23:47
• I've tried calculating the distance for the vector PB (sqrt(25-10a+a^2)) and for PA (sqrt(a+2)). Now, I tried making them both equal to each other and solve for a, but I got a different answer from that in the textbook. May 7, 2019 at 23:54
• The lengths you list are backwards. Also, neither length is $\sqrt{a+2}$. You will likely get much better responses if you transcribe the problem into your Question and show the work you have already attempted. May 8, 2019 at 0:01

The distance between two points $$(a_1, a_2)$$ and $$(b_1, b_2) \in \mathbb{R}^2$$ is given by:

$$\sqrt{(b_1 - a_1)^2 + (b_2 - a_2)^2}$$

For the first part, we have

$$|\overrightarrow {PA}| = |\overrightarrow {PB}|$$

$$\sqrt{a^2 + 2^2} = \sqrt{(5-a)^2}$$

$$a^2 + 4 = 25 -10a + a^2$$

$$a = 2.1$$

So $$P$$ is the point $$(2.1,0)$$.

For the second part, we have

$$|\overrightarrow{QA}| = |\overrightarrow{QB}|$$

$$\sqrt{(b-2)^2} = \sqrt{b^2 + 5^2}$$

$$b^2 -4b + 4 = b^2 + 25$$

$$b = \frac{-21}{4} = -5.25$$

So $$Q$$ is the point $$(0,-5.25)$$.