Question

chapter ii.5
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question 1
take $v = \langle - 10,-5\
angle$. write out $v_{perp}$.

Explanation:

Step1: Recall perpendicular vector property

For a two - dimensional vector $\vec{V}=(a,b)$, a vector $\vec{V}_{\perp}=(x,y)$ is perpendicular to it if their dot - product $\vec{V}\cdot\vec{V}_{\perp}=ax + by=0$. Let $\vec{V}=(- 10,-5)$ and $\vec{V}_{\perp}=(x,y)$. Then $-10x-5y = 0$, which simplifies to $y=-2x$. We can choose $x = 1$, then $y=-2$.

Step2: Write the perpendicular vector

A vector perpendicular to $\vec{V}=(-10,-5)$ is $\vec{V}_{\perp}=(1, - 2)$. In general, all vectors of the form $k(1,-2)$ where $k\in\mathbb{R}$ are perpendicular to $\vec{V}$. We can choose the simplest non - zero case when $k = 1$.

Answer:

$(1,-2)$