Question

what is known about θ, the angle between two nonzero vectors u and v, under each condition (see figure)? (a) u·v = 0 〇 u and v are orthogonal and θ = \frac{pi}{2} 〇 0leq\thetaleqpi 〇 \frac{pi}{2}<\thetaleqpi 〇 0<\thetaleq\frac{pi}{2} 〇 0leq\theta<\frac{pi}{2} (b) u·v>0 〇 0<\thetaleq\frac{pi}{2} 〇 \frac{pi}{2}<\thetaleqpi 〇 \frac{pi}{2}leq\theta<pi 〇 0leq\theta<\frac{pi}{2} 〇 u and v are orthogonal and θ = \frac{pi}{2} (c) u·v<0 〇 0leq\theta<\frac{pi}{2} 〇 \frac{pi}{2}<\thetaleqpi 〇 0<\thetaleqpi 〇 \frac{pi}{2}leq\theta<pi 〇 0leq\thetaleqpi

Explanation:

Step1: Recall dot - product formula

The dot - product of two non - zero vectors $\mathbf{u}$ and $\mathbf{v}$ is given by $\mathbf{u}\cdot\mathbf{v}=\vert\mathbf{u}\vert\vert\mathbf{v}\vert\cos\theta$, where $\theta$ is the angle between $\mathbf{u}$ and $\mathbf{v}$ with $0\leq\theta\leq\pi$.

Step2: Analyze case (a)

If $\mathbf{u}\cdot\mathbf{v} = 0$, then $\vert\mathbf{u}\vert\vert\mathbf{v}\vert\cos\theta=0$. Since $\mathbf{u}$ and $\mathbf{v}$ are non - zero vectors, $\vert\mathbf{u}\vert
eq0$ and $\vert\mathbf{v}\vert
eq0$. So, $\cos\theta = 0$. In the range $0\leq\theta\leq\pi$, when $\cos\theta = 0$, $\theta=\frac{\pi}{2}$, and the vectors are orthogonal.

Step3: Analyze case (b)

If $\mathbf{u}\cdot\mathbf{v}>0$, then $\vert\mathbf{u}\vert\vert\mathbf{v}\vert\cos\theta>0$. Since $\vert\mathbf{u}\vert>0$ and $\vert\mathbf{v}\vert>0$, $\cos\theta>0$. In the range $0\leq\theta\leq\pi$, when $\cos\theta>0$, $0\leq\theta<\frac{\pi}{2}$.

Step4: Analyze case (c)

If $\mathbf{u}\cdot\mathbf{v}<0$, then $\vert\mathbf{u}\vert\vert\mathbf{v}\vert\cos\theta<0$. Since $\vert\mathbf{u}\vert>0$ and $\vert\mathbf{v}\vert>0$, $\cos\theta<0$. In the range $0\leq\theta\leq\pi$, when $\cos\theta<0$, $\frac{\pi}{2}<\theta\leq\pi$.

Answer:

(a) $\theta=\frac{\pi}{2}$, $\mathbf{u}$ and $\mathbf{v}$ are orthogonal
(b) $0\leq\theta<\frac{\pi}{2}$
(c) $\frac{\pi}{2}<\theta\leq\pi$