Q. 60 - Calculus

Question

Let \(θ\) be the angle between nonzero vectors \(u\) and \(v\).

Prove each of the following:

(a) \(θ\) is acute if and only if \(u · v > 0\)

(b) \(θ\) is right if and only if \(u · v = 0\)

(c) \(θ\) is obtuse if and only if \(u · v < 0\)

Step-by-Step Solution

Answer

Part (a) It is proven that \(θ\) is acute if and only if \(u · v > 0\).

Part (b) It is proven that \(θ\) is right if and only if \(u · v = 0\). 

Part (c) It is proven that \(θ\) is obtuse if and only if \(u · v < 0\).

Part (a) Step 1. Given Information

It is given that \(θ\) is the angle between nonzero vectors \(u\) and \(v\).

We have to prove that \(θ\) is acute if and only if \(u · v > 0\).

Part (a) Step 2. Prove

To prove \(θ\) is acute if and only if \(u · v > 0\). Let \(u=\left<u_{1},v_{1} \right>\) and \(v=\left<u_{2},v_{2} \right>\) and \(θ\) be the angle between vectors \(u\) and \(v\) is \(cos\theta =\frac{u\cdot v}{\left\|u \right\|\left\|v \right\|}\).

As it is given that \(θ\) is acute which means \(θ\) lies between \(\left (0,\frac{\pi }{2}  \right )\).

So, \(cos\theta>0\) which is possible when \(u · v > 0\). 

Hence proved.

Part (b) Step 1. Prove

To prove that \(θ\) is right if and only if \(u · v = 0\). Let \(u=\left<u_{1},v_{1} \right>\) and \(v=\left<u_{2},v_{2} \right>\) and \(θ\) be the angle between vectors \(u\) and \(v\) is \(cos\theta =\frac{u\cdot v}{\left\|u \right\|\left\|v \right\|}\). 

So, when \(\theta =90^{\circ }\) 

\(cos90^{\circ } =\frac{u\cdot v}{\left\|u \right\|\left\|v \right\|}\)

\(0=\frac{u\cdot v}{\left\|u \right\|\left\|v \right\|}\) 

So, \(0=\frac{u\cdot v}{\left\|u \right\|\left\|v \right\|}\) if and only if \(u · v = 0\). 

Hence proved.

Part (c) Step 1. Prove

To prove that \(θ\) is obtuse if and only if \(u · v < 0\). Let \(u=\left<u_{1},v_{1} \right>\) and \(v=\left<u_{2},v_{2} \right>\) and \(θ\) be the angle between vectors \(u\) and \(v\) is \(cos\theta =\frac{u\cdot v}{\left\|u \right\|\left\|v \right\|}\). 

As it is given that \(θ\) is obtuse which means \(θ\) lies between \(\left (\frac{\pi }{2},\pi  \right )\).

So, \(cos\theta<0\) which is possible when \(u · v < 0\). 

Hence proved.