Multiplying Vectors — Question 2

Question 2

A negative dot product locates an obtuse angle

Dimensionless vectors are $\vec a=(1,2,2)$ and $\vec b=(2,-1,-2)$ in an orthonormal spatial frame. The dot product satisfies $\vec a\cdot\vec b=ab\cos\theta$, with $0^\circ\leq\theta\leq180^\circ$.

1. Find the dot product, both magnitudes and the angle between the vectors.

2. Find the signed scalar projection of a onto the direction of b, defined as $a\cos\theta$.

3. If b is reversed, find the new angle and projection. Why is taking the absolute value of the dot product too early incorrect?

Question 2 - Solution

A negative dot product locates an obtuse angle

(a) Combine the component and angle definitions.

$$\vec a\cdot\vec b=2-2-4=-4,\quad a=b=3,\quad\boxed{\theta=\cos^{-1}(-4/9)=116.388^\circ}$$ $\vec a\cdot\vec b=2-2-4=-4$, and $a=b=\sqrt9=3$. Hence $\cos\theta=-4/9$ and $\theta=\cos^{-1}(-4/9)\approx116.388^\circ$. The negative product signals an obtuse angle.

(b) Divide by the length of the projection direction.

$$a_{\parallel b}=\frac{\vec a\cdot\vec b}{b}=-\frac43,\qquad|a_{\parallel b}|=\frac43$$ The signed scalar projection is $(\vec a\cdot\vec b)/b=-4/3$. Its negative sign says that the projection points opposite to positive b. The length of that projection is $4/3$, but its signed value retains more information.

(c) Reverse the reference direction.

$$\theta'=180^\circ-\theta=63.612^\circ,\quad a_{\parallel(-b)}=+\frac43$$ The product becomes $+4$, so the new angle is $63.612^\circ=180^\circ-\theta$ and the signed projection is $+4/3$. Taking $|\vec a\cdot\vec b|$ at the outset would confuse the original obtuse angle with its acute supplement.