Question

In Problems 7–16,

(a) find the dot product $v·w$;

(b) find the angle between v and w;

(c) state whether the vectors are parallel, orthogonal, or neither.

role="math" localid="1646803954117" $$\begin{gathered} v=\sqrt{3}i-j \\ w=i+j \end{gathered}$$

Short answer

Part a.$v·w=\sqrt{3}-1$

Part b. The angle between the vectors is $75°$.

Part c. The vectors are neither parallel nor orthogonal.

Step-by-step solution

Part (a) Step 1. Given Information

Given two vectors $v=\sqrt{3}i-j$and $w=i+j$.

We need to find their dot product, the angle between the two vectors, and the nature of the two vectors.

Part (a) Step 2. Find the dot product 

The dot product for the given vectors is given as

$$\begin{gathered} v·w=(\sqrt{3}i-j)(i+j) \\ v·w=\sqrt{3}·1+(-1)·1 \\ v·w=\sqrt{3}-1 \end{gathered}$$

Part (b) Step 1. Find v and w

For the vector $v=\sqrt{3}i-j$the magnitude is given as

$$\begin{gathered} \left|v\right|=\sqrt{{\left(\sqrt{3}\right)}^2+{(-1)}^2} \\ \left|v\right|=\sqrt{3+1} \\ \left|v\right|=\sqrt{4} \\ \left|v\right|=2 \end{gathered}$$

And for the vector $w=i+j$the magnitude is given as

$$\begin{gathered} \left|w\right|=\sqrt{1^2+1^2} \\ \left|w\right|=\sqrt{1+1} \\ \left|w\right|=\sqrt{2} \end{gathered}$$

Part (b) Step 2. Find the angle  

The angle between the vectorsv andw is given by the formula

$\cos \theta =\frac{v·w}{\left|v\right|\left|w\right|}$

On substituting the found values we get

$$\begin{gathered} \cos \theta =\frac{\sqrt{3}-1}{2·\sqrt{2}} \\ \cos \theta =\frac{\sqrt{3}-1}{2\sqrt{2}} \\ \cos \theta \approx 0.2588 \end{gathered}$$

And so the angle $\theta$is given as

$$\begin{gathered} \theta ={\cos }^{-1}(0.2588) \\ \theta =75° \end{gathered}$$

Part (c) Step 1. Identify the nature of the vectors  

As the angle between the two vectors is $75°$, so the given two vectors v andw are neither parallel nor orthogonal.