Why is the dot product of two unit vectors equal to zero?

Question

Can someone explain why the scalar product of $\hat{i} \cdot \hat{j}=0$ and the cross product of $\hat{i} \times \hat{j}=\hat{k}.$ Here we define $\hat{i}=(1,0,0), \hat{j}=(0,1,0), \hat{k}=(0,0,1)$.

Thanks for your help.

Answer 1

Let $a$ and $b$ be any two nonzero vectors in $\mathbb{R}^n$. That is, $a=(a_1,a_2,a_3, \cdots,a_n)$ and $b=(b_1,b_2,b_3,\cdots,b_n)$. Then by definition of calar product, we have $$a.b=a_1b_1+a_2b_2+a_3b_3+\cdots+ a_nb_n.$$ In your case,if $n=3$, we have $i=(1,0,0)$ and $j=(0,1,0)$ and thier scalar product is given by $$i.j=1\cdot 0+0\cdot 1+0\cdot 0=0.$$

For vector product, refer the definition.

Answer 2

Think about the definition of the dot product: $$\vec{u}\cdot\vec{v} = (\vec{u}_\hat{i} \times \vec{v}_\hat{i}) + (\vec{u}_\hat{j} \times \vec{v}_\hat{j}) + (\vec{u}_\hat{k} \times \vec{v}_\hat{k}) $$

i.e.: $$(2,5,6)\cdot(1,5,6)=(2\times1)+(5\times5)+(6\times6)=2+25+36=63$$

Now, let's see what happens if we use unit vectors: $$(1, 0)\cdot(0, 1) = (1\times0)+(0\times1) = 0 + 0 = 0$$

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Another way you can think of the dot product of $\vec{i} \cdot \vec{j}$ to be, "How much of $\hat{i}$ is being projected onto $\hat{j}$?

That is, if you took $\hat{i}$ and flattened it onto $\hat{j}$, how much of the length of $\hat{j}$ would be covered by this squashed version of $\hat{i}$? In this case, since they are perpendicular, the piece of $\hat{i}$ projected onto $\hat{j}$ would be zero.