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From my previous question, I looked at the solution:

Let $v_1, v_2 \in \mathbb{R}^3$ such that $v_1, v_2$ unit vectors and orthogonal. The plane P is defined by $P = \text{span } \{v_1, v_2\}$. Prove that for any $\overrightarrow{x} \in \mathbb{R}^3$ we have $\text{proj}_{P}(x) = (x \cdot v_1)v_1 + (x \cdot v_2)v_2$ Soln:

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I am unsure of the step taken at the underlined part.

So we have:

$$v_1 \cdot (x - \frac{x \cdot n}{||n||^2}n) = \ldots + \bigg ( \frac{x \cdot n}{||n||^2}n\bigg ) \cdot v_1$$

How can they claim:

$$\bigg ( \frac{x \cdot n}{||n||^2}n\bigg ) \cdot v_1 = \bigg ( \frac{x \cdot n}{||n||^2}v_1\bigg ) \cdot n$$

??

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    This follows from properties of the inner product in $\mathbb{R}^3$: $$ (\lambda v)\cdot w=v\cdot(\lambda w)=(\lambda w)\cdot v $$ where $\lambda\in\mathbb{R}$ and $v,w\in \mathbb{R}^3$. – 2017-02-06

1 Answers 1

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As we know -

$y \cdot (x_1 - x_2) = y \cdot x_1 - y \cdot x_2$

So in your first doubt both terms in brackets multiplied with $\vec{v_1}$ using dot product.

In your second doubt -

Term with $\vec{n_1} \cdot \vec{n_1} = 0$

Similarity $\vec{v_1} \cdot \vec{v_1} = 0$