The properties of the scalar product

Let $ \vec{u}$ and $ \vec{v}$ be two vectors.

We represent $||\vec{u}|| $ and $||\vec{v}|| $ as the respective magnitudes of $ \vec{u}$ and $ \vec{v}$, and $ (\vec{u} , \vec{v})$ the forming angle by these two vectors.

We call the scalar product $ \vec{u}.\vec{v} $, the real number resulting from:

$$ \vec{u}.\vec{v} = ||\vec{u}|| \times ||\vec{v}|| \times cos(\vec{u}, \vec{v}) $$

This is the magnitude of the orthogonal projection of the vector $ \vec{u}$ on the vector $ \vec{v}$, multiplied by the magnitude of the vector $ \vec{v}$.

Orthogonal projection of the vector u on the vector v

Commutative law

$$ \forall (\vec{u}, \vec{v}), $$

$$ \vec{u} . \vec{v} = \vec{v}. \vec{u} $$

Orthogonal vectors

$$ \forall (\vec{u}, \vec{v}) \neq \vec{0},$$

$$ \vec{u} \ and \ \vec{v} \ orthogonal \ vectors \Longleftrightarrow \vec{u}. \vec{v} = 0 $$

Squared vector

$$ \forall \vec{u},$$

$$ \vec{u}.\vec{u} = || \vec{u} ||^2 $$

Coordinates product

$$ \forall \left [\vec{u}\begin{pmatrix} x\\ y\\z \end{pmatrix} , \vec{v}\begin{pmatrix} x'\\ y'\\z' \end{pmatrix} \right], $$

$$ \vec{u}. \vec{v} = xx' + yy' +zz' $$

Bilinearity

$$ \forall \lambda \in \hspace{0.05em} \mathbb{R}, \ \forall (\vec{u}, \vec{v}),$$

$$ (\lambda\vec{u}).\vec{v} = \vec{u}.(\lambda\vec{v}) = |\lambda| \times \vec{u}. \vec{v}$$

Furthermore, performing the scalar product $ (\lambda\vec{u}).(\mu\vec{v})$, we do obtain:

$$ \forall (\lambda, \mu) \in \hspace{0.05em} \mathbb{R}^2, \ \forall (\vec{u}, \vec{v}),$$

$$ (\lambda\vec{u}).(\mu\vec{v}) = |\lambda| \times |\mu| \times \vec{u}. \vec{v} $$

Distributive law in relation to the addition

$$ \forall (\vec{u}, \vec{v}, \vec{w}), $$

$$ \vec{u}.( \vec{v} + \vec{w}) = \vec{u}.\vec{v} + \vec{u}.\vec{w} $$

And also the distributive law to the left:

$$ \forall (\vec{u}, \vec{v}, \vec{w}), $$

$$ (\vec{u} + \vec{v}) . \vec{w} = \vec{u}.\vec{w} + \vec{v}.\vec{w} $$

Remarkable identities

These are the same formulas as remarkable identities.

$$ \forall (\vec{u}, \vec{v}), $$

$$ (\vec{u} + \vec{v})^2 = || \vec{u} ||^2 + 2 \vec{u}.\vec{v} + || \vec{v} ||^2 $$

$$ (\vec{u} - \vec{v})^2 = || \vec{u} ||^2 - 2 \vec{u}.\vec{v} + || \vec{v} ||^2 $$

$$ || \vec{u} ||^2 - || \vec{v} ||^2= (\vec{u} + \vec{v}) (\vec{u} - \vec{v}) $$

Expression depending on norms

$$ \forall (\vec{u}, \vec{v}), $$

$$ \vec{u}.\vec{v} =\frac{1}{2} \left( || \vec{u} + \vec{v} ||^2 - || \vec{u} ||^2 - || \vec{v} ||^2 \right ) $$

$$ \vec{u}.\vec{v} =\frac{1}{2} \left( || \vec{u} ||^2 + || \vec{v} ||^2 - || \vec{u} -\vec{v} ||^2 \right ) $$

Recap table of the properties of the scalar product

Click on the title to access to the recap table.

Demonstrations

Commutative law

Let $\vec{u} $ and $\vec{v} $ be two vectors.

Performing both scalar products $ \vec{u}.\vec{v} $ and $ \vec{v}.\vec{u} $, we do have:

$$ \Biggl \{ \begin{align*} \vec{u}.\vec{v} = ||\vec{u}|| \times ||\vec{v}|| \times cos(\vec{u}, \vec{v}) \\ \vec{v}.\vec{u} = ||\vec{v}|| \times ||\vec{u}|| \times cos(\vec{v}, \vec{u}) \end{align*} $$

Let us call $\alpha $ = $ (\vec{u}, \vec{v}) $, the forming angle by $ \vec{u} $ and $\vec{v} $, then:

$$ \Biggl \{ \begin{align*} cos(\alpha ) = cos(\vec{u}, \vec{v}) \\ cos( - \alpha ) = cos(\vec{v}, \vec{u}) \end{align*} $$

But, the cosine function is pair and: $ \forall \alpha \in \mathbb{R}, \ cos(\alpha ) = cos(-\alpha ) $.

From this we can deduce that: $ cos(\vec{u}, \vec{v}) = cos(\vec{v}, \vec{u}) $.

And finally,

$$ \forall (\vec{u}, \vec{v}), $$

$$ \vec{u}.\vec{v} =\vec{v}.\vec{u} $$

Orthogonal vectors

Let $\vec{u} $ and $\vec{v} $ be two vectors, different from the zero vector.

From left to right implication

Let us assume that $\vec{u} $ and $\vec{v} $ are orthogonal.

Then, by the definition of the scalar product,

$$ \vec{u}.\vec{v} = ||\vec{u}|| \times ||\vec{v}|| \times cos(\vec{u}, \vec{v}) $$

But, if $\vec{u} $ ans $\vec{v} $ are orhtogonal, then $cos(\vec{u}, \vec{v}) = cos \left(\frac{\pi}{2}\right) = 0 $.

Thus, their scalar product will be zero, and:

$$ \forall (\vec{u}, \vec{v}) \neq \vec{0},$$

$$ \vec{u} \ et \ \vec{v} \ orthogonal \ vectors \Longrightarrow \vec{u}. \vec{v} = 0 $$

Reciprocal

Reciprocally, being said that the two vectors $\vec{u} $ and $\vec{v} $ are non-zero vectors, so:

$$ \vec{u}. \vec{v} = 0 \Longrightarrow cos(\vec{u}, \vec{v}) = 0 $$

And,

$$ \forall k \in \mathbb{N}, \ cos(\vec{u}, \vec{v}) = 0 \Longrightarrow \Biggl\{ (\vec{u}, \vec{v}) = \frac{k\pi}{2} \Biggr \} $$

And as a result, $\vec{u} $ and $\vec{v} $ are necessarily orthogonal vectors.

$$\vec{u}. \vec{v} = 0 \Longrightarrow \vec{u} \ et \ \vec{v} \ orthogonal \ vectors $$

Conclusion

$$ \forall (\vec{u}, \vec{v}) \neq \vec{0},$$

$$ \vec{u} \ et \ \vec{v} \ orthogonal \ vectors \Longleftrightarrow \vec{u}. \vec{v} = 0 $$

Squared vector

Let $\vec{u} $ be a vector.

Then, by the definition of the scalar product,

$$ \vec{u}.\vec{u} = ||\vec{u}|| \times ||\vec{u}|| \times cos(\vec{u}, \vec{u}) $$

But $cos(\vec{u}, \vec{u}) = cos \left(0\right) = 1 $.

Thus, only magnitudes remains in the product.

And finally:

$$ \forall \vec{u},$$

$$ \vec{u}.\vec{u} = || \vec{u} ||^2 $$

Coordinates product

Let $\vec{u}\begin{pmatrix} x\\ y\\z \end{pmatrix}$ and $\vec{v}\begin{pmatrix} x'\\ y'\\z' \end{pmatrix}$ be two vectors represented in an orthonormal coordinate system $ (O, \vec{i}, \vec{j}, \vec{k})$ and such as the following figure:

So, we can express $\vec{u} $ and $\vec{v} $ in the following formula:

$$ \Biggl \{ \begin{align*} \vec{u} = x \vec{i} + y \vec{j} +z \vec{k} \\ \vec{v} = x' \vec{i} + y' \vec{j} +z' \vec{k} \end{align*} $$

Then, the scalar product $\vec{u}.\vec{v} $ is worth:

$$ \vec{u}.\vec{v} = (x \vec{i} + y \vec{j} +z \vec{k}) . ( x' \vec{i} + y' \vec{j} +z' \vec{k}) $$

$$ \vec{u}.\vec{v} = xx'\vec{i}.\vec{i} + xy'\vec{i}.\vec{j} + xz'\vec{i}.\vec{k} + yx'\vec{j}.\vec{i} + yy''\vec{j}.\vec{j} + yz'\vec{j}.\vec{k} + zx'\vec{k}.\vec{i} + zy'\vec{k}.\vec{j} + zz'\vec{k}.\vec{k} $$

$$ \vec{u}.\vec{v} = xx' || \vec{i} ||^2 + yy''|| \vec{j} ||^2 + zz'|| \vec{k} ||^2 + xy'\vec{i}.\vec{j} + xz'\vec{i}.\vec{k} + yx'\vec{j}.\vec{i} + yz'\vec{j}.\vec{k} + zx'\vec{k}.\vec{i} + zy'\vec{k}.\vec{j} \qquad (1) $$

The three vectors $\vec{i},\vec{j}, \vec{k} $ are by hypothesis our unit vectors, then:

$$ || \vec{i} ||^2 = || \vec{j} ||^2 = || \vec{k} ||^2 = 1 $$

Furthermore, being in an orthognal reference frame, the three vectors $\vec{i},\vec{j}, \vec{k} $ are orthogonal, so their scalar product is zero:

$$ \vec{i}.\vec{j} = \vec{i}.\vec{k}= \vec{j}.\vec{k} = 0 $$

And by the commutative law of the scalar product, we do also have:

$$ \vec{j}.\vec{i} = \vec{k}.\vec{i}= \vec{k}.\vec{j} = 0 $$

Now, rewriting $(1) $,

$$ \vec{u}.\vec{v} = xx' + yy'' + zz' + \hspace{0.1em} \underbrace { xy'\vec{i}.\vec{j} + xz'\vec{i}.\vec{k} + yx'\vec{j}.\vec{i} + yz'\vec{j}.\vec{k} + zx'\vec{k}.\vec{i} + zy'\vec{k}.\vec{j} } _\text{ $= \hspace{0.1em} 0$} $$

Et finalement,

$$ \forall \left [\vec{u}\begin{pmatrix} x\\ y\\z \end{pmatrix} , \vec{v}\begin{pmatrix} x'\\ y'\\z' \end{pmatrix} \right], $$

$$ \vec{u}. \vec{v} = xx' + yy' +zz' $$

Bilinearity

Let $(\lambda, \mu) \in \hspace{0.05em} \mathbb{R}^2$ be two real numbers, and $\vec{u}\begin{pmatrix} x\\ y\\z \end{pmatrix}, \ \vec{v}\begin{pmatrix} x'\\ y'\\z' \end{pmatrix}$ two vectors.

Performing the scalar product $ (\lambda\vec{u}).\vec{v}$, we do have:

$$ (\lambda\vec{u}).\vec{v} = || \lambda \vec{u} || \times || \vec{v} || \times cos(\lambda \vec{u}, \vec{v}) $$

The norm $ || \lambda \vec{u} || $ of thevector $ \vec{\lambda u}\begin{pmatrix} \lambda x\\ \lambda y\\ \lambda z \end{pmatrix} $ is worth:

$$ || \lambda \vec{u} || = \sqrt{ (\lambda x)^2 + (\lambda y)^2 + (\lambda z)^2 }$$

$$ || \lambda \vec{u} || = \sqrt{ \lambda^2 (x^2 + y^2 + z^2) }$$

$$ || \lambda \vec{u} || = |\lambda| \times || \vec{u} ||$$

So,

$$ (\lambda\vec{u}).\vec{v} = |\lambda| \times || \vec{u} || \times || \vec{v} || \times cos(\lambda \vec{u}, \vec{v}) $$

Regarding the angle $ (\lambda \vec{u}, \vec{v}) $, it is no different from $ ( \vec{u}, \vec{v}) $, because $ ( \lambda \vec{u}) $ is only the extension of the vector $ \vec{u}$.

$$ (\lambda\vec{u}).\vec{v} = |\lambda| \times || \vec{u} || \times || \vec{v} || \times cos( \vec{u}, \vec{v}) $$

And as result,

$$ \forall \lambda \in \hspace{0.05em} \mathbb{R}, \ \forall (\vec{u}, \vec{v}),$$

$$ (\lambda\vec{u}).\vec{v} = |\lambda| \times \vec{u}. \vec{v}$$

By performing the same reasoning on the right-handed vector, we do obtain right linearity:

$$ \forall \lambda \in \hspace{0.05em} \mathbb{R}, \ \forall (\vec{u}, \vec{v}),$$

$$ \vec{u}.(\lambda\vec{v}) = |\lambda| \times \vec{u}. \vec{v}$$

And we then obtain bilinearity:

$$ \forall \lambda \in \hspace{0.05em} \mathbb{R}, \ \forall (\vec{u}, \vec{v}),$$

$$ (\lambda\vec{u}).\vec{v} = \vec{u}.(\lambda\vec{v}) = |\lambda| \times \vec{u}. \vec{v}$$

Furthermore, performing the scalar product $ (\lambda\vec{u}).(\mu\vec{v})$, we do obtain:

$$ \forall (\lambda, \mu) \in \hspace{0.05em} \mathbb{R}^2, \ \forall (\vec{u}, \vec{v}),$$

$$ (\lambda\vec{u}).(\mu\vec{v}) = |\lambda| \times |\mu| \times \vec{u}. \vec{v} $$

Distributive law in relation to the addition

Let $\vec{u}\begin{pmatrix} x\\ y\\z \end{pmatrix}$, $\vec{v}\begin{pmatrix} x'\\ y'\\z' \end{pmatrix}$ and $\vec{w}\begin{pmatrix} x''\\ y''\\z'' \end{pmatrix}$ be three vectors.

Let us express $\vec{u} $, $\vec{v} $ and $\vec{w} $ under their coordinates form:

$$ \begin{Bmatrix} \vec{u} = x \vec{i} + y \vec{j} +z \vec{k} \\ \vec{v} = x' \vec{i} + y' \vec{j} +z' \vec{k} \\ \vec{w} = x'' \vec{i} + y'' \vec{j} +z'' \vec{k} \end{Bmatrix} $$

Distributive law to the right

So, the scalar product $\vec{u}.( \vec{v} + \vec{w}) $ is worth:

$$ \vec{u}.( \vec{v} + \vec{w}) = (x \vec{i} + y \vec{j} +z \vec{k}) . ( x' \vec{i} + y' \vec{j} +z' \vec{k} + x'' \vec{i} + y'' \vec{j} +z'' \vec{k}) $$

$$ \vec{u}.( \vec{v} + \vec{w}) = (x \vec{i} + y \vec{j} +z \vec{k}) . \Bigl( (x' + x'') \vec{i} + (y' + y'') \vec{j} + (z' + z'') \vec{k} \Bigr) $$

$$ \vec{u}.( \vec{v} + \vec{w}) = (x(x' + x'')) \vec{i}.\vec{i} + (x(y' + y'')) \vec{i}.\vec{j} + (x(z' + z'')) \vec{i}.\vec{k} + (y(x' + x'')) \vec{j}.\vec{i} + (y(y' + y'')) \vec{j}.\vec{j} + (y(z' + z'')) \vec{j}.\vec{k} + (z(x' + x'')) \vec{k}.\vec{i} + (z(y' + y'')) \vec{k}.\vec{j} + (z(z' + z'')) \vec{k}.\vec{k} $$

$$ \vec{u}.( \vec{v} + \vec{w}) = (x(x' + x'')) ||\vec{i}||^2 + (y(y' + y'')) ||\vec{j}||^2 + (z(z' + z'')) ||\vec{k}||^2 + (x(y' + y'')) \vec{i}.\vec{j} + (y(x' + x''))+ \vec{j}.\vec{i} + (y(z' + z'')) \vec{j}.\vec{k} + (z(x' + x'')) \vec{k}.\vec{i} + (z(y' + y'')) \vec{k}.\vec{j} + (x(z' + z'')) \vec{i}.\vec{k} \qquad(2) $$

The three vectors $\vec{i},\vec{j}, \vec{k} $ are by hypothesis our unit vectors, so:

$$ || \vec{i} ||^2 = || \vec{j} ||^2 = || \vec{k} ||^2 = 1 $$

Furthermore, being in an orthognal reference frame, the three vectors $\vec{i},\vec{j}, \vec{k} $ are orthogonal, so their scalar product is zero:

$$ \vec{i}.\vec{j} = \vec{i}.\vec{k}= \vec{j}.\vec{k} = 0 $$

And by the commutative law of the scalar product, we do also have:

$$ \vec{j}.\vec{i} = \vec{k}.\vec{i}= \vec{k}.\vec{j} = 0 $$

Now, rewriting $ (2) $,

$$ \vec{u}.( \vec{v} + \vec{w}) = (x(x' + x'')) + (y(y' + y'')) + (z(z' + z'')) + \hspace{0.1em} \underbrace{ (x(y' + y'')) \vec{i}.\vec{j} + (y(x' + x''))+ \vec{j}.\vec{i} + (y(z' + z'')) \vec{j}.\vec{k} + (z(x' + x'')) \vec{k}.\vec{i} + (z(y' + y'')) \vec{k}.\vec{j} + (x(z' + z'')) \vec{i}.\vec{k} } _\text{ $= \hspace{0.1em} 0$} $$

$$ \vec{u}.( \vec{v} + \vec{w}) = (x(x' + x'')) + (y(y' + y'')) + (z(z' + z'')) $$

$$ \vec{u}.( \vec{v} + \vec{w}) = xx' + xx'' + yy' + yy'' + zz' + zz'' \qquad (3) $$

And now rearranging $ (3) $ :

$$ \vec{u}.( \vec{v} + \vec{w}) = xx' + yy' + zz' + xx'' + yy'' + zz'' $$

We recognize $ \vec{u}.\vec{v} $ and $ \vec{u}.\vec{w} $ under their coordinates product form :

$$ \vec{u}.( \vec{v} + \vec{w}) = \hspace{0.1em} \underbrace { xx' + yy' + zz' } _\text{ $= \hspace{0.1em} \vec{u}.\vec{v} $} \hspace{0.1em} + \hspace{0.1em} \underbrace { xx'' + yy'' + zz'' } _\text{ $= \hspace{0.1em} \vec{u}.\vec{w} $} $$

AS a result we do have,

$$ \forall (\vec{u}, \vec{v}, \vec{w}), $$

$$ \vec{u}.( \vec{v} + \vec{w}) = \vec{u}.\vec{v} + \vec{u}.\vec{w} $$

Distributive law to the left

And thanks to the commutative law of the scalar product, we do have the same thing to the left:

$$ (\vec{u} + \vec{v}) . \vec{w} = \vec{w} . (\vec{u} + \vec{v}) $$

Now developping the expression with the distributive law to the right, we do have:

$$ (\vec{u} + \vec{v}) . \vec{w} = \vec{u}.\vec{w} + \vec{v}.\vec{w} $$

And finally,

$$ \forall (\vec{u}, \vec{v}, \vec{w}), $$

$$ (\vec{u} + \vec{v}) . \vec{w} = \vec{u}.\vec{w} + \vec{v}.\vec{w} $$

Remarkable identities

We saw above that the scalar product is distributive, that is the property that will be used in this part.

Calculation of $ (\vec{u} + \vec{v})^2 $

$$ (\vec{u} + \vec{v})^2 = (\vec{u} + \vec{v}).(\vec{u} + \vec{v}) $$

$$ (\vec{u} + \vec{v})^2 =\vec{u}. \vec{u} + \vec{u}.\vec{v} + \vec{u}.\vec{v} - \vec{v}. \vec{v} $$

$$ (\vec{u} + \vec{v})^2 = || \vec{u} ||^2 + 2 \vec{u}.\vec{v} + || \vec{v} ||^2 $$

Calculation of $ (\vec{u} - \vec{v})^2 $

This is the same calculation but with a $(-)$ sign in the double product.

$$ (\vec{u} \textcolor{#8E5B5B}{-} \vec{v})^2 = (\vec{u} \textcolor{#8E5B5B}{-} \vec{v}).(\vec{u} \textcolor{#8E5B5B}{-} \vec{v}) $$

$$ (\vec{u} \textcolor{#8E5B5B}{-} \vec{v})^2 =\vec{u}. \vec{u} \textcolor{#8E5B5B}{-} \vec{u}.\vec{v} \textcolor{#8E5B5B}{-} \vec{u}.\vec{v} - \vec{v}. \vec{v} $$

$$ (\vec{u} \textcolor{#8E5B5B}{-} \vec{v})^2 = || \vec{u} ||^2 \textcolor{#8E5B5B}{-} 2 \vec{u}.\vec{v} + || \vec{v} ||^2 $$

Calculation of $ (\vec{u} + \vec{v}). (\vec{u} - \vec{v}) $

$$ (\vec{u} + \vec{v}). (\vec{u} - \vec{v}) = \vec{u}. \vec{u} - \vec{u}. \vec{v} + \vec{v}. \vec{u} - \vec{v}. \vec{v} $$

We saw above that the scalar product is commutative, so $ \vec{u}. \vec{v} = \vec{v}. \vec{u} $, and:

$$ (\vec{u} + \vec{v}). (\vec{u} - \vec{v}) = || \vec{u} ||^2 \hspace{0.1em} \underbrace{ - \ \vec{u}. \vec{v} + \vec{u}. \vec{v} } _\text{ $= \hspace{0.1em} 0 $} \hspace{0.1em} - || \vec{v} ||^2$$

$$ || \vec{u} ||^2 - || \vec{v} ||^2= (\vec{u} + \vec{v}) (\vec{u} - \vec{v}) $$

As a result, we obtain the same formulas as the classical remarkable identities.

Expression depending on norms

We saw above that:

$$ \Biggl \{ \begin{align*} (\vec{u} + \vec{v})^2 = || \vec{u} ||^2 + 2 \vec{u}.\vec{v} + || \vec{v} ||^2 \qquad (4) \\ (\vec{u} - \vec{v})^2 = || \vec{u} ||^2 - 2 \vec{u}.\vec{v} + || \vec{v} ||^2 \qquad (5) \end{align*} $$

Working with the expression $ (4) $, we do have:

$$ (\vec{u} + \vec{v})^2 = || \vec{u} ||^2 + 2 \vec{u}.\vec{v} + || \vec{v} ||^2 \qquad (4) $$

But thanks to the squared vectors property, we do know that:

$$ \forall \vec{u}, \ (\vec{u})^2 = || \vec{u} ||^2 $$

Then $ (4) $ becomes now:

$$ || \vec{u} + \vec{v} ||^2 = || \vec{u} ||^2 + 2 \vec{u}.\vec{v} + || \vec{v} ||^2$$

$$ 2 \vec{u}.\vec{v} = || \vec{u} + \vec{v} ||^2 - || \vec{u} ||^2 - || \vec{v} ||^2$$

And finally,

$$ \vec{u}.\vec{v} =\frac{1}{2} \left( || \vec{u} + \vec{v} ||^2 - || \vec{u} ||^2 - || \vec{v} ||^2 \right ) $$

In the same way, working with the expression $ (5) $:

$$ (\vec{u} \textcolor{#8E5B5B}{-} \vec{v})^2 = || \vec{u} ||^2 \textcolor{#8E5B5B}{-} 2 \vec{u}.\vec{v} + || \vec{v} ||^2 $$

$$ ||\vec{u} - \vec{v}||^2 = || \vec{u} ||^2 - 2 \vec{u}.\vec{v} + || \vec{v} ||^2 $$

$$ 2 \vec{u}.\vec{v} = || \vec{u} ||^2 + || \vec{v} ||^2 - ||\vec{u} - \vec{v}||^2 $$

$$ \vec{u}.\vec{v} =\frac{1}{2} \left( || \vec{u} ||^2 + || \vec{v} ||^2 - || \vec{u} -\vec{v} ||^2 \right ) $$

Recap table of the properties of the scalar product

Examples

Calculate a distance from a point to a plane

Let $(\mathcal{P})$ be a plane in space, passing through a point $A(x_0, y_0, z_0)$ and orthogonal to a vector $\vec{n}\begin{pmatrix} a\\ b \\c \end{pmatrix}$(with $a, b, c $ all three non-zero).

Distance from a point to a plane by the scalar product

We need to determine the distance $[HM]$ from point $M(x, y, z)$ to the plane $(\mathcal{P})$.

As both vectors $\vec{n}$ and $\overrightarrow{HA}$ are orthogonal, we do have:

$$ \vec{n} \ . \ \overrightarrow{HA} = 0$$

$$ \vec{n} \ . \ (\overrightarrow{HM} + \ \overrightarrow{MA}) = 0$$

$$ \vec{n} \ . \ \overrightarrow{HM} + \ \vec{n} \ . \ \overrightarrow{MA} = 0$$

$$ || \vec{n} || \times || \overrightarrow{HM} || + \vec{n} \ . \ \overrightarrow{MA} = 0$$

$$ || \overrightarrow{HM} || = \frac{\vec{n} \ . \ \overrightarrow{AM}}{ || \vec{n} ||} $$

$$ HM = \left | \frac{\vec{n} \ . \ \overrightarrow{AM}}{ || \vec{n} ||} \right| $$

Therefore, knowing the coordinates of point $M$, we can apply the scalar product calculation by the coordinates product:

$$ HM = \left | \frac{a(x -x_0) + b(y-y_0) + c(z-z_0) }{ \sqrt{a^2 + b^2 + c^2}} \right| $$

The properties of the scalar product

Demonstrations

From left to right implication

Reciprocal

Conclusion

Examples

Calculate a distance from a point to a plane