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The dot product of two vectors

x=[x1] and y = [y1]
[x2] [y2]
[ ⋮ ] [ ⋮ ]
[xn] [yn]
in R" is defined by x.y=x1y1+x2y2+...+xnyn. The vectors 7 and y are called perpendicular if y=0. Any vector in Rº perpendicular to
[-5]
[-4]
[-7]
can be written in the form
[ ] a + [ ] t
[ ] [ ]
[ ] [ ]

User Sintaxi
by
7.9k points

1 Answer

3 votes

Vectors x and y are perpendicular if their dot product is zero. A vector perpendicular to
\( \begin{bmatrix} -5 \\ -4 \\ -7 \end{bmatrix} \) can be expressed as
\( \begin{bmatrix} -4 \\ 5 \\ 0 \end{bmatrix}t + \begin{bmatrix} -7 \\ 0 \\ 1 \end{bmatrix}s \).

The dot product of two vectors
\( x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \) and
\( y = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix} \) in
\( \mathbb{R}^n \) is defined as
\( x \cdot y = x_1y_1 + x_2y_2 + \ldots + xnyn \).

Two vectors x and y are called perpendicular if their dot product is zero, i.e.,
\( x \cdot y = 0 \). In this case,
\( x \cdot y = x_1y_1 + x_2y_2 + \ldots + xnyn = 0 \).

Now, consider the vector
\( y = \begin{bmatrix} -5 \\ -4 \\ -7 \end{bmatrix} \) . A vector a is perpendicular to y if
\( a \cdot y = 0 \). Let
\( a = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} \). Then, the condition for perpendicularity is:


\[ a \cdot y = a_1(-5) + a_2(-4) + a_3(-7) = 0 \]

This simplifies to
\( -5a_1 - 4a_2 - 7a_3 = 0 \).

The solution to this equation gives the set of all vectors a perpendicular to y in
\( \mathbb{R}^3 \). It is given by:
\[ a = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} = \begin{bmatrix} -4 \\ 5 \\ 0 \end{bmatrix}t + \begin{bmatrix} -7 \\ 0 \\ 1 \end{bmatrix}s \]where t and s are scalar parameters.

The question probable may be:

The dot product of two vectors
\( x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \) and
\( y = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix} \) in R" is defined by
\( x \cdot y = x_1y_1 + x_2y_2 + \ldots + xnyn \). The vectors 7 and y are called perpendicular if y=0. Any vector in Rº perpendicular to
\( \begin{bmatrix} -5 \\ -4 \\ -7 \end{bmatrix} \) can be written in what the form?

User Charmelle
by
7.5k points