what are the parallel vector to line y=3

� 2008  Rasmus ehf and J�hann �sak

Vectors

Lesson 5

Vectors and straight lines

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Example 1:

Discover i vector that is parallel to the line y = 3x + ii and a second vector that is perpendicular to the line

We begin by finding two points that lie on the line.

    Choose 10 = 0 and observe the respective y value.
y = three�0 + 2 = ii. The point (0, 2) lies on the line.

 Cull ten = one and discover y.

    y = 3�1 + 2 = 5. The point (i, 5) lies on the line.

Detect a vector that joins these 2 points and can call it .  has the same direction equally the line and is called a direction vector. If we rotate the vector by 90� we get a vector that is perpendicular to the line. This is called  a Normal vector and is labelled . The diagram shows these 2 vectors.

We can rewrite the equation of the line in the example y = 3x + 2  in the form 3x � y + two = 0 past moving y over the equal sign and arranging the terms so that the term in x comes first, so the term in y and finally the abiding term. Notation that the coefficients of x and y, 3 and �i are the same every bit the coordinates of the normal vector .

We will now show that this is true for all straight lines.

We use the general grade for a directly line, ax + by + c = 0.

Detect two points on the line, start by choosing  x = 0 and finding y and and so by choosing y = 0 and finding x.

a� 0 + past + c = 0

y = �c/b if x = 0

ax + b� 0 + c = 0

ten = �c/a if y = 0

The points (0, �c/b) and (�c/a, 0) prevarication on the line.

The management vector is therefore and the normal vector is .

Information technology�s customary, and easier, to piece of work with whole numbers rather than fractions so nosotros multiply the coordinates of the vectors by ab/c. By doing this nosotros change just the length of the vectors, not the direction. The vectors are nonetheless parallel or perpendicular to the line.

The new management vector will be and the new normal vector will be

To find a direction vector or a normal vector for a straight line all we have to do is write the equation in the general class. We tin can and so read direct from the equation.

The general equation of a straight line: ax + by + c = 0.

A normal vector is

A direction vector is

We tin use these vectors to find the angle betwixt straight lines past using the scalar production of the vectors.  The angle between the lines is the aforementioned every bit the bending between their direction vectors. Elementary geometry shows that the angle between the lines is too equal to the angle between their normal vectors.

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Example two:

Find the angle v� betwixt the lines l₁ , with equation y = 3x + two and the line l₂ with equation y = x + 4 (see diagram).

Rearranging the equations to the general form we get: 3x � y + 2 = 0 and 10 � y + 2 = 0.

The normal vectors are and

The lengths of the normal vectors are:

We now have all the data nosotros need to use the scalar product to find the angle 5� betwixt the normal vectors, which is equal to the angle betwixt the lines.

5� ≈ 26.6�

Discover a vector that starts in the point ( ane , 2 ) and is perpendicular to the vector

If we give the end point of the vector coordinates (10, y) then we can see that all vectors of the form are perpendicular to the given vector.

The scalar production of perpendicular vectors is 0 which means we can write the post-obit equation:

3(x � i ) + (�1)(y � 2 ) = 0

3x � 3 � y + 2 = 0

3x � y � 1 = 0

This is the equation of a straight line which goes through the indicate  (one, 2) with a normal vector

This gives united states a method for finding the equation of a directly line if nosotros know ane point on the line and a normal vector.

The equation of a straight line that goes through the point  (x₁, y₁) and has a normal vector is:

a(ten - ten₁ ) + b(ten - y₁ ) = 0

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Example iii:

Detect the equation of 2 straight lines l₁ and l₂ that intersect in the bespeak ( 3 , iii ). 50_i is parallel and l_two is perpendicular to the line 3x �y + 2 = 0 .

The normal vector  of l_one is the same every bit the normal  of  3x �y + 2 = 0, that is .

  We can put this data directly into the equation

a(x � x_one ) + b(10 � y₁ ) = 0

three(x � 3 ) + (�1)(y � 3 ) = 0

3x � 9 � y + 3 = 0

3x � y � 6 = 0 (equation of the line l_ane)

The normal vector for fifty_two is the aforementioned as the direction vector of  3x �y + ii = 0 , that is .

  Again nosotros tin put these values into the bones line equation

a(10 � x₁ ) + b(x � y₁ ) = 0

1(x � 3 ) + three(y � 3 ) = 0

x � 3 + 3y � 9 = 0

10 + 3y � 12 = 0 (equation of the line l₂)

Finally, let�south look at the diagram.

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Example 4:

Find the altitude betwixt the parallel lines 3x � y + 2 = 0 and 3x � y � 6 = 0 (see the diagram).

Nosotros cull any point, for instance (ane, v), on the line 3x � y + 2 = 0 and notice the shortest distance of the betoken from the other line
3x � y � half dozen = 0. This will requite us the required distance betwixt the two lines.

The length of the normal vector is not usually the required distance.

From the diagram we tin can encounter that in this example the normal vector, , is longer than the altitude between the two lines. Nosotros therefore need to notice a number t that, when  multiplied by  the , gives us a vector that is exactly the right length.
Phone call the cease point of the vector (on the second line) (x, y). We can at present write the post-obit vector equation:

This gives us two equations that can be solved for  ten and y.

3t = 10 � one          and �t = y � 5

ten = 3t + ane y = �t + five

The point ( x, y ) lies on the second line so we can put these x and y values into the equation of the line.

three x � y � 6 = 0

three( 3t + 1 ) � (�t + five) � 6 = 0

9t + three + t � 5 � 6 = 0

10t = 8

t = 0,8

In example 2 we found that the length of  was .

The altitude between the ii lines is therefore t� or the required distance is 0.8�

We will at present do the above case with messages instead of numbers. This leads to a very useful formula for the distance of a indicate from a line.

The general equation of a straight line is ax + by + c = 0 and nosotros choose a point with coordinates (ten_i, y₁).
Now nosotros can write the following equation:

We solve this vector equation for x and y.

ta = 10 � x_one           and tb = y � y _ane

10 = ta + 10 ₁ y = tb + y _ane

Putting these values into the equation and solving for t:

a x + b y + c = 0

a( ta + 10 ₁ ) + b( tb + y ₁ ) + c = 0

ta² + ax₁ + tb² + by₁ + c = 0

ta² + tb^two = �ax₁ � by_one � c

t(a^two + b²) = �ax_one � past₁ � c

The length of the normal vector is given by | |² = aⁱⁱ + b^two then we can rewrite the equation every bit:

t�||^two = �ax₁ � past_i � c

If we dissever through by | |ⁱⁱ nosotros go a value for t, which we tin can multiply by the length of the normal vector . We need to call up that length cannot be negative so nosotros use the accented value sign.

t = |�ax₁ � by₁ � c| /||²

If we multiply by | | we become the following formula:

The distance of the point (x₁, y₁) from the line ax + by + c = 0 is

or

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Example 5:

We are now going to reflect the point (�3, 5) in the line 3x � y � half-dozen = 0 and find the point of reflection.

The program is as follows. Outset find a vector that is perpendicular to the line 3x � y � 6 = 0 going from the point (�iii, 5) to the point
P = (10, y), which lies on the line. This volition be the normal vector t  (see the diagram). We discover the reflected indicate S by calculation the vector t  to the position vector of P.

We brainstorm by finding the value of t with the same method equally in the previous case:

Solving these ii equations for 10 and y we become:

3t = x + iii        and �t = y � 5

ten = 3t � 3 y = �t + 5

Putting these values for x and y into the equation we can find t.

3 x � y � 6 = 0

3( 3t � iii ) � ( �t + v ) � 6 = 0

9t � 9 + t � v � 6 = 0

10t = 20

t = 2

This gives us the vector and we can detect the coordinates of the position vector of P.

Now we add two  ( t ) to this and become the position vector for S.

The coordinates of the bespeak of reflection: Southward = (9, i)

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Example 6:

Find the equations of 2 lines l_one and l_two that intersect in the point ( 3 , 4 ) and are parallel to the vectors and respectively.

And so find the equation of the angle bisector of the angle between the two lines.

The normal vectors of l₁ and 50₂ are and

The equation of  l_i is therefore:

1(x � three ) � 3(y � iv ) = 0

ten � three � 3y + 12 = 0

10 � 3y + ix = 0

and the equation of l_ii is

iii(x � three ) � 1(y � 4 ) = 0

3x � nine � y + 4 = 0

3x � y � 5 = 0.

All points (x, y) on the bending bisector are equidistant from the lines  l₁ and l₂. We tin therefore find the equation of the angle bisector by finding the distance of (x, y) from each line and equating them.

�x + 3y � ix = �(�3x + y + v)

� because of the absolute value

This gives usa ii equations.

�x + 3x + 3y � y � 9 � five = 0    or �ten � 3x + 3y + y � 9 + 5 = 0

2x + 2y � 14 = 0 �4x + 4y � iv = 0

10 + y � 7 = 0 x � y + one = 0

There are two answers considering in that location are 2 angles between the lines and therefore 2 angle bisectors (see diagram).

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Try Quiz five on Vectors.
Remember to employ the checklist to keep rail of your work .

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Source: http://www.rasmus.is/uk/t/F/Su58k05.htm