Hi there.
Our space given in the previous session a
From this point, a given point to strike a
Finding the right equation and still a point-and-
the direction perpendicular to this direction We learned to find the equation of the plane.
As an example we solved even find out the truth the.
This example also here once
Repeat the air more thoroughly We will proceed to catch.
Flats will solve examples related.
Zero point given us a pin.
Components of a two or three.
From this point a given direction.
Component one one one.
Figures easy to get jobs, accounts easy whether he simply selected figures.
As in the first process in this way
from the zero point defined so that p and in accordance with the
E, find the equation of straight line.
Secondly, a gene C and D in space given point.
On the right of this point, it We want to know whether.
This figure was drawn with the value given shape.
I do not want to go into detail.
given a given p is zero.
Here we will find the truth.
As we learned the correct equation.
a vector variable x.
the point where x is zero, the basic point vector.
the direction vector.
the variable t.
Or you can also call parameters.
Argument.
Get in here and u given x is zero unless x the usa vector, x is zero.
Where x is zero.
One, two, three components plus t times with given direction.
One by one.
This means that the following is happening, of course, any a
floating-point, point A of the vector multiple thereof, t is solid.
A few times saying that the variable as a parameter of the problem is completed.
This vector equation x, y and z components
If we divide the x component of the vector on the left a first component of x plus t.
The second component y equals two plus t.
The third component Z plus, plus t.
z equals three plus t.
This is a very simple way bi emerges.
We answer these questions immediately.
C and D on the right?
It's meant to be on the right coordinates of points
Did we find the equation makes Does not provide?
Taking the three C-point coordinates C one was two.
While x t three what?
y is equal to two years we put two here.
Two plus t.
T what we'll find here?
z is equal to one.
T what we'll find here?
See the following situation here encounter.
We have an unknown.
In contrast we have three equation.
This we have always been accustomed to since middle school two two equations do not like unknowns.
An unknown number of bi units.
If the number of equations to three.
Thus each equation t We'll find value.
In the first two from the equation t We'll find.
The second equation is the left and right dilemma simplifies each other.
t we find zero.
The third equation, we find t minus two.
EUR value than all three of these have of course t can not be.
So these values are contradictory.
Therefore, there is no such point b C providing the correct equation.
Doğalo, so that the point C On the right is not defined.
If we do the same thing to point D for x three, four y, z for
we see that when we put the five t from the first equation equals two.
The second equation again t equals two.
T is equal to two third of the equation again.
So a single value for t can be found.
These three results are consistent with each other.
Therefore, point D i t is equal to that of the provides the correct equation.
Therefore, the right of point D above.
However, conflicting results point C On the right is not given.
In a second embodiment of a plane equation We want to find.
This is still at a point in the plane, the components which is a two to three
and a perpendicular vector minus one dot given two.
Of the plane defined by these two information We want to find the equation.
In the second case bi pin point us partially granted.
Provided the first and second components The third component, but has been left ambiguous.
And the question we define as follows.
This pin point where the plane defined
What can be found on the alpha should it be?
This plane equation in two ways, both way we can.
Bi increasingly recognize one parent with inner product x minus x is zero on this plane
x zero, given x from the zero point is any vector which.
All of these vectors with variable x to n is steep.
When we wrote this plane equation we obtain anyway.
Simple geometry like that have defined plane.
To do so, the process to make Let's write the components.
x old x-zero components x.
a first component of x is zero.
y minus x minus x is zero because of two we find.
BI in the z minus three.
N and n is an inner product components minus one or two.
We are writing to bring it on.
Means the inner product of the first component of their between second components
among them, their third component among them is to hit the balls.
Do this by x minus one minus that in so that in the minus one
minus y z minus three minus two plus two times.
Structure the right of this equation results 're getting.
How we organize?
We collect the unknown on the left.
x minus y plus z on the left going two.
The number of known numbers on the right We collect.
Here are minus one.
Plus two minus six minus five occurs.
This former five five get the right glance We are.
This is completely a point and a perpendicular plane The definition given vector.
The second way to search for information from b we use.
We know that the equation of the plane equation, Less Ax
By equals plus plus Cz D structure will be.
Because x, y and z are all the first forces will appear.
Wherein n and A, B, C coefficients given by perpendicular vector.
About our problem minus the two given.
Thus a B to a C. minus when positioned as two
general equation of the plane D everything outside is clear.
Therefore, we have a known.
But we have the information yet we use.
x zero point on this plane.
Thus the components of x is zero, coordinates when positioned
instead of x, y, minus was minus two instead a following
Two plus three times z, z that the three According six.
Yet we find in D equals five.
The result of this first method confirms.
And we're going to cut a little more.
Because the inner product has already been made here state.
For this reason we use this for intermediate values We concluded a little more quickly.
The first point in the second case this pin given the second coordinates.
Not given to third alpha.
This alpha want to find.
When?
This pin point on this plane is the.
If the point P is above the plane of the plane equation, right provides.
Plane equation.
Wherein x is between five and as D is Instead of a Y instead
As you can see by our sheep instead of two alpha these interests are now on the left side.
Right side of the five already evident.
See also here, here, by the alpha-soluble minus one occurs.
When you bring it right from the first two terms of six happens.
So from six to two alpha equals alpha There are equals three.
We also want to give a few assignments.
With this assignment, always show a way I put together.
In many cases the result of the promise.
E alone to find your result here, I did not.
Because the second assignment of these results obtained in two ways We can.
I want to compare.
Now a plane, a point to a line
As we can give direction and the two dot can give.
We know it from the plane.
Since points A and B given us the By combining the right of two points
The basic structure of the response vectors We'll find.
So as the EU interpretative bi
Follow the steps in the previous example The equations are correct.
Here again, one should pay attention to the EU.
Result in this example does not change much
but from the coordinates of the second point We will first remove.
It also reviewing vectors We have emphasized.
Now the following question immediately comes to mind here.
Did we choose x to zero to B. we choose?
Second assignment, want to highlight it.
Also if you choose to B. Even if you choose the same You can find the results.
B lonely little attention to the following details I need to.
Once you have selected as the base point of the first, the main points
Once you get u as u te Use the variable.
B is the base point, taking the base point, fixed point
When using this to take the p variable taken.
Because otherwise every two lines
When comparing the different parameters I need to use.
Otherwise it points in the right Download the free kayamaz.
These two lines of the equation found that the
Want a B. Even if you have to want to choose the same t and s are correct
By a simple transformation of You can show.
A second example where the plane on 're doing.
Bi-plane with a point and perpendicular vector As we can define the whole plane
points perpendicular to the perpendicular vector also We can also define two intersecting with true.
Here are the intersecting points of the first k, Verifies that intersects the first right.
Right this second.
Given to the direction of the first DC.
the second line has been correctly
and that these two intersect at the point We assume.
If we find that the point of intersection plane perpendicular vector
the course of these two direction vectors We just found by multiplying the vector.
Now let's put them into practice, this thoughts.
So this problem so that a point and of the other
smoothing the transformed vector We're going.
T and s in this way given line We are writing with arguments.
Again, as I said earlier, they We take the same is not true.
Then this is the right of the first and second We are connected to each other makes.
However, the point here from another are independent.
To ensure this, two different values we need to take.
For these two lines intersect,
This is the right of the equation We need to synchronize.
Means to synchronize the first x at the Build tab to the x, first the
and similarly to the second of y's y's z is also to synchronize.
First right of x t is a plus.
We are writing here.
T is equal to one plus the second right the first component zero plus p.
Similarly, the y components
t is zero and zero plus two first right plus two t,
t is equal to two, of course, remains only the y component of the second.
He is also a minus p.
One minus p.
When sync third components t will be equal to one plus three.
Minus one plus two p.
Now here we used a gene are faced with the situation.
Since middle school, because we still
two equations to work with two unknown If you're used to.
But here are two unknowns.
t and s.
However there three equations.
So these three equations the two unknown should provide.
The traditional way is it.
Any two of these equations we get.
Here we solve the two unknowns.
We took third place in the equation.
In the third equation means that are consistent is this
There are solutions, and the two towards each other cuts.
If you violate third equation results If the data, so contradictory, results
If then the two vectors of two lines means do not overlap.
It is not in the anti-plane.
Two lines are parallel to the plane.
Thank intersect.
However, in the space of two lines parallel and that intersect
b contrary to their third possibilities is.
That solution is found from two equations solution
sağlamıyos third violation of this condition income counterparts.
In this case, of course, knowing coefficients was set.
From these two equations with two unknowns t is zero and p equals one is found.
Let's Put to provide zero to t, the an equal pp, a.
Put the left of zero to t in the second equation side is zero.
On the right a minus p.
s also a gene.
So you do not provide.
Take these values in the third equation t see when positioned
Let's put a zero instead of the left side plus becomes zero.
Let's put the water on the right side rather than a negative two would be a plus.
It also gives an equals.
Hence the left of this equation is also a.
It is also a right.
So, the first two with the third equation is consistent, compatible and
t is zero and s is equal to one, each of the three They also provide equations.
These values of x at any place by sheep,
t is equal to zero, we find, for example by sheep x is zero.
would have a zero value of x.
Or take a thinkin 's second See the first number in the equation is zero plus one.
The second number is a zero of a minus.
A minus one plus two.
Again we find the same value.
These difficulties are perhaps watch here You may be taking.
I hope you're not attracted.
Because the slowly growing very slowly I'm telling.
But they do need to do your own in your head.
Now, as a second step.
We have learned that the intersection of these two lines.
We will set on the two lines We find the plane and
vector perpendicular to these two vectors for this We need to find.
It's very easy.
A direction vector of the first DC two or three.
A negative response vectors of the second DC one or two.
We cross product of two vectors, the If we take
a vector perpendicular to both these interests.
This is the plane perpendicular vector.
As you will recall vector multiplication were doing.
i, j, k are writing.
In the second line of the first vector 're writing.
In the third line of the second vector We are writing this we calculate the determinant.
i, j, k is also no need to put.
i component of two times two plus three.
j component begins with a minus b.
You'll recall when opening determinants plus minus
plus he is going alternating signs.
Cons are taking this column.
We take this line.
Means that once back three times two minus a
There are also negative at the beginning of the two minus three bi j hence.
Where k is the column of the third component and we throw line.
Minus one or minus two minus one or minus two.
When they're strong, and of course a seven minus three turns.
So the problem is returned to the previous one.
In the following meanings:An x is zero cut-off point.
We found that in the first step.
n found here.
E this problem to the basic definition of the plane turned.
A point and the plane perpendicular vector.
We also show that the inner product or runner up
and we find its way easily obtained this equation.
