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junior college 1 | H2 Maths
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Gd morning! Can someone kindly help me with (ii) please? Thanks so much:)
Date Posted: 4 years ago
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Good afternoon Sonia! Here are my workings for this question.
We first convert the equation form into the same form as line 1. Then, we are given that lines 1 and 2 intersect at P. Yes, we are given (not asked to prove), so we solve the three simultaneous equations in three unknowns a, lambda and mu and solve for all three of them. We substitute either lambda or mu back into line 1 or line 2 for the coordinates of P.
We first convert the equation form into the same form as line 1. Then, we are given that lines 1 and 2 intersect at P. Yes, we are given (not asked to prove), so we solve the three simultaneous equations in three unknowns a, lambda and mu and solve for all three of them. We substitute either lambda or mu back into line 1 or line 2 for the coordinates of P.
Date Posted:
4 years ago
Hi Eric! For (i), i got r.(2,-1,0)=0
is this correct or wrong?
I used the scalar product eqn form
is this correct or wrong?
I used the scalar product eqn form
I took (4,8,3) x (20,40,19) to get the normal vector n
Then r.n=a.n
Then r.n=a.n
Good afternoon Sonia!
Let M be (4, 8, 3) and N be (20, 40, 19)
When you cross multiply those two vectors, you are basically finding the vector which is perpendicular to both OM and ON, which would be
(8 x 19 - 3 x 40, 3 x 20 - 4 x 19, 4 x 40 - 8 x 20)
= (-8. 4, 0)
= -4 (2. -1, 0)
This is basically the direction vector of a normal to a plane containing OM and ON (which will of course contain MN as well).
However, this would fit the vector equation of a plane better than the vector equation of a line, since this expression is for the equations of planes.
For simply the equations of lines, we just do our regular finding of "gradient" and a point, just like the case in 2D where you find y2-y1 divided by x2-x1 and then substituting a point to the line.
Let M be (4, 8, 3) and N be (20, 40, 19)
When you cross multiply those two vectors, you are basically finding the vector which is perpendicular to both OM and ON, which would be
(8 x 19 - 3 x 40, 3 x 20 - 4 x 19, 4 x 40 - 8 x 20)
= (-8. 4, 0)
= -4 (2. -1, 0)
This is basically the direction vector of a normal to a plane containing OM and ON (which will of course contain MN as well).
However, this would fit the vector equation of a plane better than the vector equation of a line, since this expression is for the equations of planes.
For simply the equations of lines, we just do our regular finding of "gradient" and a point, just like the case in 2D where you find y2-y1 divided by x2-x1 and then substituting a point to the line.
So your form of r . (2 -1 0) is more appropriate for the vector equation of a plane rather than the vector equation of a line.
Thank you Eric for the prompt and clear explanation! One more question, for the “gradient” you’re talking about, does it matter which point to minus which point? Or can we just take the subtraction of two points regardless of which comes first?
Thank you!
Thank you!
Sonia, technically we are expressing our answers as a direction vector, because the direction vector tell us the direction or angling of our line.
Sure, a vector and its negative vector are not going in the same direction, but aren’t they sharing the same line after all?
If you recall, we can use y2 - y1 divided by x2 - x1 or y1 - y2 divided by x1 - x2. Both will get us the same value.
Here, we are more interested in the angle of the vector rather than the actual direction. So, it does not matter if your vector points forward or backward, since your vector equation can be reexpressed again.
Like let’s say
r = (0 0 0) + k (-4 2 0).
I can reconfigure this to say that
r = (0 0 0) + h (2 -1 0)
where h = -2k is a variable constant.
At first glance they appear different, but the set of values that h and are practically infinite, so they are actually the same.
This is why we can take any two points in any order to do the subtraction.
Sure, a vector and its negative vector are not going in the same direction, but aren’t they sharing the same line after all?
If you recall, we can use y2 - y1 divided by x2 - x1 or y1 - y2 divided by x1 - x2. Both will get us the same value.
Here, we are more interested in the angle of the vector rather than the actual direction. So, it does not matter if your vector points forward or backward, since your vector equation can be reexpressed again.
Like let’s say
r = (0 0 0) + k (-4 2 0).
I can reconfigure this to say that
r = (0 0 0) + h (2 -1 0)
where h = -2k is a variable constant.
At first glance they appear different, but the set of values that h and are practically infinite, so they are actually the same.
This is why we can take any two points in any order to do the subtraction.
Also note that for perpendicular lines in 3 dimensions, there is no such thing as “product of gradients of two perpendicular lines is -1”, since we are talking about 3 dimensions here and there can be many perpendicular lines to a given line in a 3D configuration.
For example, the y-axis is perpendicular to the x-axis, but so does the z-axis! Yet the y-axis and the z-axis are not in the same direction.
For example, the y-axis is perpendicular to the x-axis, but so does the z-axis! Yet the y-axis and the z-axis are not in the same direction.
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