Mathematics: Post your doubts here!

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smartangel

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Jaf said:
This is, in essence, an incorrect question to begin with (I think).
The distance of a point from a line is never constant. What can be asked is the distance of a point from another point on a line. You're probably missing some essential detail of the question. I could be wrong, but I'm pretty sure I'm not.
Click to expand...
this question is from the black P3 book ...Pure mathematics 2 and 3 by Hugh neil and douglas quadling..you can chk it out if you have the book!
umm... i really dont know if its a correct question or not.. i actually dont get any of the questions of the exercise..here"s another one

Q2) find the distance of (1,0,0) from the line r= t(12i-3j-4k).. the answer is ((69)^0.5)/13
 
MHHaider707

MHHaider707

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smartangel said:
this question is from the black P3 book ...Pure mathematics 2 and 3 by Hugh neil and douglas quadling..you can chk it out if you have the book!
umm... i really dont know if its a correct question or not.. i actually dont get any of the questions of the exercise..here"s another one

Q2) find the distance of (1,0,0) from the line r= t(12i-3j-4k).. the answer is ((69)^0.5)/13
Click to expand...

Hey how did u do that?
 
ffaadyy

ffaadyy

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smartangel said:
i need help with vectors..Distance of a point from a line! this question is from the CIE endorsed book and not frm the pastpapers!
Q) Find the distance of (3,4) from the straight line 3x+ 4y=0
the answer is 5.
Click to expand...


Jaf said:
This is, in essence, an incorrect question to begin with (I think).
The distance of a point from a line is never constant. What can be asked is the distance of a point from another point on a line. You're probably missing some essential detail of the question. I could be wrong, but I'm pretty sure I'm not.
Click to expand...


This is how we'll do this question.


3x + 4y = 0

First of all, we'll be needing to find a set of coordinates on this line. We can find them by a fairly easy method.


3x + 4y = 0
4y = -3x
y = -3x/4


Keep 'x=4' in the above equation so that we get 'y=-3'. Now if we put these points back in the equation '3x + 4y = 0', the answer is '0' which proves that these points lie on the line. Similarly, we can put any two points in the equation '3x + 4y = 0' so that the final answer is equal to '0'. For e.g, we can also take the points 'x=1/3' and 'y=-1/4'. Next, we'll write a vector equation for this line.

r = (4)(-3) + t (3)(4) - (3)(4)
r = (4+3t-3)(-3+4t-4)
r = (1+3t)(-7+4t)

As the point '3,4' must be perpendicular from the straight line, we'll take out its cross product with the vector equation of the line and equate it to '0' in order to find the value of 't'.

(1+3t)(-7+4t) . (3)(4) = 0
3 + 9t - 28 + 16t = 0
-25 + 25t = 0
t = 1

Put this value of 't' back in the equation 'r = (1+3t)(-7+4t)'.

r = (1+3t)(-7+4t)
r = [(1 + 3(1)] [(-7 + 4(1)]
r = (4)(-3)

Use the formula '[(x)^2 + ( y )^2]^(1/2)' to find the distance.

[(x)^2 + ( y )^2]^(1/2)
[(4)^2 + (-3)^2]^(1/2)
(25)^(1/2)
5

Therefore, the distance is 5 units.
 
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unique840

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ffaadyy said:
This is how we'll do this question.


3x + 4y = 0

First of all, we'll be needing to find a set of coordinates on this line. We can find them by a fairly easy method.


3x + 4y = 0
4y = -3x
y = -3x/4


Keep 'x=4' in the above equation so that we get 'y=-3'. Now if we put these points back in the equation '3x + 4y = 0', the answer is '0' which proves that these points lie on the line. Similarly, we can put any two points in the equation '3x + 4y = 0' so that the final answer is equal to '0'. For e.g, we can also take the points 'x=1/3' and 'y=-1/4'. Next, we'll write a vector equation for this line.

r = (4)(-3) + t (3)(4) - (3)(4)
r = (4+3t-3)(-3+4t-4)
r = (1+3t)(-7+4t)

As the point '3,4' must be perpendicular from the straight line, we'll take out its cross product with the vector equation of the line and equate it to '0' in order to find the value of 't'.

(1+3t)(-7+4t) . (3)(4) = 0
3 + 9t - 28 + 16t = 0
-25 + 25t = 0
t = 1

Put this value of 't' back in the equation 'r = (1+3t)(-7+4t)'.

r = (1+3t)(-7+4t)
r = [(1 + 3(1)] [(-7 + 4(1)]
r = (4)(-3)

Use the formula '[(x)^2 + (y)^2]^(1/2)' to find the distance.

[(x)^2 + (y)^2]^(1/2)
[(4)^2 + (-3)^2]^(1/2)
(25)^(1/2)
5

Therefore, the distance is 5 units.
Click to expand...
kia baat hai apki... hum ques ko galat keh rahe hain n apne to solve bhi krdia :p
 
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unique840

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q) The diagram shows the curve y = x sin x, for 0 ≤ x ≤ π. The point Q (pie/2) , (pie/2) lies on the curve.
(i) Show that the normal to the curve at Q passes through the point (π, 0).
i got the normal but how to prove that it passes through the point (pie,0)?
 
Spunkyfarhan

Spunkyfarhan

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ffaadyy said:
This is how we'll do this question.


3x + 4y = 0

First of all, we'll be needing to find a set of coordinates on this line. We can find them by a fairly easy method.


3x + 4y = 0
4y = -3x
y = -3x/4


Keep 'x=4' in the above equation so that we get 'y=-3'. Now if we put these points back in the equation '3x + 4y = 0', the answer is '0' which proves that these points lie on the line. Similarly, we can put any two points in the equation '3x + 4y = 0' so that the final answer is equal to '0'. For e.g, we can also take the points 'x=1/3' and 'y=-1/4'. Next, we'll write a vector equation for this line.

r = (4)(-3) + t (3)(4) - (3)(4)
r = (4+3t-3)(-3+4t-4)
r = (1+3t)(-7+4t)

As the point '3,4' must be perpendicular from the straight line, we'll take out its cross product with the vector equation of the line and equate it to '0' in order to find the value of 't'.

(1+3t)(-7+4t) . (3)(4) = 0
3 + 9t - 28 + 16t = 0
-25 + 25t = 0
t = 1

Put this value of 't' back in the equation 'r = (1+3t)(-7+4t)'.

r = (1+3t)(-7+4t)
r = [(1 + 3(1)] [(-7 + 4(1)]
r = (4)(-3)

Use the formula '[(x)^2 + ( y )^2]^(1/2)' to find the distance.

[(x)^2 + ( y )^2]^(1/2)
[(4)^2 + (-3)^2]^(1/2)
(25)^(1/2)
5

Therefore, the distance is 5 units.
Click to expand...
Iz it for O Levels also?
 
ffaadyy

ffaadyy

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unique840 said:
q) The diagram shows the curve y = x sin x, for 0 ≤ x ≤ π. The point Q (pie/2) , (pie/2) lies on the curve.
(i) Show that the normal to the curve at Q passes through the point (π, 0).
i got the normal but how to prove that it passes through the point (pie,0)?
Click to expand...


I presume you'll simply be substituting the point 'π, 0' in the equation of the normal which you've found out and show that both the sides are equal. The equation of the normal to the curve is y + x = π. Now if you substitute the point (π, 0) in this equation, the final answer is 'π = π'.
 
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smartangel

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ffaadyy said:
This is how we'll do this question.


3x + 4y = 0

First of all, we'll be needing to find a set of coordinates on this line. We can find them by a fairly easy method.


3x + 4y = 0
4y = -3x
y = -3x/4


Keep 'x=4' in the above equation so that we get 'y=-3'. Now if we put these points back in the equation '3x + 4y = 0', the answer is '0' which proves that these points lie on the line. Similarly, we can put any two points in the equation '3x + 4y = 0' so that the final answer is equal to '0'. For e.g, we can also take the points 'x=1/3' and 'y=-1/4'. Next, we'll write a vector equation for this line.

r = (4)(-3) + t (3)(4) - (3)(4)
r = (4+3t-3)(-3+4t-4)
r = (1+3t)(-7+4t)

As the point '3,4' must be perpendicular from the straight line, we'll take out its cross product with the vector equation of the line and equate it to '0' in order to find the value of 't'.

(1+3t)(-7+4t) . (3)(4) = 0
3 + 9t - 28 + 16t = 0
-25 + 25t = 0
t = 1

Put this value of 't' back in the equation 'r = (1+3t)(-7+4t)'.

r = (1+3t)(-7+4t)
r = [(1 + 3(1)] [(-7 + 4(1)]
r = (4)(-3)

Use the formula '[(x)^2 + ( y )^2]^(1/2)' to find the distance.

[(x)^2 + ( y )^2]^(1/2)
[(4)^2 + (-3)^2]^(1/2)
(25)^(1/2)
5

Therefore, the distance is 5 units.
Click to expand...
thanks so much!!!!:D
 
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