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its A2 vectors!A2 or As vectors?
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its A2 vectors!A2 or As vectors?
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!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.
how can u do this ques when the vector equation is without a direction vector. A2 vectors always have direction vectorsits A2 vectors!
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
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.
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.
kia baat hai apki... hum ques ko galat keh rahe hain n apne to solve bhi krdiaThis 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 + ^2]^(1/2)' to find the distance.
[(x)^2 + ^2]^(1/2)
[(4)^2 + (-3)^2]^(1/2)
(25)^(1/2)
5
Therefore, the distance is 5 units.
kia baat hai apki... hum ques ko galat keh rahe hain n apne to solve bhi krdia
how can u do this ques when the vector equation is without a direction vector. A2 vectors always have direction vectors
yar itna kon sochta haiFrom 3x + 4y = 0, the direction vector is (3)(4).
Iz it for O Levels also?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.
Iz it for O Levels also?
Broo! Plz gimme ur email id !The formula used to find the distance is indeed from O Levels.
once again i request you to compile p1 and s1 also,pleaseeeeeCompiled P3 Attached .....
Compiled P4 (M1) Attached...
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)?
thanks so much!!!!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.
Could u help me with some CIE Pure math and statistics a2 concepts?
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