Mathematics: Post your doubts here!

[smartangel]

A2 or As vectors?

its A2 vectors!

 

[smartangel]

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.

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

 

[unique840]

its A2 vectors!

how can u do this ques when the vector equation is without a direction vector. A2 vectors always have direction vectors

 

[MHHaider707]

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

Hey how did u do that?

 

[ffaadyy]

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.

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.

 

[unique840]

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 + ^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

 

[ffaadyy]

kia baat hai apki... hum ques ko galat keh rahe hain n apne to solve bhi krdia

 

[ffaadyy]

how can u do this ques when the vector equation is without a direction vector. A2 vectors always have direction vectors

From 3x + 4y = 0, the direction vector is (3)(4).

 

[unique840]

From 3x + 4y = 0, the direction vector is (3)(4).

yar itna kon sochta hai

 

[unique840]

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)?

 

[Jaf]

Oops sorry. :/

 

[Spunkyfarhan]

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?

 

[ffaadyy]

Iz it for O Levels also?

The formula used to find the distance is indeed from O Levels.

 

[Spunkyfarhan]

The formula used to find the distance is indeed from O Levels.

Broo! Plz gimme ur email id !

 

[Chaitanya Agrawal]

Compiled P3 Attached .....

Compiled P4 (M1) Attached...

once again i request you to compile p1 and s1 also,pleaseeeee

 

[ffaadyy]

Broo! Plz gimme ur email id !

[email protected]

 

[ffaadyy]

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)?

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'.

 

[smartangel]

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.

thanks so much!!!!

 

[Zephyro]

Can anyone here help me understand iteration?

 

[Zephyro]

This is how you'll do q7.

View attachment 6397

View attachment 6398

View attachment 6399

Could u help me with some CIE Pure math and statistics a2 concepts?