Another P3 question Watch

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lexazver203
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The point A(24,6,0), B(30,12,12), C(18,6,36) are refered to cartesian axes, origin O.

(a) Find a vector equation for the line passing through the points A and B. (easy done)

The pint P lieso n the line passing though A and B
(b)show that CP (with an arrow on top going from c to p) can be expressed as
(6+t)i+tj+(2t-36)k where t is a parameter
In the end i got the answer but not sure how i got there.

(c) Given that CP is perpendicular to AB, find the coordinates of P.
I generally struggle on this type of questions i.e. where u have to find a point on the line. so i cannot do it.

(d) Hence, or otherwise, find the area of the triangle ABC, giving your answer to 3 significant figures. No idea hotw to tackle it.
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IntegralAnomaly
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(Original post by lexazver203)
The point A(24,6,0), B(30,12,12), C(18,6,36) are refered to cartesian axes, origin O.

(a) Find a vector equation for the line passing through the points A and B. (easy done)

The pint P lieso n the line passing though A and B
(b)show that CP (with an arrow on top going from c to p) can be expressed as
(6+t)i+tj+(2t-36)k where t is a parameter
In the end i got the answer but not sure how i got there.

(c) Given that CP is perpendicular to AB, find the coordinates of P.
I generally struggle on this type of questions i.e. where u have to find a point on the line. so i cannot do it.

(d) Hence, or otherwise, find the area of the triangle ABC, giving your answer to 3 significant figures. No idea hotw to tackle it.
u've found the line in the form of r=ai+bj+ck+t(di+ej+fk) right?Now express this is r=(a+td)i+(b+et)j+(c+ft)k.Now since 0P lies ANYWERE on this line then this is also the position vector of 0P,u know the position vector of C so using CP=CO+OP u can find CP in terms of t.
If CP is perpendicular to AB then u know that CP.AB=0,from this u can find the value of t and hence the position vector of CP.But CP=CO+OP,so now u can find OP.
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IntegralAnomaly
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for the last part:
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lexazver203
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(Original post by IntegralAnomaly)
u've found the line in the form of r=ai+bj+ck+t(di+ej+fk) right?Now express this is r=(a+td)i+(b+et)j+(c+ft)k.Now since 0P lies ANYWERE on this line then this is also the position vector of 0P,u know the position vector of C so using CP=CO+OP u can find CP in terms of t.
If CP is perpendicular to AB then u know that CP.AB=0,from this u can find the value of t and hence the position vector of CP.But CP=CO+OP,so now u can find OP.
How would you go on multiplying CP and AB i mean which parts do u use and which u dont ? Im not very clear on the whole vector topic.
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MalaysianDude
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(Original post by lexazver203)
The point A(24,6,0), B(30,12,12), C(18,6,36) are refered to cartesian axes, origin O.

(a) Find a vector equation for the line passing through the points A and B. (easy done)

The pint P lieso n the line passing though A and B
(b)show that CP (with an arrow on top going from c to p) can be expressed as
(6+t)i+tj+(2t-36)k where t is a parameter
In the end i got the answer but not sure how i got there.

(c) Given that CP is perpendicular to AB, find the coordinates of P.
I generally struggle on this type of questions i.e. where u have to find a point on the line. so i cannot do it.

(d) Hence, or otherwise, find the area of the triangle ABC, giving your answer to 3 significant figures. No idea hotw to tackle it.
a)Equation
24i + 6j + k(i+j+2k)

b)
take k as t
thus
=t(i+j+2k) + 24i + 6j
=(t+24)i + (6+t)j + 2tk
thus
CP = (t+24)i + (6+t)j + 2tk - (18i+6j+36k)
CP = (6+t)i+ tj + (2t-36)k

c)
CP.AB = 0
(6+t)i+ tj + (2t-36)k . (6i + 6j + 12k) = 0
you`ll get
36 + 6t + 6t + 24t - 432 = 0
t = 11

put t in the equation of p
thus,coordinates
P(35,17,22)

d)find angle between A and B
A.B = |A||B|CosX
192 = 24.73 x 65.73 x Cos n
Cos n = 0.1181
n= 83.22
then just use formula
A = 1/2 (abSinC)
A = 1/2 (24.73 x 65.73 x Sin 83.22)
A = 807.1 unit²

There ya go..
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IntegralAnomaly
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(Original post by lexazver203)
How would you go on multiplying CP and AB i mean which parts do u use and which u dont ? Im not very clear on the whole vector topic.
Not multiply,dot them.Sorry if i did'nt make it clear.
Use AB.AC=|AB||AC|cosθ,but because the two lines are perpendicular then θ=90° and so Cosθ=0 → AB.AC=0.
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lexazver203
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(Original post by MalaysianDude)
a)Equation
24i + 6j + k(i+j+2k)

b)
take k as t
thus
=t(i+j+2k) + 24i + 6j
=(t+24)i + (6+t)j + 2tk
thus
CP = (t+24)i + (6+t)j + 2tk - (18i+6j+36k)
CP = (6+t)i+ tj + (2t-36)k

c)
CP.AB = 0
(6+t)i+ tj + (2t-36)k . (6i + 6j + 12k) = 0
you`ll get
36 + 6t + 6t + 24t - 432 = 0
t = 11

put t in the equation of p
thus,coordinates
P(35,17,22)

d)find angle between A and B
then just use formula
A = 1/2 (abSinC)
you can get it
too lazy to do it


Thanx a lot for that just a quick question where it says (6+t)i+ tj + (2t-36)k . (6i + 6j + 12k) = 0 do you just use the second part of the equation of the line when doind a.b???
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MalaysianDude
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i just edited my answers
i hope i got the angle right though..
hehehe
go check out for the answers above there

and yes i think i just used the second part
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