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1)

ABC is any triangle and G its centroid. Prove that if O is any point in space

a) OG=(OA + OB + OC)/3

b) a^2 + b^2 + c^2 + 9OG^2 = 3(OA^2 + OB^2 + OC^2) where a = |BC|etc

2)

a)write the eq of plane x-2y-2x=27 in r.n=d form.

b) what is dist of origin from plane and show that (6,-12, -12) is the point which is the reflection of the origin.

c) P(-3,2,1). what are the direction cosines of the reflection of the line OP in the plane

ABC is any triangle and G its centroid. Prove that if O is any point in space

a) OG=(OA + OB + OC)/3

b) a^2 + b^2 + c^2 + 9OG^2 = 3(OA^2 + OB^2 + OC^2) where a = |BC|etc

2)

a)write the eq of plane x-2y-2x=27 in r.n=d form.

b) what is dist of origin from plane and show that (6,-12, -12) is the point which is the reflection of the origin.

c) P(-3,2,1). what are the direction cosines of the reflection of the line OP in the plane

(1)

(a)

See http://www.netcomuk.co.uk/~jenolive/centroid.html

(b)

a^2 + b^2 + c^2 + 9|OG|^2

= |BC|^2 + |AC|^2 + |AB|^2 + 9|OG|^2

= |OC - OB|^2 + |OC - OA|^2 + |OB - OA|^2 + |OA + OB + OC|^2

= (|OC|^2 - 2OC.OB + |OB|^2)

+ (|OC|^2 - 2OC.OA + |OA|^2)

+ (|OB|^2 - 2OB.OA + |OA|^2)

+ (|OA|^2 + |OB|^2 + |OC|^2 + 2OA.OB + 2OA.OC + 2OB.OC)

= 3(|OA|^2 + |OB|^2 + |OC|^2)

(2)

(a) r.n = 27, where n = (1, -2, -2).

(b) The shortest path from the origin to the plane is along the line r = sn (where s is a scalar parameter). That line intersects the plane where s * n.n = 27; ie, where s = 27/9 = 3; ie, at 3n = (3, -6, -6). The distance between the origin and the plane is |3n| = 9. The origin reflects in the plane to O' = 6n = (6, -12, -12).

(c) The perpendicular path from P to the plane is along the line r = sn + (-3, 2, 1). That line intersects the plane where s * n.n + (-3, 2, 1).n = 27; ie, where s = 36/9 = 4; ie, at 4n + (-3, 2, 1). P reflects in the plane to P' = 8n + (-3, 2, 1). Then O'P' = (8n + (-3, 2, 1)) - 6n = 2n + (-3, 2, 1) = (2, -4, -4) + (-3, 2, 1) = (-1, -2, -3). [Check: this has the same length as (-3, 2, 1).] Then O'P'/|O'P'| = (-1, -2, -3)/sqrt(14), so the direction cosines of O'P' are -1/sqrt(14), -2/sqrt(14) and -3/sqrt(14).

(a)

See http://www.netcomuk.co.uk/~jenolive/centroid.html

(b)

a^2 + b^2 + c^2 + 9|OG|^2

= |BC|^2 + |AC|^2 + |AB|^2 + 9|OG|^2

= |OC - OB|^2 + |OC - OA|^2 + |OB - OA|^2 + |OA + OB + OC|^2

= (|OC|^2 - 2OC.OB + |OB|^2)

+ (|OC|^2 - 2OC.OA + |OA|^2)

+ (|OB|^2 - 2OB.OA + |OA|^2)

+ (|OA|^2 + |OB|^2 + |OC|^2 + 2OA.OB + 2OA.OC + 2OB.OC)

= 3(|OA|^2 + |OB|^2 + |OC|^2)

(2)

(a) r.n = 27, where n = (1, -2, -2).

(b) The shortest path from the origin to the plane is along the line r = sn (where s is a scalar parameter). That line intersects the plane where s * n.n = 27; ie, where s = 27/9 = 3; ie, at 3n = (3, -6, -6). The distance between the origin and the plane is |3n| = 9. The origin reflects in the plane to O' = 6n = (6, -12, -12).

(c) The perpendicular path from P to the plane is along the line r = sn + (-3, 2, 1). That line intersects the plane where s * n.n + (-3, 2, 1).n = 27; ie, where s = 36/9 = 4; ie, at 4n + (-3, 2, 1). P reflects in the plane to P' = 8n + (-3, 2, 1). Then O'P' = (8n + (-3, 2, 1)) - 6n = 2n + (-3, 2, 1) = (2, -4, -4) + (-3, 2, 1) = (-1, -2, -3). [Check: this has the same length as (-3, 2, 1).] Then O'P'/|O'P'| = (-1, -2, -3)/sqrt(14), so the direction cosines of O'P' are -1/sqrt(14), -2/sqrt(14) and -3/sqrt(14).

1)

a) G is the centroid of the triangle ABC and G is a pont on the line CM such that the length GM is 1/3 of the length CM. M is the mid-point of the line AB.

OG=(OA + OB + OC)/3

3OG - OA - OB - OC = 0

(OG - OA) + (OG - OB) + (OG - OC) = 0

(AO + OG) + (BO + OG) + (CO + OG) = 0

AG + BG + CG = 0

2MG + CG = 0 (AG + BG =2MG)

now |MG| is 1/3 |CM| or,

|MG| = ½|CG|

and MG is in opposite sign to CG

i.e. CG = 2GM

:. 0 + 0 = 0

========

b) a^2 + b^2 + c^2 + 9OG^2

a² = |BC|BC| = BC.BC

b² = |AC||AC| = AC.AC

c² = |AB||AB| = AB.AB

9OG² = OA² + OB² + OC² + 2OA.OB + 2OA.OC + 2OB.OC

a² + b² + c² + 9OG² = BC.BC + AC.AC + AB.AB + OA² + OB² + OC² + 2OA.OB + 2OA.OC + 2OB.OC

= OA² + OB² + OC² + BC.BC + AC.AC + AB.AB + 2OA.OB + 2OA.OC + 2OB.OC ---------------------------------(1)

dealing only with,

BC.BC + AC.AC + AB.AB + 2OA.OB + 2OA.OC + 2OB.OC ------------------------------------------------(2)

Some of the vector triangles,

OB = OA + AB -> AB = OB - OA

AC = AO + OC

AC = AB + BC -> BC = AC - AB

substituting for AB, AC, BC in (2),

(AC - AB).(AC - AB) + (AO + OC).(AO + OC) + (OB - OA).(OB - OA) + 2OA.OB + 2OA.OC + 2OB.OC

AC² -2AC.AB + AB² + AO² + 2AO.OC + OC² + OB² - 2OB.OA + OA² + 2OA.OB + 2OA.OC + 2OB.OC

cancelling, - 2OB.OA with + 2OA.OB, + 2AO.OC with + 2OA.OC gives,

2OA² +OB² + OC² + AC² - 2AC.AB + AB² + 2OB.OC

dealing only with,

AC² - 2AC.AB + AB² + 2OB.OC

AC(AC - AB) + AB(AB - AC) + 2OB.OC

AC(BA + AC) + AB(CA + AB) +2OB.OC

AC.BC + AB.CB + 2OB.OC

BC(AC - AB) + 2OB.OC

BC(BA + AC) + 2OB.OC

BC² + 2OB.OC

(BO + OC)² + 2OB.OC

BO² + 2BO.OC + OC² + 2OB.OC

OB² + OC² (2BO.OC cancels with 2OB.OC) ----------------------------------(3)

substituting for (3) in (2) gives

2OA² +OB² + OC² + OB² + OC²

2OA² + 2OB² + 2OC² ------------------------------------------------------------------(4)

substituting for (4) in (1) gives,

OA² + OB² + OC² + 2OA² + 2OB² + 2OC²

3OA² + 3OB² + 3OC²

3(OA² + OB² + OC²)

===============

a) G is the centroid of the triangle ABC and G is a pont on the line CM such that the length GM is 1/3 of the length CM. M is the mid-point of the line AB.

OG=(OA + OB + OC)/3

3OG - OA - OB - OC = 0

(OG - OA) + (OG - OB) + (OG - OC) = 0

(AO + OG) + (BO + OG) + (CO + OG) = 0

AG + BG + CG = 0

2MG + CG = 0 (AG + BG =2MG)

now |MG| is 1/3 |CM| or,

|MG| = ½|CG|

and MG is in opposite sign to CG

i.e. CG = 2GM

:. 0 + 0 = 0

========

b) a^2 + b^2 + c^2 + 9OG^2

a² = |BC|BC| = BC.BC

b² = |AC||AC| = AC.AC

c² = |AB||AB| = AB.AB

9OG² = OA² + OB² + OC² + 2OA.OB + 2OA.OC + 2OB.OC

a² + b² + c² + 9OG² = BC.BC + AC.AC + AB.AB + OA² + OB² + OC² + 2OA.OB + 2OA.OC + 2OB.OC

= OA² + OB² + OC² + BC.BC + AC.AC + AB.AB + 2OA.OB + 2OA.OC + 2OB.OC ---------------------------------(1)

dealing only with,

BC.BC + AC.AC + AB.AB + 2OA.OB + 2OA.OC + 2OB.OC ------------------------------------------------(2)

Some of the vector triangles,

OB = OA + AB -> AB = OB - OA

AC = AO + OC

AC = AB + BC -> BC = AC - AB

substituting for AB, AC, BC in (2),

(AC - AB).(AC - AB) + (AO + OC).(AO + OC) + (OB - OA).(OB - OA) + 2OA.OB + 2OA.OC + 2OB.OC

AC² -2AC.AB + AB² + AO² + 2AO.OC + OC² + OB² - 2OB.OA + OA² + 2OA.OB + 2OA.OC + 2OB.OC

cancelling, - 2OB.OA with + 2OA.OB, + 2AO.OC with + 2OA.OC gives,

2OA² +OB² + OC² + AC² - 2AC.AB + AB² + 2OB.OC

dealing only with,

AC² - 2AC.AB + AB² + 2OB.OC

AC(AC - AB) + AB(AB - AC) + 2OB.OC

AC(BA + AC) + AB(CA + AB) +2OB.OC

AC.BC + AB.CB + 2OB.OC

BC(AC - AB) + 2OB.OC

BC(BA + AC) + 2OB.OC

BC² + 2OB.OC

(BO + OC)² + 2OB.OC

BO² + 2BO.OC + OC² + 2OB.OC

OB² + OC² (2BO.OC cancels with 2OB.OC) ----------------------------------(3)

substituting for (3) in (2) gives

2OA² +OB² + OC² + OB² + OC²

2OA² + 2OB² + 2OC² ------------------------------------------------------------------(4)

substituting for (4) in (1) gives,

OA² + OB² + OC² + 2OA² + 2OB² + 2OC²

3OA² + 3OB² + 3OC²

3(OA² + OB² + OC²)

===============

- Mechanics 1 Help Please - Position vectors
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- Aqa A level physics Electric field question (1)
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