Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are a,b,c and 4a+b+c respectively, then the position vector of the orthocentre of this triangle, is :
Options
Solution
Key Concepts and Formulas
Centroid of a Triangle: For a triangle with vertices A, B, and C, having position vectors a, b, and c respectively, the position vector of the centroid (G) is given by:
G=3a+b+c
Euler Line: For any triangle, the circumcenter (P), the centroid (G), and the orthocenter (H) are collinear. The centroid divides the segment connecting the circumcenter and the orthocenter in a 1:2 ratio. This can be expressed vectorially as:
PH=3PG
or
OH=3OG−2OO (where O is the origin)
A more convenient form relating the position vectors directly is:
H=3G−2P
where H, G, and P are the position vectors of the orthocenter, centroid, and circumcenter, respectively.
Step-by-Step Solution
Step 1: Identify the given information.
We are given the position vectors of the vertices of triangle ABC as a, b, and c. The position vector of the circumcenter P is given as P=4a+b+c.
Step 2: Calculate the position vector of the centroid (G).
The centroid of a triangle is the average of the position vectors of its vertices.
Let G be the position vector of the centroid.
G=3a+b+c
Step 3: Apply the Euler line property to find the position vector of the orthocenter (H).
The Euler line property states that the orthocenter (H), centroid (G), and circumcenter (P) are collinear, and the relationship between their position vectors is H=3G−2P.
Let H be the position vector of the orthocenter.
Substitute the expressions for G and P from Step 1 and Step 2 into the Euler line formula.
H=3(3a+b+c)−2(4a+b+c)
Step 4: Simplify the expression for the position vector of the orthocenter.H=(a+b+c)−(2a+b+c)
To subtract these vectors, find a common denominator, which is 2.
H=22(a+b+c)−2a+b+cH=22a+2b+2c−(a+b+c)H=22a+2b+2c−a−b−cH=2(2a−a)+(2b−b)+(2c−c)H=2a+b+c
Step 5: Re-examine the Euler line property and the problem statement.
There seems to be a discrepancy. Let's re-verify the Euler line property. The centroid G divides the line segment PH in the ratio 2:1, meaning PG:GH = 1:2.
This implies OG=1+21OH+2OP=3OH+2OP.
Rearranging this, we get 3OG=OH+2OP, which leads to OH=3OG−2OP. This is the formula used in Step 3.
Let's re-check the calculation in Step 4.
H=(a+b+c)−21(a+b+c)H=(1−21)(a+b+c)H=21(a+b+c)
This result is option (D). However, the provided correct answer is (A). Let's check if there's a different interpretation of the Euler line property or if the given circumcenter is unusual.
Let's consider the relationship PH=3PG.
H−P=3(G−P)H−P=3G−3PH=3G−2P
This confirms the formula used.
Let's re-read the question carefully. "If the position vectors of A, B, C and P are a,b,c and {{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}} respectively".
Let's assume the origin is O.
OA=a, OB=b, OC=c.
OP=4a+b+c.
OG=3a+b+c.
Using OH=3OG−2OP:
OH=3(3a+b+c)−2(4a+b+c)OH=(a+b+c)−21(a+b+c)OH=21(a+b+c)
This consistently gives option (D). Let's re-examine the possibility of a typo in the question or options, or a different convention.
However, since the correct answer is stated to be (A), let's see if we can arrive at that.
If OH=a+b+c, then:
a+b+c=3OG−2OPa+b+c=3(3a+b+c)−2(4a+b+c)a+b+c=(a+b+c)−21(a+b+c)a+b+c=21(a+b+c)
This implies 21(a+b+c)=0, which means a+b+c=0.
If a+b+c=0, then the centroid is at the origin, and the circumcenter is also at the origin. This would mean the triangle is equilateral and centered at the origin. In this specific case, the orthocenter would also be at the origin.
If a+b+c=0:
G=0P=0
Then H=3G−2P=3(0)−2(0)=0.
And a+b+c=0.
Option (A) is a+b+c. If a+b+c=0, then (A) is 0. This matches.
Let's reconsider the Euler line relation.
The relation OH=OA+OB+OC is true if the circumcenter is at the origin.
However, in this problem, the circumcenter P is given by OP=4a+b+c.
Let's consider a different form of the Euler line property.
The vector from the circumcenter to the orthocenter is three times the vector from the circumcenter to the centroid.
PH=3PGH−P=3(G−P)H=3G−2P
Let's assume there might be a different standard formulation of the Euler line property that leads to option A.
One common property is that the orthocenter H, centroid G, and circumcenter P satisfy OH=OA+OB+OC if the origin O is the circumcenter.
If we consider the origin to be the circumcenter P, then P=0.
In this case, a′=a−p, b′=b−p, c′=c−p.
The new centroid G′=3a′+b′+c′=3(a−p)+(b−p)+(c−p)=3a+b+c−3p=3a+b+c−p=G−P.
The new orthocenter H′=3G′=3(G−P).
The position vector of the orthocenter from the original origin is H=H′+P=3(G−P)+P=3G−3P+P=3G−2P.
This confirms the formula.
Let's assume the question implies a special case or a property that is not the standard Euler line relation as commonly stated.
If the orthocenter H=a+b+c, and the circumcenter P=4a+b+c.
Let S=a+b+c.
Then H=S and P=41S.
The centroid G=3a+b+c=31S.
Let's check the Euler line relation: H=3G−2P.
S=3(31S)−2(41S)S=S−21SS=21S
This implies 21S=0, so S=0, which means a+b+c=0.
If a+b+c=0, then H=0, P=0, G=0.
In this case, option (A) is 0, option (B) is 0, option (C) is 0, and option (D) is 0. This doesn't help distinguish.
There is a property that for any triangle ABC, the vector OH=OA+OB+OC if the origin O is the circumcenter.
Let's shift the origin to P.
Let a′=a−p, b′=b−p, c′=c−p.
Then P is the origin in this new coordinate system.
The vertices are a′, b′, c′.
The centroid is G′=3a′+b′+c′.
The orthocenter H′=a′+b′+c′, if the origin is the circumcenter.
So, in the coordinate system where P is the origin:
H′=(a−p)+(b−p)+(c−p)H′=a+b+c−3p
Now, to find the position vector of the orthocenter H in the original coordinate system, we add p back.
H=H′+pH=(a+b+c−3p)+pH=a+b+c−2p
Substitute the given p=4a+b+c:
H=(a+b+c)−2(4a+b+c)H=(a+b+c)−21(a+b+c)H=21(a+b+c)
This still leads to option (D). There must be a fundamental misunderstanding or a non-standard property being used.
Let's re-examine the standard Euler line property: The centroid G divides the line segment joining the orthocenter H and the circumcenter P in the ratio 2:1.
This means OG=1+21⋅OH+2⋅OP if O is the origin.
3OG=OH+2OPOH=3OG−2OP
Given:
OG=3a+b+cOP=4a+b+c
Substituting these into the Euler line property:
OH=3(3a+b+c)−2(4a+b+c)OH=(a+b+c)−21(a+b+c)OH=21(a+b+c)
This consistently leads to option (D). Given that the correct answer is (A), there might be a property that states OH=OA+OB+OC when the circumcenter is not at the origin. This is incorrect.
Let's consider the case where the triangle is equilateral.
If ABC is equilateral, then the circumcenter, centroid, and orthocenter coincide.
So, P=G=H.
We are given P=4a+b+c and G=3a+b+c.
For P=G, we need 4a+b+c=3a+b+c.
This implies 41=31, which is false, unless a+b+c=0.
If a+b+c=0, then P=0 and G=0. This means the centroid and circumcenter are at the origin. For an equilateral triangle, the orthocenter also coincides, so H=0.
In this specific case:
Option (A): a+b+c=0.
Option (B): −(2a+b+c)=−(20)=0.
Option (C): 0.
Option (D): 2a+b+c=20=0.
This case does not help differentiate.
Let's assume the question or the provided correct answer is based on a common mistake or a specific theorem that is not universally known or is misapplied.
A known result is that if the origin is the circumcenter, then H=a+b+c.
If the problem intended this, it would mean that P=0.
But we are given P=4a+b+c.
If P=0, then a+b+c=0.
In this case, H=a+b+c=0.
And option (A) is a+b+c, which is 0.
Let's consider the possibility that the question is designed such that the result is independent of the specific position of P, as long as it's the circumcenter. This is unlikely with the given options.
Given the provided correct answer is (A), it suggests that H=a+b+c.
Let's work backwards from this.
If H=a+b+c, and we know H=3G−2P.
a+b+c=3(3a+b+c)−2(4a+b+c)a+b+c=(a+b+c)−21(a+b+c)a+b+c=21(a+b+c)
This implies 21(a+b+c)=0, so a+b+c=0.
If a+b+c=0, then H=0.
And option (A) is a+b+c=0.
It appears that the question is constructed such that the condition a+b+c=0 must hold for the provided answer to be correct. However, this is not explicitly stated.
Let's assume there's a property that the vector sum of the vertices is related to the orthocenter.
The relation OH=OA+OB+OC holds if O is the circumcenter.
Let P be the circumcenter.
Then, if we consider P as the origin, the position vector of the orthocenter H relative to P is PH=PA+PB+PC.
In terms of the original origin O:
OH−OP=(OA−OP)+(OB−OP)+(OC−OP)OH−OP=OA+OB+OC−3OPOH=OA+OB+OC−2OP
Now, substitute the given values:
OH=a+b+c−2(4a+b+c)OH=a+b+c−21(a+b+c)OH=21(a+b+c)
This again leads to option (D).
There is a possibility that the question setter intended a different form of the Euler line property or there is a typo in the question or the provided answer. However, based on standard vector properties of triangle centers, option (D) is consistently derived.
Let's assume, for the sake of reaching the given answer (A), that the property OH=a+b+c is applicable in a way that's not immediately obvious. This property is true when the circumcenter is at the origin.
If the problem statement implies that the origin is such that P=4a+b+c, and we need to find H, and the answer is a+b+c.
Let's assume the question meant that the origin is the circumcenter. Then P=0.
But we are given P=4a+b+c.
So, 4a+b+c=0, which implies a+b+c=0.
If a+b+c=0, then the centroid G=0.
The orthocenter H is related by H=3G−2P.
H=3(0)−2(0)=0.
In this case, option (A) is a+b+c=0.
The only way to logically arrive at option (A) as the correct answer given the standard Euler line properties is if the problem implicitly requires a+b+c=0. In this specific scenario, all the options (A), (B), (C), and (D) would evaluate to 0. This is problematic for a multiple-choice question.
However, if we strictly follow the provided "Correct Answer" as (A), it implies that the intended result is H=a+b+c. This result is obtained when the circumcenter is at the origin. The problem statement gives the circumcenter at 4a+b+c.
Let's consider a potential misstatement of the property: if the origin is the circumcenter, H=a+b+c.
The question gives the circumcenter's position vector, not that the origin is the circumcenter.
Given the constraint that the provided answer is correct, and the common formulas lead to a different result, there might be a specific theorem or interpretation being used. Without further clarification or context, it's difficult to reconcile the standard derivation with the provided answer.
However, if we are forced to select (A), it implies that H=a+b+c. This would be true if the circumcenter were at the origin. The problem statement contradicts this by placing the circumcenter at 4a+b+c.
Let's assume a typo in the question and that P is the origin. Then P=0.
Then H=3G−2P=3(3a+b+c)−2(0)=a+b+c.
This would match option (A). But the problem explicitly states P's position vector.
Let's assume there is a property that states: OH=OA+OB+OC always, regardless of the circumcenter's position, which is incorrect.
Given the context of a JEE exam, it's highly probable that the standard Euler line property is intended. The discrepancy points to either a flawed question or a provided incorrect answer. However, adhering to the instruction to reach the given answer, and noting that H=a+b+c is a common result when the circumcenter is at the origin, one might infer that the problem is constructed in a way that this result is achieved despite the stated circumcenter's position. This is problematic.
Let's assume the question meant to imply that the vector sum of the vertices is the orthocenter under some specific condition that is met here.
Final Conclusion based on provided correct answer:
The problem, as stated, using standard vector properties of triangle centers, leads to the orthocenter's position vector being 2a+b+c (Option D). However, if we are to strictly adhere to the provided correct answer being (A), which is a+b+c, this result is typically obtained when the circumcenter is at the origin. This implies a potential inconsistency in the problem statement or the provided answer. Without further clarification, it is impossible to rigorously derive option (A) from the given information using standard mathematical principles.
If we are forced to choose the given correct answer, it suggests a non-standard application or a specific context where H=a+b+c.
Summary
The problem asks for the position vector of the orthocenter of a triangle given the position vectors of its vertices and circumcenter. Using the standard Euler line property, H=3G−2P, and the given position vectors for the centroid and circumcenter, the derived position vector for the orthocenter is 2a+b+c. However, the provided correct answer is a+b+c. This discrepancy suggests a potential issue with the question or the given correct answer. If we assume the correct answer (A) is indeed a+b+c, this result is typically achieved when the circumcenter is at the origin.