Answer
396.9k+ views
Hint: Here we need to find the following relation between the given vectors. We will first draw the triangle based on the given information. We will consider the LHS of the given equation and simplify it as per the diagram. We will then use the relation between orthocenter, centroid and the circumcenter and form an equation. We will simplify it further and prove \[\overline {QA} + \overline {QB} + \overline {QC} = \overline {QP} \].
Complete step-by-step answer:
First, we will draw the triangle with orthocenter, centroid and the circumcenter. Let \[C'\] be the centroid of the triangle.
Here we need to prove \[\overline {QA} + \overline {QB} + \overline {QC} = \overline {QP} \].
Now, we will consider the left hand side of the given equation.
We can write these vectors as
\[\overline {QA} + \overline {QB} + \overline {QC} = \overline A - \overline Q + \overline B - \overline Q + \overline C - \overline Q \]
On adding and subtracting the like vectors, we get
$\Rightarrow$ \[\overline {QA} + \overline {QB} + \overline {QC} = \overline A + \overline B + \overline C - 3\overline Q \]
Now, multiplying and dividing by 3, we get
$\Rightarrow$ \[\overline {QA} + \overline {QB} + \overline {QC} = 3\left( {\dfrac{{\overline A + \overline B + \overline C }}{3} - \overline Q } \right)\]
We know the relation \[\dfrac{{\overline A + \overline B + \overline C }}{3} = \overline {C'} \] because \[P\] is the orthocenter.
On substituting this value here, we get
$\Rightarrow$ \[\overline {QA} + \overline {QB} + \overline {QC} = 3\left( {\overline {C'} - \overline Q } \right)\]
We can write these vectors as
$\Rightarrow$ \[\overline {QA} + \overline {QB} + \overline {QC} = 3\overline {QC'} \] ……… \[\left( 1 \right)\]
We know that the centroid, orthocenter and circumcenter of a triangle are always collinear and centroid divides the orthocenter and circumcenter in \[1:2\] .
From the diagram we can write
\[\overline {C'} = \dfrac{{2\overline Q + \overline P }}{3}\]
On cross multiplying the terms, we get
\[ \Rightarrow 3\overline {C'} = 2\overline Q + \overline P \]
On further simplification, we get
\[ \Rightarrow 3\overline {C'} - 2\overline Q = \overline P \]
Now, we will subtract \[\overline Q \] from both sides. Therefore, we get
\[ \Rightarrow 3\overline {C'} - 2\overline Q - \overline Q = \overline P - \overline Q \]
On subtracting the like terms, we get
\[ \Rightarrow 3\left( {\overline {C'} - \overline Q } \right) = \overline P - \overline Q \]
On further simplification, we get
\[ \Rightarrow 3\overline {QC'} = \overline {QP} \]
Now, we will substitute this value in equation \[\left( 1 \right)\], therefore, we get
\[ \Rightarrow \overline {QA} + \overline {QB} + \overline {QC} = \overline {QP} \]
Hence, we have proved the following relation between the vectors.
Note: We need to remember that the centroid, orthocenter and circumcenter of a triangle are always collinear i.e. they are on the same straight line. Orthocenter is defined as a point that is formed by the intersection of all three altitudes of the triangle and circumcenter of the triangle is defined as the point which is formed when we draw a circumcircle of the triangle.
Complete step-by-step answer:
First, we will draw the triangle with orthocenter, centroid and the circumcenter. Let \[C'\] be the centroid of the triangle.
![seo images](https://www.vedantu.com/question-sets/2e1d25e4-88cf-4063-826c-01d3c81d9e345086982535334700728.png)
Here we need to prove \[\overline {QA} + \overline {QB} + \overline {QC} = \overline {QP} \].
Now, we will consider the left hand side of the given equation.
We can write these vectors as
\[\overline {QA} + \overline {QB} + \overline {QC} = \overline A - \overline Q + \overline B - \overline Q + \overline C - \overline Q \]
On adding and subtracting the like vectors, we get
$\Rightarrow$ \[\overline {QA} + \overline {QB} + \overline {QC} = \overline A + \overline B + \overline C - 3\overline Q \]
Now, multiplying and dividing by 3, we get
$\Rightarrow$ \[\overline {QA} + \overline {QB} + \overline {QC} = 3\left( {\dfrac{{\overline A + \overline B + \overline C }}{3} - \overline Q } \right)\]
We know the relation \[\dfrac{{\overline A + \overline B + \overline C }}{3} = \overline {C'} \] because \[P\] is the orthocenter.
On substituting this value here, we get
$\Rightarrow$ \[\overline {QA} + \overline {QB} + \overline {QC} = 3\left( {\overline {C'} - \overline Q } \right)\]
We can write these vectors as
$\Rightarrow$ \[\overline {QA} + \overline {QB} + \overline {QC} = 3\overline {QC'} \] ……… \[\left( 1 \right)\]
We know that the centroid, orthocenter and circumcenter of a triangle are always collinear and centroid divides the orthocenter and circumcenter in \[1:2\] .
![seo images](https://www.vedantu.com/question-sets/0ad85773-81f1-4ca0-9de7-ebfa92ab0f7a7873413634411403631.png)
From the diagram we can write
\[\overline {C'} = \dfrac{{2\overline Q + \overline P }}{3}\]
On cross multiplying the terms, we get
\[ \Rightarrow 3\overline {C'} = 2\overline Q + \overline P \]
On further simplification, we get
\[ \Rightarrow 3\overline {C'} - 2\overline Q = \overline P \]
Now, we will subtract \[\overline Q \] from both sides. Therefore, we get
\[ \Rightarrow 3\overline {C'} - 2\overline Q - \overline Q = \overline P - \overline Q \]
On subtracting the like terms, we get
\[ \Rightarrow 3\left( {\overline {C'} - \overline Q } \right) = \overline P - \overline Q \]
On further simplification, we get
\[ \Rightarrow 3\overline {QC'} = \overline {QP} \]
Now, we will substitute this value in equation \[\left( 1 \right)\], therefore, we get
\[ \Rightarrow \overline {QA} + \overline {QB} + \overline {QC} = \overline {QP} \]
Hence, we have proved the following relation between the vectors.
Note: We need to remember that the centroid, orthocenter and circumcenter of a triangle are always collinear i.e. they are on the same straight line. Orthocenter is defined as a point that is formed by the intersection of all three altitudes of the triangle and circumcenter of the triangle is defined as the point which is formed when we draw a circumcircle of the triangle.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)