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Vector Algebra question

2016 · 10 Apr · Shift 1 · Q32
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Vector Algebra question

2016 · 10 Apr · Shift 1 · Q32

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are a→,b→,c→\overrightarrow a ,\overrightarrow b ,\overrightarrow ca,b,c and a→+b→+c→4{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}4a+b+c​ respectively, then the position vector of the orthocentre of this triangle, is :
  1. A
    a→+b→+c→{\overrightarrow a + \overrightarrow b + \overrightarrow c }a+b+c
  2. B
    −(a→+b→+c→2)- \left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}} \right)−(2a+b+c​)
  3. C
    0→\overrightarrow 00
  4. D
    (a→+b→+c→2)\left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}} \right)(2a+b+c​)
View written solutionFree

Correct answer: D

  1. Use the standard vector relation for a triangle

For any triangle, if OOO is the circumcentre and HHH is the orthocentre, then with respect to any origin,

OH→=OA→+OB→+OC→.\overrightarrow{OH}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}.OH=OA+OB+OC.

In position vector form, if the position vectors of A,B,C,O,HA,B,C,O,HA,B,C,O,H are a⃗,b⃗,c⃗,h⃗\vec a,\vec b,\vec c,\vec ha,b,c,h respectively, then

h⃗=a⃗+b⃗+c⃗−2o⃗,\vec h = \vec a+\vec b+\vec c-2\vec o,h=a+b+c−2o,

because

OH→=h⃗−o⃗,\overrightarrow{OH}=\vec h-\vec o,OH=h−o,

and

OA→+OB→+OC→=(a⃗−o⃗)+(b⃗−o⃗)+(c⃗−o⃗).\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=(\vec a-\vec o)+(\vec b-\vec o)+(\vec c-\vec o).OA+OB+OC=(a−o)+(b−o)+(c−o).

So,

h⃗−o⃗=(a⃗+b⃗+c⃗)−3o⃗\vec h-\vec o=(\vec a+\vec b+\vec c)-3\vec oh−o=(a+b+c)−3o

which gives

h⃗=a⃗+b⃗+c⃗−2o⃗.\vec h=\vec a+\vec b+\vec c-2\vec o.h=a+b+c−2o.
  1. Substitute the circumcentre position vector

Given the circumcentre PPP has position vector

p⃗=a⃗+b⃗+c⃗4.\vec p=\frac{\vec a+\vec b+\vec c}{4}.p​=4a+b+c​.

Thus,

h⃗=a⃗+b⃗+c⃗−2p⃗.\vec h=\vec a+\vec b+\vec c-2\vec p.h=a+b+c−2p​.

Substitute p⃗\vec pp​:

h⃗=a⃗+b⃗+c⃗−2(a⃗+b⃗+c⃗4).\vec h=\vec a+\vec b+\vec c-2\left(\frac{\vec a+\vec b+\vec c}{4}\right).h=a+b+c−2(4a+b+c​).
  1. Simplify
h⃗=a⃗+b⃗+c⃗−a⃗+b⃗+c⃗2\vec h=\vec a+\vec b+\vec c-\frac{\vec a+\vec b+\vec c}{2}h=a+b+c−2a+b+c​ h⃗=a⃗+b⃗+c⃗2.\vec h=\frac{\vec a+\vec b+\vec c}{2}.h=2a+b+c​.
  1. Match with the options

This is exactly Option D:

a⃗+b⃗+c⃗2\boxed{\frac{\vec a+\vec b+\vec c}{2}}2a+b+c​​
  1. Comparison with stored answer

Stored correct answer: D

Our derived answer is also D, so they agree.

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