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If the position vectors of the vertices A, B and C of a ΔABC are respectively 4\hat{i}+7\hat{j}+8\hat{k},\: 2\hat{i}+3\hat{j}+4\hat{k} and 2\hat{i}+5\hat{j}+7\hat{k} then the position vector of the point, where the bisector of ∠A meets BC is :

  • Option 1)

    \frac{1}{2}\left ( 4\hat{i}+8\hat{j}+11\hat{k} \right )

  • Option 2)

    \frac{1}{3}\left ( 6\hat{i}+11\hat{j}+15\hat{k} \right )

  • Option 3)

    \frac{1}{3}\left ( 6\hat{i}+13\hat{j}+18\hat{k} \right )

  • Option 4)

    \frac{1}{4}\left ( 8\hat{i}+14\hat{j}+19\hat{k} \right )

 

Answers (1)

best_answer

As we learned,

 

Section Formula -

1)    Internal Division

\left ( \frac{mx_{2}+nx_{1}}{m+n}, \frac{my_{2}+ny_{1}}{m+n}, \frac{mz_{2}+nz_{1}}{m+n} \right )

2)    External Division

\left ( \frac{mx_{2}-nx_{1}}{m-n}, \frac{my_{2}-ny_{1}}{m-n}, \frac{mz_{2}-nz_{1}}{m-n} \right )

3)    Mid Point Formula

\left ( \frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2} \right )

- wherein

 

 

Let angular bisector of A meet side BC at

P(x,y,z)

By Angular Bisector theorem, AB: AC=BP:PC

BP:PC = :2:1=m:n

B\equiv (2,3,4)    C\equiv(2,5,7)

Put values P\left ( x,y,z \right )=\left ( \frac{6}{3},\frac{13}{3},\frac{18}{3} \right )


Option 1)

\frac{1}{2}\left ( 4\hat{i}+8\hat{j}+11\hat{k} \right )

Option 2)

\frac{1}{3}\left ( 6\hat{i}+11\hat{j}+15\hat{k} \right )

Option 3)

\frac{1}{3}\left ( 6\hat{i}+13\hat{j}+18\hat{k} \right )

Option 4)

\frac{1}{4}\left ( 8\hat{i}+14\hat{j}+19\hat{k} \right )

Posted by

Himanshu

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