Question

If 4i + 7j + 8k, 2i + 3j+ 4k and 2i + 5j + 7k are the position vector of the vertices A,B and C respectively of triangle ABC. the position vector of the point where the bisector of angle A meets BC is

Answer 1

Therefore the position vector of the point is $2i+133j+6k$ and $2i+7j+10k$

Step-by-step explanation:

Given;

$A=4i+7j+8k$,$B=2i+3j+4k$ and $C=2i+5j+7k$ are the point vector of triangle 'ABC' and

The bisector of $\angle A$ meets 'BC' which is known as 'D'

we know the angle bisector theorem,which states that,

An angle bisector of an angle of a triangle divides the opposite side in two segments that are proportional to the other two sides of the triangle.

So,

$BD:DC::AB:AC$

Now,

Vector $AB=$P.V of $B$ - P.V of $A$$=(2i+3j+4k)-(4i+7j+8k)$$=-2i-4j-4k$

Vector $AC=$P.V of $C$ - P.V of $A$$=(2i+ 5j + 7k)-(4i+7j+8k)=-2i-2j-k$

$\left |AB\right |=\sqrt{(-2)^{2}+(-4)^{2}+(-4)^{2} }$$=\sqrt{16+16+4} =6$

$\left |AC\right |=\sqrt{(-2)^{2}+(-2)^{2} +(-1)^{2} } =\sqrt{4+4+1} =3$

So as for,

$BD:DC::AB:AC$

We have,

$\frac{BD}{DC} =\frac{6}{3}=\frac{2}{1}$

So,

'D' divides the side BC in the ratio $2:1$ internally if 'AD' is the internal angle bisector. and $2:1$ externally if 'AD' is the external angle bisector.

So, position vector of 'D' can be obtained by using section formula;

P.V of  $D=\frac{2\times(P.V of C)+1\times(P.V of B)}{2+1}$ and  $\frac{2\times(P.V of C)-1\times(P.V of B)}{2-1}$

P.V of  $D=\frac{2\times(2i+ 5j + 7k)+1\times(2i+3j+4k)}{2+1}$ and  $\frac{2\times(2i+ 5j + 7k)-1\times(2i+3j+4k)}{2-1}$

P.V of $D=2i+133j+6k$ and $2i+7j+10k$

So the position vector of the point is $2i+133j+6k$ and $2i+7j+10k$