Find the ratio in which the point

Question:

Find the ratio in which the point C(5, 9, -14) divides the join of A(2, -3, 4) and B(3, 1, -2).

Solution:

Let the ratio be k:1 in which point R divides point P and point Q.

Using $\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}, \frac{m z_{2}+n z_{1}}{m+n}\right)$, we get,

Here $m$ and $n$ are $k$ and 1 . The point which this formula gives is already given, i.e. $R(5,9,-14)$ and the joining points are $P(2,-3,4)$ and $Q(3,1,-2)$.

$(5,9,-14)=\left(\frac{\mathrm{k} \times 3+1 \times 2}{\mathrm{k}+1}, \frac{\mathrm{k} \times 1+1 \times-3}{\mathrm{k}+1}, \frac{\mathrm{k} \times-2+1 \times 4}{\mathrm{k}+1}\right)$

Taking any point and finding the value of k, we get

$5=\frac{\mathrm{k} \times 3+1 \times 2}{\mathrm{k}+1}$

$5 k+5=3 k+2$

$2 k=-3$

$\mathrm{K}=-\frac{3}{2}$

Since, the ratio is $-3: 2$. hence the division is external division.

The external division ratio is $3: 2$.

 

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