The position vectors of point A,B and C are 9i-10j , 4i+2j and Ki-2j respectively.Find the value of k if the point A,B and c are collinear. Help me solve this.thank u
Hello, sharmala1
The position vectors of points A, B and C are 9i-10j , 4i+2j and Ki-2j respectively.
Find the value of k if the points A,B and c are collinear.
$\displaystyle \overrightarrow{BA} \:=\:\langle 5,\,\text{-}12\rangle $
$\displaystyle \overrightarrow{BC} \:=\:\langle (K\!-\!4),\,\text{-}4\rangle$
If $\displaystyle A,B,C$ are collinear, then: .$\displaystyle \overrightarrow{BA} \parallel \overrightarrow{BC}$
$\displaystyle \text{That is: }\:a\cdot\overrightarrow{BC} \:=\:\overrightarrow{BA}\,\text{ for some real number }a \ne0.$
We have: .$\displaystyle a\cdot\langle(K\!-\!4),\,\text{-}4\rangle \;=\;\langle5,\,\text{-}12\rangle$
. . . . . . . . .$\displaystyle \langle a(K\!-\!4),\,\text{-}4a\rangle \;=\;\langle 5,\,\text{-}12\rangle$
Hence: .$\displaystyle \begin{Bmatrix}a(K\!-\!4) &=& 5 & [1] \\ \text{-}4a &=& \text{-}12 & [2] \end{Bmatrix}$
From [2]: .$\displaystyle a\,=\,3$
Subtitute into [1]: .$\displaystyle 3(K-4) \:=\:5 \quad\Rightarrow\quad K-4\:=\:\tfrac{5}{3}$
Therefore: .$\displaystyle K \:=\:\frac{17}{3}$