Question

Show that the points A(1,-2,-8),B(5,0,-2),and C(11,3,7)are collinear,and find the ratio with B devides AC.

Answer 1

The given points are A(1,-2,-8),B(5,0,-2),and C(11,3,7).
∴$\vec{AB}$=(5-1)$\hat{i}$+(0+2)$\hat{j}$+(-2+8)$\hat{k}$=4$\hat{i}$+2$\hat{j}$+6$\hat{k}$
$\vec{BC}$=(11-5)$\hat{i}$+(3-0)$\hat{j}$+(7+2)$\hat{k}$=6$\hat{i}$+3$\hat{j}$+9$\hat{k}$
$\vec{AC}$=(11-1)$\hat{i}$+(3+2)$\hat{j}$+(7+8)$\hat{k}$=10$\hat{i}$+5$\hat{j}$+15$\hat{k}$
|$\vec{AB}$|=$\sqrt{42+22+62}$ =$\sqrt{16+4+36}$=$\sqrt{56}$=$2\sqrt{14}$
|$\vec{BC}$|=$\sqrt{62+32+92}$ =$\sqrt{36+9+81}$=$\sqrt{126}$=$3\sqrt{14}$
|$\vec{AC}$|=$\sqrt{102+52+152}$ =$\sqrt{100+25+225}$=$\sqrt{350}$=$5\sqrt{14}$
∴|$\vec{AC}$|=|$\vec{AB}$|+|$\vec{BC}$|
Thus,the given points A,B,and C are collinear.
Now,let point B devide Ac in the ratio λ:1.Then,we have:
$\vec{OB}$=\lambda$$\vec{OC}+$\frac{\vec{OA}}{(\lambda+1)}$
$\Rightarrow$5$\hat{i}$-2$\hat{k}$=$\frac{\lambda(11\hat{i}+3\hat j+7\hat k)+(\hat i-2\hat j-8\hat k)}{\lambda +1}$
$\Rightarrow$(λ+1)($\hat{i}$-2$\hat{k}$)=11λ$\hat{i}$+3λ$\hat{j}$+7λ$\hat{k}$+$\hat{i}$-2$\hat{j}$-8$\hat{k}$
$\Rightarrow$5(λ+1)$\hat{i}$-2(λ+1)$\hat{k}$=(11λ+1)$\hat{i}$+(3λ-2)$\hat{j}$+(7λ-8)$\hat{k}$
On equating the corresponding components,we get:
$\Rightarrow$5(λ+1)=11λ+1
$\Rightarrow$5λ+5=11λ+1
$\Rightarrow$6λ=4
$\Rightarrow$λ=$\frac{4}{6}$=$\frac{2}{3}$
Hence,point B devides AC in ratio 2:3.