If three points $A, B$ and $C$ have position vectors $(1, x, 3),(3,4,7)$ and $(y,-2,-5)$ respectively and if…
If three points $A, B$ and $C$ have position vectors $(1, x, 3),(3,4,7)$ and $(y,-2,-5)$ respectively and if they are collinear, then $(x, y)$ is
- $(2,-3)$
- $(-2,3)$
- $(-2,-3)$
- $(2,-3)$
Solution
Given that
$
\begin{aligned}
\mathbf{A} & =\hat{\mathbf{i}}+x \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{B}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}, \\
\mathbf{C} & =y \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} \\
\mathbf{A B} & =2 \hat{\mathbf{i}}+(4-x) \hat{\mathbf{j}}+4 \hat{\mathbf{k}},
\end{aligned}
$
and $\mathbf{B C}=(y-3) \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-12 \hat{\mathbf{k}}$
$\Rightarrow$ Since, A, B, C are collinear, then
$
\begin{aligned}
& \quad \mathbf{A B}=t \mathrm{BC} \\
& 2 \hat{\mathbf{i}}+(4-x) \hat{\mathbf{j}}+4 \hat{\mathbf{k}}=t\{(y-3) \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-12 \hat{\mathbf{k}}\} \\
& \Rightarrow 2 \hat{\mathbf{i}}+(4-x) \hat{\mathbf{j}}+4 \hat{\mathbf{k}}=t(y-3) \hat{\mathbf{i}}-6 t \hat{\mathbf{j}}-12 t \hat{\mathbf{k}}
\end{aligned}
$
Equating the coefficient, of $\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$
$
\begin{aligned}
t(y-3) & =2, \\
4-x & =-6 t, \text { and } 4=-12 t \\
t & =-\frac{1}{3}
\end{aligned}
$
$\begin{aligned} & \therefore 4-x=-6\left(-\frac{1}{3}\right)=2 \Rightarrow x=2 \\ & \text { and } \quad-\frac{1}{3}(y-3)=2 \\ & y-3=-6 \Rightarrow y=-3 \\ & \text { Then }(x, y)=(2,-3) \\ & \end{aligned}$
Asked in: AP EAMCET 2002
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