If the points $\mathrm{P}(4,5, \mathrm{x}), \mathrm{Q}(3, \mathrm{y}, 4)$ and $\mathrm{R}(5,8,0)$ are…
If the points $\mathrm{P}(4,5, \mathrm{x}), \mathrm{Q}(3, \mathrm{y}, 4)$ and $\mathrm{R}(5,8,0)$ are collinear, then the value of $x+y$ is
- 6
- 7
- 4
- 5
Solution
$\begin{aligned}
& \overline{\mathrm{PQ}}=-\hat{\mathrm{i}}+(\mathrm{y}-5) \hat{\mathrm{j}}+(4-\mathrm{x}) \hat{\mathrm{k}} \\
& \overline{\mathrm{PR}}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}=\mathrm{x} \hat{\mathrm{k}}
\end{aligned}$
Since points $\mathrm{P}$. Q, R are collinear
$\begin{aligned}
& \overline{\mathrm{PQ}}=\mathrm{a} \overline{\mathrm{PR}} \\
& \therefore-\hat{\mathrm{i}}+(\mathrm{y}-5) \hat{\mathrm{j}}+(4-\mathrm{x}) \hat{\mathrm{k}}=\mathrm{a}(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}=\mathrm{x} \hat{\mathrm{k}}) \\
& \therefore \mathrm{a}=-1,3 \mathrm{a}=\mathrm{y}-5,-\mathrm{ax}=4-\mathrm{x} \\
& \therefore \mathrm{a}=-1 \Rightarrow-3=\mathrm{y}-5 \quad \text { i.e. } \mathrm{y}=2 \text { and } \mathrm{x}=4-\mathrm{x} \Rightarrow \mathrm{x}=2 \Rightarrow \\
& \mathrm{x}+\mathrm{y}=4
\end{aligned}$
Asked in: MHT CET 2021 (22 Sep Shift 2)
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