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Let $A(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})$ and $B(13 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+9 \hat{\mathbf{k}})$ be two points on a line $L$. $C$ and $D$ be the points on $L$ on either side of $A$ at distance of 9 and 6 units respectively and $C$ lies between $A$ and $B$. Then position vectors of $C$ and $D$ are respectively
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Verified Answer
The correct answer is:
$9 \hat{i}-2 \hat{j}+5 \hat{k},-\hat{i}+3 \hat{j}-5 \hat{k}$
$\begin{gathered}\text { Given, } A(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) \text { and } B(13 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+9 \hat{\mathbf{k}}) \\ A B=\sqrt{10^2+5^2+10^2}=15\end{gathered}$
$\begin{aligned}
A C & =9, D A=6 \\
\because \quad \mathbf{C} & =\frac{9(13 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+9 \hat{\mathbf{k}})+6(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})}{9+6} \\
\quad \mathbf{C} & =\frac{135 \hat{\mathbf{i}}-30 \hat{\mathbf{j}}+75 \hat{\mathbf{k}}}{15}=9 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}
\end{aligned}$
Let position vector of $D$ is $x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$
$\begin{aligned}
& \therefore 3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}=\frac{6 \times(9 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})+9(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})}{6+9} \\
& \Rightarrow \quad 3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}=\frac{2(9 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})+3(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})}{5} \\
& \Rightarrow \quad 15 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}=(18+3 x) \hat{\mathbf{i}}+(-4+3 y) \hat{\mathbf{j}}+(10+3 z) \hat{\mathbf{k}} \\
& \therefore \quad 15=18+3 x \Rightarrow x=-1 \\
& \quad 5=-4+3 y \Rightarrow y=3 \\
& \text { and }-5=10+3 z \Rightarrow \quad z=-5
\end{aligned}$
$\therefore$ Position vector of $D$ is $-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$
$\begin{aligned}
A C & =9, D A=6 \\
\because \quad \mathbf{C} & =\frac{9(13 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+9 \hat{\mathbf{k}})+6(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})}{9+6} \\
\quad \mathbf{C} & =\frac{135 \hat{\mathbf{i}}-30 \hat{\mathbf{j}}+75 \hat{\mathbf{k}}}{15}=9 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}
\end{aligned}$
Let position vector of $D$ is $x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$
$\begin{aligned}
& \therefore 3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}=\frac{6 \times(9 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})+9(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})}{6+9} \\
& \Rightarrow \quad 3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}=\frac{2(9 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})+3(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})}{5} \\
& \Rightarrow \quad 15 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}=(18+3 x) \hat{\mathbf{i}}+(-4+3 y) \hat{\mathbf{j}}+(10+3 z) \hat{\mathbf{k}} \\
& \therefore \quad 15=18+3 x \Rightarrow x=-1 \\
& \quad 5=-4+3 y \Rightarrow y=3 \\
& \text { and }-5=10+3 z \Rightarrow \quad z=-5
\end{aligned}$
$\therefore$ Position vector of $D$ is $-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$
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