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If the line joining the points $A(2,3,-1)$ and $B(3,5,-3)$ is perpendicular to the line joining $C(1,2,3)$ and $D(3, y, 7)$, then $y=$
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Verified Answer
The correct answer is:
5
$$
\begin{aligned}
& \text {Given, } A=(2,3-1) \\
& \qquad B=(3,5-3) \\
& \text { DR's of } \mathbf{A B}=(3-2,5-3,-3+1)=(1,2,-2) \\
& \text { and } C=(1,2,3) \\
& \qquad D=(3, y, 7)
\end{aligned}
$$
$$
\therefore \mathrm{DR}^{\prime} \mathrm{s} \text { of } \mathbf{C D}=(3-1, y-2,7-3)=(2, y-2,4)
$$
Given, $\mathbf{A B} \perp \mathbf{C D}$
$$
\begin{aligned}
& \Rightarrow a_1 a_2+b_1 b_2+c_1 c_2=0 \\
& \Rightarrow(1)(2)+2(y-2)+(-2)(4)=0 \\
& \Rightarrow 2 y-10=0 \\
& \Rightarrow y=5
\end{aligned}
$$
Hence, option (c) is correct.
\begin{aligned}
& \text {Given, } A=(2,3-1) \\
& \qquad B=(3,5-3) \\
& \text { DR's of } \mathbf{A B}=(3-2,5-3,-3+1)=(1,2,-2) \\
& \text { and } C=(1,2,3) \\
& \qquad D=(3, y, 7)
\end{aligned}
$$
$$
\therefore \mathrm{DR}^{\prime} \mathrm{s} \text { of } \mathbf{C D}=(3-1, y-2,7-3)=(2, y-2,4)
$$
Given, $\mathbf{A B} \perp \mathbf{C D}$
$$
\begin{aligned}
& \Rightarrow a_1 a_2+b_1 b_2+c_1 c_2=0 \\
& \Rightarrow(1)(2)+2(y-2)+(-2)(4)=0 \\
& \Rightarrow 2 y-10=0 \\
& \Rightarrow y=5
\end{aligned}
$$
Hence, option (c) is correct.
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