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If $(2,-1,2)$ and $(K, 3,5)$ are the triads of direction ratios of two lines and the angle between them is $45^{\circ}$, then the value of $K$ is
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Verified Answer
The correct answer is:
$4$
Given direction ratios are $(2,-1,2)$ and $(K, 3,5)$
Let $\theta$ be the angle between two lines, then
$\begin{aligned}
& \cos \theta=\frac{(2,-1,2)(K, 3,5)}{\sqrt{4+1+4} \sqrt{K^2+9+25}} \\
& \cos 45^{\circ}=\frac{2 K-3+10}{\sqrt{9} \sqrt{K^2+34}} \\
& \frac{1}{\sqrt{2}}=\frac{2 K+7}{3 \times \sqrt{K^2+9+25}} \\
& 3 \sqrt{K^2+34}=\sqrt{2}(2 K+7) \\
& 9\left(K^2+34\right)=2(2 K+7)^2 \\
& 9 K^2+306=8 K^2+56 K+98 \\
& K^2-56 K+208=0 \Rightarrow K=52,4
\end{aligned}$
$\therefore \quad$ According to the given options, $K=4$
Let $\theta$ be the angle between two lines, then
$\begin{aligned}
& \cos \theta=\frac{(2,-1,2)(K, 3,5)}{\sqrt{4+1+4} \sqrt{K^2+9+25}} \\
& \cos 45^{\circ}=\frac{2 K-3+10}{\sqrt{9} \sqrt{K^2+34}} \\
& \frac{1}{\sqrt{2}}=\frac{2 K+7}{3 \times \sqrt{K^2+9+25}} \\
& 3 \sqrt{K^2+34}=\sqrt{2}(2 K+7) \\
& 9\left(K^2+34\right)=2(2 K+7)^2 \\
& 9 K^2+306=8 K^2+56 K+98 \\
& K^2-56 K+208=0 \Rightarrow K=52,4
\end{aligned}$
$\therefore \quad$ According to the given options, $K=4$
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