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If $A(3,4,5), B(4,6,3), C(-1,2,4)$ and $D(1,0,5)$ are such that the angle between the lines $D C$ and $A B$ is $\theta$, then $\cos \theta$ is equal to
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Verified Answer
The correct answer is:
$\frac{4}{9}$
Given points are $A(3,4,5), B(4,6,3), C(-1,2,4)$ and $D(1,0,5)$.
Now DR's of $D C=(-2,2,-1)$
DR's of $A B=(1,2,-2)$
Let, $\theta$ be the angle between $A B$ and $D C$.
$$
\begin{aligned}
\therefore \quad \cos \theta & =\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}} \\
= & \frac{-2 \times 1+2 \times 2-1 \times(-2)}{\sqrt{(-2)^2+(2)^2+(-1)^2} \sqrt{(1)^2+(2)^2+(-2)^2}} \\
& =\frac{-2+4+2}{\sqrt{4+4+1} \sqrt{1+4+4}}=\frac{4}{3 \times 3}=\frac{4}{9}
\end{aligned}
$$
Now DR's of $D C=(-2,2,-1)$
DR's of $A B=(1,2,-2)$
Let, $\theta$ be the angle between $A B$ and $D C$.
$$
\begin{aligned}
\therefore \quad \cos \theta & =\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}} \\
= & \frac{-2 \times 1+2 \times 2-1 \times(-2)}{\sqrt{(-2)^2+(2)^2+(-1)^2} \sqrt{(1)^2+(2)^2+(-2)^2}} \\
& =\frac{-2+4+2}{\sqrt{4+4+1} \sqrt{1+4+4}}=\frac{4}{3 \times 3}=\frac{4}{9}
\end{aligned}
$$
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