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प्रश्न
Write the angle between the lines 2x = 3y = −z and 6x = −y = −4z.
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उत्तर
We have ,
2x = 3y = −z
6x = −y = −4z
The given lines can be re-written as
\[\frac{x}{\frac{1}{2}} = \frac{y}{\frac{1}{3}} = \frac{z}{- 1} \text { and } \frac{x}{\frac{1}{6}} = \frac{y}{- 1} = \frac{z}{- \frac{1}{4}}\]
\[ \Rightarrow \frac{x}{3} = \frac{y}{2} = \frac{z}{- 6} \text{ and } \frac{x}{2} = \frac{y}{- 12} = \frac{z}{- 3}\]
These lines are parallel to vectors
\[\vec{b_1} = 3 \hat{i} + 2 \hat{j} - 6 \hat{k} \text{ and } \vec{b_2} = 2 \hat{i} - 12 \text{j} - 3 \hat{k} \]
Let \[\theta\] be the angle between these lines.
Now,
\[\cos \theta = \frac{\vec{b_1} . \vec{b_2}}{\left| \vec{b_1} \right| \left| \vec{b_2} \right|}\]
\[ = \frac{\left( 3 \hat{i} + 2 \hat{j} - 6 \hat{k} \right) . \left( 2 \hat{i} - 12 \hat{j} - 3 \hat{k} \right)}{\sqrt{3^2 + 2^2 + \left( - 6 \right)^2} \sqrt{2^2 + \left( - 12 \right)^2 + \left( - 3 \right)^2}}\]
\[ = \frac{6 - 24 + 18}{\sqrt{9 + 4 + 36} \ \sqrt{4 + 144 + 9}}\]
\[ = 0\]
\[ \Rightarrow \theta = \frac{\pi}{2}\]
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