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# Solution for Find the Acute Angle Between the Lines Whose Direction Ratios Are Proportional to 2 : 3 : 6 and 1 : 2 : 2. - CBSE (Science) Class 12 - Mathematics

ConceptDirection Cosines and Direction Ratios of a Line

#### Question

Find the acute angle between the lines whose direction ratios are proportional to 2 : 3 : 6 and 1 : 2 : 2.

#### Solution

$\text { Let } \vec{a} \text{ be a vector parallel to the vector with direction ratios } 2, 3, 6 .$

$\Rightarrow \vec{a} = 2 \hat{i} + 3 \hat{j}+ 6 \hat{k} .$

$\text{ Let } \vec{b} \text { be a vector parallel to the vector with direction ratios }1, 2, 2 .$

$\Rightarrow \vec{b} = \hat{i} + 2 \hat{j} + 2 \hat{k}$

$\text { Let } \theta \text{ be the angle between the the given vectors }.$

$Now,$

$\cos \theta = \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| \left| \vec{b} \right|}$

$= \frac{\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right) . \left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right)}{\left| 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right|\left| \hat{i} + 2 \hat{j} + 2 \hat{k} \right|}$

$= \frac{2 + 6 + 12}{\sqrt{4 + 9 + 36} \sqrt{1 + 4 + 4}}$

$= \frac{20}{21}$

$\Rightarrow \theta = \cos^{- 1} \left( \frac{20}{21} \right)$

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Solution Find the Acute Angle Between the Lines Whose Direction Ratios Are Proportional to 2 : 3 : 6 and 1 : 2 : 2. Concept: Direction Cosines and Direction Ratios of a Line.
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