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# Find the Angle Between the Vectors Whose Direction Cosines Are Proportional to 2, 3, −6 and 3, −4, 5. - CBSE (Arts) Class 12 - Mathematics

ConceptDirection Cosines and Direction Ratios of a Line

#### Question

Find the angle between the vectors whose direction cosines are proportional to 2, 3, −6 and 3, −4, 5.

#### Solution

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

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

$\ \text { Let } \vec{b} \text { be a vector with direction ratios } 3, - 4, 5 .$

$\Rightarrow \vec{b} = 3 \hat{i} - 4 \hat{j} + 5 \hat{k}$

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

$\text{ Now, }$

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

$= \frac{\left( 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \right) . \left( 3 \hat{i} - 4 \hat{j} + 5 \hat{k} \right)}{\left| 2 \hat{i} + 3 \hat{j} - 6 \hat{k} \right|\left| 3 \hat{i} - 4 \hat{j} + 5 \hat{k} \right|}$

$= \frac{6 - 12 - 30}{\sqrt{4 + 9 + 36} \sqrt{9 + 16 + 25}}$

$= \frac{- 36}{\sqrt{49} \sqrt{50}}$

$= \frac{- 36}{35\sqrt{2}}$

$\text{ Rationalising the result, we get }$

$\cos \theta = - \frac{18\sqrt{2}}{35}$

$\therefore \theta = \cos^{- 1} \left( - \frac{18\sqrt{2}}{35} \right)$

$\ \text { Thus, the angle between the given vectors measures }\cos^{- 1} \left( - \frac{18\sqrt{2}}{35} \right) .$

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#### Video TutorialsVIEW ALL [3]

Solution Find the Angle Between the Vectors Whose Direction Cosines Are Proportional to 2, 3, −6 and 3, −4, 5. Concept: Direction Cosines and Direction Ratios of a Line.
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