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Solution for Find the Angle Between the Vectors with Direction Ratios Proportional to 1, −2, 1 and 4, 3, 2. - CBSE (Science) Class 12 - Mathematics

ConceptDirection Cosines and Direction Ratios of a Line

Question

Find the angle between the vectors with direction ratios proportional to 1, −2, 1 and 4, 3, 2.

Solution

$\text{Let}\ \vec{a}\ \text{be a vector with direction ratios}\ 1, - 2, 1 .$

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

$\text{Let} \ \vec{b}\ \text{be a vector with direction ratios} \ 4, 3, 2 .$

$\Rightarrow \vec{b} = 4\hat{ i} + 3 \hat{j} + 2 \hat{k} .$

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

$Now,$

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

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

$= \frac{4 - 6 + 2}{\sqrt{1 + 4 + 1} \sqrt{16 + 9 + 4}}$

$= \frac{0}{\sqrt{6} \sqrt{29}}$

$= 0$

$\therefore \theta = \frac{\pi}{2}$

$\text{Thus, the angle between the given vectors measures }\frac{\pi}{2} .$

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Solution Find the Angle Between the Vectors with Direction Ratios Proportional to 1, −2, 1 and 4, 3, 2. Concept: Direction Cosines and Direction Ratios of a Line.
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