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# Solution for If → α = 3 ^ I + 4 ^ J + 5 ^ K and → β = 2 ^ I + ^ J − 4 ^ K , Then Express → β in the Form of → β = → β 1 + → β 2 , Where → β 1 is Parallel to → α and → β 2 is Perpendicular to → α - CBSE (Science) Class 12 - Mathematics

ConceptBasic Concepts of Vector Algebra

#### Question

If $\vec{\alpha} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \text{ and } \vec{\beta} = 2 \hat{i} + \hat{j} - 4 \hat{k} ,$ then express $\vec{\beta}$ in the form of  $\vec{\beta} = \vec{\beta_1} + \vec{\beta_2} ,$  where $\vec{\beta_1}$ is parallel to $\vec{\alpha} \text{ and } \vec{\beta_2}$  is perpendicular to $\vec{\alpha}$

#### Solution

$\text{ Given that } \vec{\alpha} =3 \hat{i} + 4 \hat{j} +5 \hat{k} \text{ and } \vec{\beta} =2 \hat{i} + \hat{j} - 4 \hat{k}$
$\hat{ Also },$
$\vec{\beta} = \vec{\beta_1} + \vec{\beta_2}$
$\Rightarrow \vec{\beta_2} = \vec{\beta} - \vec{\beta}_1 . . . \left( 1 \right)$
$\text{ Since } \vec{\beta}_1 \text{ is parallel to } \vec{\alpha} ,$
$\vec{\beta_1} = t \vec{\alpha}$
$\Rightarrow \vec{\beta_1} = t \left( 3 \hat{i} + 4 \hat{j} +5 \hat{k} \right) = 3t \hat{i} + 4t \hat{j} +5t \hat{k} ...(2)$
$\text{ Substituting the values of } \vec{\beta_1} \text{ and } \vec{\alpha} \text{ in } (1), \text{ we get }$
$\vec{\beta_2} = 2 \hat{i} + \hat{j} - 4 \hat{k} - \left( 3t \hat{i} + 4t \hat{j} +5t \hat{k} \right) = \left( 2 - 3t \right) \hat{i} + \left( 1 - 4t \right) \hat{j} + \left( - 4 - 5t \right) \hat{k} . . . \left( 3 \right)$
$\text{ Since } \vec{\beta_2} \text{ is perpendicular to } \vec{\alpha} ,$
$\vec{\beta_2} . \vec{\alpha} = 0$
$\Rightarrow \left[ \left( 2 - 3t \right) \text{i} + \left( 1 - 4t \right) \hat{j} + \left( - 4 - 5t \right) \hat{k} \right] . \left( 3 \hat{i} + 4 \hat{j} +5 \hat{k} \right) = 0$
$\Rightarrow 3 \left( 2 - 3t \right) + 4 \left( 1 - 4t \right) + 5 \left( - 4 - 5t \right) = 0$
$\Rightarrow 6 - 9t + 4 - 16t - 20 - 25t = 0$
$\Rightarrow - 50t = 10$
$\Rightarrow t = \frac{- 1}{5}$
$\text{ From } (2) \text{ and } (3), \text{ we get }$
$\vec{\beta_1} = \frac{- 1}{5} \left( 3 \hat{i} + 4 \hat{j} +5 \hat{k} \right)$
$\vec{\beta_2} = \frac{13}{5} \hat{i} + \frac{9}{5} \hat{j} - 3 hat{k} = \frac{1}{5}\left( 13 \hat{i} + 9 \hat{j} - 15 \hat{k} \right)$

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Solution If → α = 3 ^ I + 4 ^ J + 5 ^ K and → β = 2 ^ I + ^ J − 4 ^ K , Then Express → β in the Form of → β = → β 1 + → β 2 , Where → β 1 is Parallel to → α and → β 2 is Perpendicular to → α Concept: Basic Concepts of Vector Algebra.
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