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Decompose the Vector 6 ^ I − 3 ^ J − 6 ^ K into Vectors Which Are Parallel and Perpendicular to the Vector ^ I + ^ J + ^ K . - CBSE (Arts) Class 12 - Mathematics

ConceptMultiplication of a Vector by a Scalar

Question

Decompose the vector $6 \hat{i} - 3 \hat{j} - 6 \hat{k}$ into vectors which are parallel and perpendicular to the vector $\hat{i} + \hat{j} + \hat{k} .$

Solution

$\text{ Let } \vec{a} =6 \hat{i} - 3 \hat{j} - 6 \hat{k} \text{ and } \vec{b} = \hat{i} + \hat{j} + \hat{k}$

$\text{ and } \vec{x} \text{ and } \vec{y} \text{ be such that }$

$\vec{a} = \vec{x} + \vec{y}$

$\Rightarrow \vec{y} = \vec{a} - \vec{x} . . . \left( 1 \right)$

$\text{ Since } \vec{x} \text{ is parallel to } \vec{b} ,$

$\vec{x} = t \vec{b}$

$\Rightarrow \vec{x} = t \left( \hat{i} + \hat{j} + \hat{k} \right) = t \hat{i} + t \hat{j} + t \hat{k} ...(2)$

$\text{ Substituting the values of } \vec{x} \text{ and } \vec{a} \text{ in } (1), \text{ we get }$

$\vec{y} = 6 \hat{i} - 3 \hat{j} - 6 \hat{k} - \left( t \hat{i} + t \hat{j} + t \hat{k} \right) = \left( 6 - t \right) \hat{i} + \left( - 3 - t \right) \hat{j} + \left( - 6 - t \right) \hat{k} . . . \left( 3 \right)$

$\text{ Since } \vec{y} \text{ is perpendicular to } \vec{b} ,$

$\vec{y} . \vec{b} = 0$

$\Rightarrow \left[ \left( 6 - t \right) \hat{i} + \left( - 3 - t \right) \hat{j} + \left( - 6 - t \right) \hat{k} \right] . \left( \hat{i} + \hat{j} + \hat{k}\right) = 0$

$\Rightarrow 1 \left( 6 - t \right) + 1\left( - 3 - t \right) + 1 \left( - 6 - t \right) = 0$

$\Rightarrow - 3 - 3t = 0$

$\Rightarrow t = - 1$

$\text{ From } (2) \text{ and } (3), \text{we get}$

$\vec{x} = - \hat{i} - \hat{j} - \hat{k}$

$\vec{y} = 7 \hat{i} - 2 \hat{j} - 5 \hat{k}$

$\text{ So },$

$\vec{a} = \vec{x} + \vec{y} = \left( - \hat{i} - \hat{j} - \hat{k} \right) + \left( 7 \hat{i} - 2 \hat{j} - 5 \hat{k} \right)$

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Solution Decompose the Vector 6 ^ I − 3 ^ J − 6 ^ K into Vectors Which Are Parallel and Perpendicular to the Vector ^ I + ^ J + ^ K . Concept: Multiplication of a Vector by a Scalar.
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