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# Solution for Show that the Vectors → a = 1 7 ( 2 ^ I + 3 ^ J + 6 ^ K ) , → B = 1 7 ( 3 H a T I − 6 H a T J + 2 ^ K ) , → C = 1 7 ( 6 ^ I + 2 ^ J − 3 ^ K ) Mutually Perpendicular Unit Vectors. - CBSE (Commerce) Class 12 - Mathematics

ConceptBasic Concepts of Vector Algebra

#### Question

Show that the vectors $\vec{a} = \frac{1}{7}\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right), \vec{b} = \frac{1}{7}\left( 3 hat{i} - 6 hat{j} + 2 \hat{k} \right), \vec{c} = \frac{1}{7}\left( 6 \hat{i} + 2 \hat{j} - 3 \hat{k} \right)$ mutually perpendicular unit vectors.

#### Solution

$\text{ We have }$
$\left| \vec{a} \right| = \frac{1}{7}\sqrt{2^2 + 3^2 + 6^2} = \frac{1}{7}\sqrt{49} = \frac{7}{7} = 1$
$\left| \vec{b} \right| = \frac{1}{7}\sqrt{3^2 + \left( - 6 \right)^2 + 2^2} = \frac{1}{7}\sqrt{49} = \frac{7}{7} = 1$
$\left| \vec{c} \right| = \frac{1}{7}\sqrt{6^2 + 2^2 + \left( - 3 \right)^2} = \frac{1}{7}\sqrt{49} = \frac{7}{7} = 1$
$\text{ And }$
$\vec{a} . \vec{b}$
$= \frac{1}{7}\left( 2 \hat{i} + 3 \hat{j} + 6 hat{k} \right) . \frac{1}{7}\left( 3 hat{i} - 6 \hat{j} + 2 \hat{k} \right)$
$= \frac{1}{49}\left( 6 - 18 + 12 \right)$
$= 0$
$\vec{b} . \vec{c}$
$= \frac{1}{7}\left( 3 \hat{i} - 6 \hat{j} + 2 \hat{k} \right) . \frac{1}{7}\left( 6 \hat{i} + 2 \hat{j} - 3 \hat{k} \right)$
$= \frac{1}{49}\left( 18 - 12 - 6 \right)$
$= 0$
$\vec{c} . \vec{a}$
$= \frac{1}{7}\left( 6 \hat{i} + 2 \hat{j} - 3 \hat{k} \right) . \frac{1}{7}\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right)$
$= \frac{1}{49}\left( 12 + 6 - 18 \right)$
$= 0$
$So,\left| \vec{a} \right| = \left| \vec{b} \right| = \left| \vec{c} \right| = 1 and \vec{a} . \vec{b} = \vec{b} . \vec{c} = \vec{c} . \vec{a} = 0$
$\text{ So }, \text{ the given vectors are mutually perpendicular unit vectors. }$

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Solution Show that the Vectors → a = 1 7 ( 2 ^ I + 3 ^ J + 6 ^ K ) , → B = 1 7 ( 3 H a T I − 6 H a T J + 2 ^ K ) , → C = 1 7 ( 6 ^ I + 2 ^ J − 3 ^ K ) Mutually Perpendicular Unit Vectors. Concept: Basic Concepts of Vector Algebra.
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