Show that each of the three given vectors is a unit vectors. also Show that they are mutually $⊥$ to each other.
$\to a = \frac{1}{7} (2^i + 3^j + 6^k)$
$\to b = \frac{1}{7} (3^i - 6^j + 2^k)$
$\to c = \frac{1}{7} (6^i + 2^j - 3^k)$

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Solution

$\to a = \frac{1}{7} (2^l + 3^j + 6^k)$
$\to b = \frac{1}{7} (3^l + 6^j + 2^k)$
$\to c == \frac{1}{7} (6^l + 2^j - 3^k)$
$\to a = \sqrt{{(\frac{2}{7})}^{2} + {(\frac{3}{7})}^{2} + {(\frac{6}{7})}^{2}} = \sqrt{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}} = 1$
$\to b = \sqrt{{(\frac{3}{7})}^{2} + {(\frac{- 3}{7})}^{2} + {(\frac{2}{7})}^{2}} = \sqrt{\frac{9}{49} + \frac{36}{49} + \frac{4}{49}} = 1$
$\to e = \sqrt{{(\frac{6}{7})}^{2} + {(\frac{2}{7})}^{2} + {(\frac{- 3}{7})}^{2}} = \sqrt{\frac{36}{49} + \frac{4}{49} + \frac{9}{49}} = 1$
each is $a$ unit vector
$(\to a . \to b) = (\frac{2}{7} \times \frac{3}{7}) + (\frac{3}{7} \times \frac{- 6}{7}) + (\frac{6}{7}) (\frac{2}{7})$
$= \frac{6}{49} - \frac{- 18}{49} + \frac{12}{49}$
$(\to b . \to c) = (\frac{3}{7}) (\frac{6}{7}) + (\frac{- 6}{7}) (\frac{2}{7}) + (\frac{2}{7}) (\frac{- 3}{7})$
$= \frac{18}{49} - \frac{12}{49} - \frac{6}{49}$
$(\to c . \to a) = (\frac{6}{7}) (\frac{2}{7}) + (\frac{2}{7}) (\frac{3}{7}) + (\frac{- 3}{7}) (\frac{6}{7})$
$= \frac{12}{49} + \frac{6}{49} - \frac{18}{49} \Rightarrow 0$
$∴$ given $3$ vector are mutally perpendicular to each other.

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Similar questions

Q. Show that each of the given three vectors is a unit vector:

\frac{1}{7} (2^i + 3^j + 6^k), \frac{1}{7} (3^i - 6^j + 2^k), \frac{1}{7} (6^i + 2^j - 3^k)

.
Also, show that they are mutually perpendicular to each other.

Q. The vector

\to a = 1 / 7 (2^i + 3^j + 6^k)

\to b = 1 / 7 (3^i - 6^j + 2^k)

\to c = 1 / 7 (6^i + 2^j - 3^k)

form

Q. Which of the following is correct for the vectors

\to a = \frac{1}{7} (2^i + 3^j + 6^k), \to b = \frac{1}{7} (3^i - 6^j + 2^k)

and

\to c = \frac{1}{7} (6^i + 2^j - 3^k)

Q. Which of the following is correct for the vectors

\to a = \frac{1}{7} (3^i + 2^j + 6^k), \to b = \frac{1}{7} (3^i - 6^j + 2^k)

and

\to c = \frac{1}{7} (6^i + 2^j - 3^k)

Let $\to a = \frac{1}{7} (2^i + 3^j + 6^k), \to b = \frac{1}{7} (6^i + 2^j - 3^k), \to c = c_{1}^i + c_{2}^j + c - 3^k$ and matrix $A = ⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} \frac{2}{7} & \frac{3}{7} & \frac{6}{7} \frac{6}{7} & \frac{2}{7} & - \frac{3}{7} c_{1} & c_{2} & c_{3} \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦$
If $A A^{T} = I, t h e n \to c$ is equal to

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Show that each of the three given vectors is a unit vectors. also Show that they are mutually ⊥ to each other.→a=17(2^i+3^j+6^k)→b=17(3^i−6^j+2^k)→c=17(6^i+2^j−3^k)

Show that each of the three given vectors is a unit vectors. also Show that they are mutually $⊥$ to each other.
$\to a = \frac{1}{7} (2^i + 3^j + 6^k)$
$\to b = \frac{1}{7} (3^i - 6^j + 2^k)$
$\to c = \frac{1}{7} (6^i + 2^j - 3^k)$