Which of the following is correct for the vectorsa - 17 3- - 2- - 6k- b - 173- - 6-2k- and c-176-2-3k- -

We have, nbsp; a = 17(3î+2ĵ+6k̂) b = 17(3î 6ĵ + 2k̂) c = 17(6î + 2ĵ 3k̂) Then, nbsp; a . nbsp;b = 17(3î+2ĵ+6k̂). nbsp; 17 (3î 6ĵ + 2k̂) = 149(9 12 ...

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Byju's Answer

Standard XIII

Physics

Multiplication with Vectors

Which of the ...

Question

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

All are mutually perpendicular.

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All are mutually parallel.

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\to b

and

\to a

are mutually perpendicular.

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\to b

and

\to c

are mutually perpendicular.

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Solution

The correct option is D $\to b$ and $\to c$ are mutually perpendicular.
We have,
$\to a = \frac{1}{7} (3^i + 2^j + 6^k)$
$\to b = \frac{1}{7} (3^i - 6^j + 2^k)$
$\to c = \frac{1}{7} (6^i + 2^j - 3^k)$
Then,
$\to a . \to b = \frac{1}{7} (3^i + 2^j + 6^k) . \frac{1}{7} (3^i - 6^j + 2^k)$
$= \frac{1}{49} (9 - 12 + 12) \neq 0$
Similarly finding, $\to b . \to c$
$\to b . \to c = 0$
$∴ \to b, \to c$ are mutually perpendicular.

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Similar questions

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)

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)

The unit vector parallel to the resultant of the vectors
$\to A = 4^i + 3^j + 6^k$ and
$\to B = -^i + 3^j - 8^k$ is

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Which of the following is correct for the vectors→a=17(3^i+2^j+6^k),→b=17(3^i−6^j+2^k) and →c=17(6^i+2^j−3^k) ?

The correct option is D →b and →c are mutually perpendicular.We have, →a=17(3^i+2^j+6^k) →b=17(3^i−6^j+2^k) →c=17(6^i+2^j−3^k) Then, →a.→b=17(3^i+2^j+6^k).17(3^i−6^j+2^k) =149(9−12+12)≠0 Similarly finding, →b.→c →b.→c=0 ∴→b,→c are mutually perpendicular.

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