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

We have, nbsp; a = 17(2î+3ĵ+6k̂) b = 17(3î 6ĵ + 2k̂) c = 17(6î + 2ĵ 3k̂) Then, nbsp; a . nbsp;b = 17(2î+3ĵ+6k̂). nbsp; 17 (3î 6ĵ + 2k̂) = 149(6 18 ...

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Standard XII

Physics

Multiplication with Vectors

Which of the ...

Question

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

All are mutually parallel.

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All are mutually perpendicular.

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\to a ⊥ \to b ∥ \to c

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\to a ⊥ \to c ∥ \to b

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Solution

The correct option is B All are mutually perpendicular.

We have,
$\to a = \frac{1}{7} (2^i + 3^j + 6^k)$
$\to b = \frac{1}{7} (3^i - 6^j + 2^k)$
$\to c = \frac{1}{7} (6^i + 2^j - 3^k)$
Then,
$\to a . \to b = \frac{1}{7} (2^i + 3^j + 6^k) . \frac{1}{7} (3^i - 6^j + 2^k)$
$= \frac{1}{49} (6 - 18 + 12) = 0$
Similarly,
$\to b . \to c = \to a . \to c = 0$
$∴ \to a, \to b, \to c$ are mutually perpendicular.

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Similar questions

Q. Which of the following is correct for the vectors

\to a = \frac{1}{7} (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)

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

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(2^i+3^j+6^k),→b=17(3^i−6^j+2^k) and →c=17(6^i+2^j−3^k)?

The correct option is B All are mutually perpendicular. We have, →a=17(2^i+3^j+6^k) →b=17(3^i−6^j+2^k) →c=17(6^i+2^j−3^k) Then, →a.→b=17(2^i+3^j+6^k).17(3^i−6^j+2^k) =149(6−18+12)=0 Similarly, →b.→c=→a.→c=0 ∴→a,→b,→c are mutually perpendicular.

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