Which of the following is correct for the vectors a -1-72 -3 -6 k- b -1-73 -6 -2 k- and c -1-76 -2 -3 k- -A- All are mutually parallel-B- All are mutually perpendicular-C- a - b cD- a - c b

The correct option is B All are mutually perpendicular. nbsp;We have, nbsp; a = 17(2î+3ĵ+6k̂) b = 17(3î 6ĵ + 2k̂) c = 17(6î + 2ĵ 3k̂) Then, nbsp; a ...

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Byju's Answer

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. 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. 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)

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

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