Find the unit vectors which are perpendicular to both the vector $i + 4 j$ and $2 i + 4 j + 3 k .$

\frac{12 i - 3 j - 4 k}{13}

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\frac{- 12 i + 3 j + 4 k}{13}

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\frac{12 i + 3 j - 4 k}{13}

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\frac{- 12 i - 3 j - 4 k}{13}

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Solution

The correct options are
A $\frac{- 12 i + 3 j + 4 k}{13}$
B $\frac{12 i - 3 j - 4 k}{13}$
Unit vectors which are perpendicular to both the vector $i + 4 j$ and $2 i + 4 j + 3 k .$ are
$\frac{\pm ((i + 4 j) \times (2 i + 4 j + 3 k))}{| (i + 4 j) \times (2 i + 4 j + 3 k) |} = \pm \frac{12 i - 3 j - 4 k}{13}$

Suggest Corrections

Similar questions

\to O A = 3 i + 4 j + 12 k, \to O B = 12 i + 3 j + 4 k,

\to O C = 4 i + 12 j + 3 k .

\to O A

\to O B

\to O A = 3 i + 4 j + 12 k, \to O B = 12 i + 3 j + 4 k,

\to O C = 4 i + 12 j + 3 k .

\to O A = 3 i + 4 j + 12 k, \to O B = 12 i + 3 j + 4 k,

\to O C = 4 i + 12 j + 3 k .

\to O A

\to O C .

13 ¯ ¯ ¯ a = 3 ¯ i + 4 ¯ j + 12 ¯ ¯ ¯ k, 13 ¯ ¯ b = 4 ¯ i - 12 ¯ j + 3 ¯ ¯ ¯ k, 13 ¯ ¯ c = 12 ¯ i + 3 ¯ j - 4 ¯ ¯ ¯ k

¯ ¯ ¯ a \times ¯ ¯ b

\to c

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