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Vectors and Its Types

The unit vect...

Question

The unit vector perpendicular to both $L_{1}$ and $L_{2}$ is

\frac{- \to i + 7 \to j + 7 \to k}{\sqrt{99}}

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\frac{- \to i - 7 \to j + 5 \to k}{5 \sqrt{3}}

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\frac{- \to i + 7 \to j + 5 \to k}{5 \sqrt{3}}

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\frac{7 \to i - 7 \to j - \to k}{\sqrt{99}}

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Solution

The correct option is C $\frac{- \to i - 7 \to j + 5 \to k}{5 \sqrt{3}}$
Vector in the direction of $L_{1} =_{1} = 3 \to i + \to j + 2 \to k$
Vector in the direction of $L_{2} =_{2} = \to i + 2 \to j + 3 \to k$
$∴$ Vector perpendicular to both $L_{1}$ and $L_{2}$
$=_{1} \times_{2} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} \to i & \to j & \to k 3 & 1 & 2 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣ ∣ = - i - 7 j + 5 k$
$∴$ Required vector $=^n = \frac{- \to i - 7 \to j + 5 \to k}{\sqrt{1 + 49 + 25}} = \frac{- \to i - 7 \to j + 5 \to k}{5 \sqrt{3}}$

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Similar questions

¯ ¯¯¯¯¯¯ ¯ A B = 2 \to i - 3 \to j + \to k

¯ ¯¯¯¯¯¯ ¯ C B = \to i + \to j + \to k

¯ ¯¯¯¯¯¯¯ ¯ C D = 4 \to i - 7 \to j

¯ ¯¯¯¯¯¯¯ ¯ A D

L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}

L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}

(1, 1, 1)

(- 1, - 2, - 1)

L_{1}

L_{2}

\frac{13}{5 \sqrt{3}}

L_{1}

L_{2}

\frac{- \to i - 7 \to j + 5 \to k}{5 \sqrt{3}}

P

5 \to i + \to j + 3 \to k

\to r = (3 \to i + 7 \to j + \to k) + t (\to j + \to k)

\to i + \to j + 2 \to k

\to i + 2 \to j + \to k

\to i + \to j + \to k

A unit vector coplanar with $\to i + \to j + 3 \to k and \to i + 3 \to j + \to k$ and perpendicular to $\to i + \to j + \to k$ is

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