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Perpendicular Distance of a Point from a Line

A unit vector...

Question

A unit vector in the plane of the vectors 2i + j + k, i - j + k and orthogonal to 5i + 2j + 6k is

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Solution

The correct option is C

A vector coplanar with $¯ a, ¯ b$ and orthogonal to $¯ c i s (¯ a \times ¯ b) \times ¯ c$
$∴ R e q u i r e d u n i t v e c t o r = \pm \frac{(¯ a \times ¯ b) \times ¯ c}{∣ ∣ (¯ a \times ¯ b) \times ¯ c ∣ ∣}$

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Similar questions

2 i + j + k, i - j + k

5 i + 2 j + 6 k

3 i + 2 j + 6 k

2 i + j + k

i - j + 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

2 \hat{i} - \hat{j} + \hat{k}, \hat{i} - 3 \hat{j} - 5 \hat{k} and 3 \hat{i} - 4 \hat{j} - 4 \hat{k}

\hat{i} + \hat{j} + \hat{k}, 2 \hat{i} + 3 \hat{j} - \hat{k} and - \hat{i} - 2 \hat{j} + 2 \hat{k}

\vec{a} and \vec{b}

\vec{a} = \hat{i} - \hat{j} and \vec{b} = \hat{j} + \hat{k}

\vec{a} = 3 \hat{i} - 2 \hat{j} - 6 \hat{k} and \vec{b} = 4 \hat{i} - \hat{j} + 8 \hat{k}

\vec{a} = 2 \hat{i} - \hat{j} + 2 \hat{k} and \vec{b} = 4 \hat{i} + 4 \hat{j} - 2 \hat{k}

\vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k} and \vec{b} = \hat{i} + \hat{j} - 2 \hat{k}

\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{b} = \hat{i} - \hat{j} + \hat{k}

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