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Distance Formula

Given a = i...

Question

Given $a = i + j - k, b = - i + 2 j + k$ and $c = - i + 2 j - k .$ A unit vector perpendicular to both $a + b$ and $b + c$ is

\frac{2 i + j + k}{\sqrt{6}}

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j

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k

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\frac{i + j + k}{\sqrt{3}}

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Solution

The correct option is D $k$
Given vectors are $\to a = i + j - k, \to b = - i + 2 j + k$ and $\to c = - i + 2 j - k$
$\to a + \to b = 3 j$ and $\to b + \to c = - 2 i + 4 j$
unit vector perpendicular to both the above vectors is $\frac{(\to a + \to b) \times (\to b + \to c)}{| (\to a + \to b) \times (\to b + \to c) |}$
$\Rightarrow \frac{(3 j) \times (- 2 i + 4 j)}{| (3 j) \times (- 2 i + 4 j) |} = k$

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Similar questions

a = i + j - k, b = - i + 2 j + k

c = - i + 2 j - k

a + b

b + c

[\vec{a} \vec{b} \vec{c}]

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

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

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

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

2 \vec{a} - \vec{b} + 3 \vec{c .}

[\vec{a} \vec{b} \vec{c}]

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

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

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

\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}, \vec{b} = \hat{i} + 2 \hat{j} - 3 \hat{k}, \vec{c} = λ \hat{i} + λ \hat{j} + 5 \hat{k}

\vec{a} = \hat{i} + 2 \hat{j} - 3 \hat{k}, \vec{b} = 3 \hat{i} + λ \hat{j} + \hat{k}, \vec{c} = \hat{i} + 2 \hat{j} + 2 \hat{k}

\vec{a} = \hat{i} + 3 \hat{j}, \vec{b} = 5 \hat{k}, \vec{c} = λ \hat{i} - \hat{j}

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