Compute - i-j-k - 2i-3j-k - - -3i-j-k - 2j-k -

The correct option is C 15i+25j+15k [ ( i j+k ) ×( 2i 3j k ) nbsp;] ×[ ( 3i+j+k ) ×( 2j+k ) nbsp;] =[ i( 1 ( 3 ) nbsp;) j( ( 1 ) ...

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Applications of Dot Product

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Question

Compute $[(i - j + k) \times (2 i - 3 j - k)] \times [(- 3 i + j + k) \times (2 j + k)]$

- 5 i - 12 j + 13 k

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- 15 i + 25 j - 15 k

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- 15 i + 25 j + 15 k

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- 5 i + 12 j - 13 k

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Solution

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Similar questions

\vec{a} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}, \vec{b} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{c} = 3 \hat{i} - \hat{j} + 2 \hat{k}

\vec{a} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}, \vec{b} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{c} = 3 \hat{i} - \hat{j} - 2 \hat{k}

\vec{a} = 11 \hat{i}, \vec{b} = 2 \hat{j}, \vec{c} = 13 \hat{k}

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

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

[\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{r} \cdot (3 \hat{i} - \hat{j} + \hat{k}) = 1 and \vec{r} \cdot (\hat{i} + 4 \hat{j} - 2 \hat{k}) = 2

- 2 \hat{i} + 7 \hat{j} + 13 \hat{k}

2 \hat{i} + 7 \hat{j} - 13 \hat{k}

- 2 \hat{i} - 7 \hat{j} + 13 \hat{k}

2 \hat{i} + 7 \hat{j} + 13 \hat{k}

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}

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