Question

a particle moves 14m along the vector 6i^+2j^+3k^,then 7m along the vector 3i^-2j^+6k^.its total displacement​

Answer 1

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Answer 2

Answer:

Total displacement =15i+2j+12k

Explanation:

Note that magnitudes of  3i-2j+6k and 6i+2j+3k are both 7.

So total displacement is given by:

14*[( 6i+2j+3k)/7]+7*[(3i-2j+6k)/7]

=12i+4j+6k+3i-2j+6k

=15i+2j+12k [Where, i ,j and k are unit vectors]