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imran
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Asked: 2023-05-01T03:02:51+00:00 In:

The unit vector to the points p1(0,1,0), p2(1,0,1), p3(0,0,1) is a) (-j – k)/1.414 b) (-i – k)/1.414 c) (-i – j)/1.414 d) (-i – j – k)/1.414

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The unit vector to the points p1(0,1,0), p2(1,0,1), p3(0,0,1) is a) (-j – k)/1.414 b) (-i – k)/1.414 c) (-i – j)/1.414 d) (-i – j – k)/1.414
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    20 Answers

    1. a) (-j – k)/1.414

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    3. a) (-j – k)/1.414

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    4. a) (-j – k)/1.414

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    6. a) (-j – k)/1.414

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    7. a) (-j – k)/1.414

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    9. Answer: a

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    10. Answer: a

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    12. (-j – k)/1.414

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    13. Answer: a
      Explanation: The cross product of p1, p2, p3 is a X b = -j – k and its magnitude is 1.414. The unit normal vector is given by, (-j –
      k)/1.414.

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    15. a) (-j – k)/1.414

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    16. Answer: a
      Explanation: The cross product of p1, p2, p3 is a X b = -j – k and its magnitude is 1.414.
      The unit normal vector is given by, (-j –
      k)/1.414.

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    18. Answer: a

      Explanation: The cross product of p1, p2, p3 is a X b = -j – k and its magnitude is 1.414. The unit normal vector is given by, (-j –k)/1.414.

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    19. (a) (-j – k)/1.414

      Explanation: The cross product of p1, p2, p3 is a X b = -j – k and its magnitude is 1.414. The unit normal vector is given by, (-j – k)/1.414.

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    21. Answer: a

      Explanation: The cross product of p1, p2, p3 is a X b = -j – k and its magnitude is 1.414. The unit normal vector is given by, (-j –

      k)/1.414.

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    22. (-j – k)/1.414

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    24. (-j – k)/1.414

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    25. a) (-j – k)/1.414

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    27. (-j – k)/1.414

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    28. a) (-j – k)/1.414

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    30. Answer: a

      Explanation: The cross product of p1, p2, p3 is a X b = -j – k and its magnitude is 1.414.

      The unit normal vector is given by, (-j –

      k)/1.414.

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