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A unit vector perpendicular to the plane...

A unit vector perpendicular to the plane of a = 2i - 6j -3k, b = 4i + 3j - k is

A

`(4i + 3j - k)/(sqrt(26))`

B

`(2i - 6j - 3k)/(7)`

C

`(3i - 2j + 6k)/(7)`

D

`(2i - 3j - 6k)/(7)`

Text Solution

Verified by Experts

The correct Answer is:
C
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Knowledge Check

  • {:(I. "Unit vector perpendicular to the plane of" 2i - 6j - 3k. 4i + 3j - k, a. (5)/(sqrt(3)) (i + j + k)),(II. "Unit vector perpendicular to the plane determined by the points" (1.-1.2).(2.0.-1).(0.2.1),b. j - k),(III. "Vector perpendicular to the plane of" i - j - k. i + j + k, c. (1)/(sqrt(6)) (2i + j + k)),(IV. "Vector of length 5 and perpendicular to both" a = 2i + j - 3k, d. (1)/(7) (3i - 2j + 6k)):}

    A
    a,c,d,b
    B
    b,a,d,c
    C
    a,d,c,b
    D
    d,c,b,a

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    A
    `- 3i + 5j + 11k`
    B
    `(-3i + 5j + 11k)/(sqrt(155))`
    C
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  • The unit vector normal to the plane containing a = I - j - k and b = I + j + k is

    A
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    B
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    C
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    D
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