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Let A vector veca =alpha hati + 2hatj +...

Let A vector `veca =alpha hati + 2hatj + beta hatk` `(alpha, beta in R)`,`veca` lies in the plane of the vectors, ` vecb= hati + hatj` and `vecc= hati -hatj+4hatk`. If `veca` bisects the angle between `vecb and vecc`, then :

A

`vec a* veci +3 =0`

B

`vec a*hati + 1=0`

C

`vec a * hatk+2=0`

D

`veca*hatk +4=0`

Text Solution

Verified by Experts

The correct Answer is:
`C`
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Knowledge Check

  • If vecA = hati + 2hatj - hatk , vecB = - hati + hatj - 2hatk , then angle between vecA and vecB is

    A
    `pi/2`
    B
    0
    C
    `pi`
    D
    `pi/3`

  • If vecA 2 hati +hatj -hatk, vecB=hati +2hatj +3hatk, vecC=6hati -2hatj-6hatk angle between (vecA+vecB) and vecC will be

    A
    `30^(@)`
    B
    `45^(@)`
    C
    `60^(@)`
    D
    `90^(@)`

  • Given the vector vecA=2hati+3hatj-hatk, vecB=3hati-2hatj-2hatk & vecC=phati+phatj+phatk . Find the angle between (vecA-vecB) &vecC

    A
    `theta=cos^(-1)""((2)/(sqrt(3)))`
    B
    `theta=cos^(-1)""((sqrt(3))/(2))`
    C
    `theta=cos^(-1)""((sqrt(2))/(3))`
    D
    none of these.
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