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Let veca=hati+hatj and vecb=2hati-hatk. ...

Let `veca=hati+hatj and vecb=2hati-hatk.` Then the point of intersection of the lines `vecrxxveca=vecbxxveca and vecrxxvecb=vecaxxvecb` is

A

(3,-1,1)

B

(3,1,-1)

C

(-3,1,1)

D

(-3,-1,-1)

Text Solution

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The correct Answer is:
b
Let `vecrxxveca= vecbxxveca`
` rArr " "(vecr-vecb)xxveca= vec0 rArr vecr= vecb+ tveca`
Similarly, other line `vecr= veca+ kvecb`, where `t and k` are scalars.
Now `veca+ kvecb= vecb+ tveca`
`rArr" "t=1, k=1 " "` (equating the coefficients of `veca and vecb` )
`therefore " "vecr= veca+ vecb= hati+ hatj+ 2hati-hatk= 3hati+ hatj-hatk`
i.e., `(3, 1, -1)`.
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