A particle moves on a given line with a constant speed v. At a certain time it is at point P on its staight line path.O is fixedpoint .show that `(Opxxv)` is independent of the position P.
A
`vec(A)=(vec(B)xxvec(C))`
B
`vec(A)=2(vec(B)xxvec(C))`
C
`vec(A)=2(vec(C)xxvec(B))`
D
`|vec(B)xxvec(C)|=sqrt(3)/2`
Text Solution
Verified by Experts
The correct Answer is:
B, C
As `vec(A) bot vec(B)` and `vec(A) bot vec(C)` so `vec(A)=+-((vec(B)xxvec(C)))/(|vec(B)xxvec(C)|)` But `|vec(B)xxvec(C)|=BC sin 30^(@)=1/2` So `vec(A)=+-2(vec(B)xxvec(C))rArr vec(A)=2(vec(B)xxvec(C))` and `vec(A)=-2(vec(B)xxvec(C))=2(vec(C)xxvec(B))`
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