1.

If 2veca, 3vecb,2(veca xx vecb) are position vectors of the vectors A,B,C, of triangleABC and |veca|=|vecb|=1,vec(OA).vec(OB)=-3 (where O is the origin), then

Answer»

TRIANGLE ABC is right-angled triangle
Angle B is `90^(@)`
`A=cos^(-1)(sqrt(7/19))`
The position vector of orthocenter is `2(veca XX vecb)`

SOLUTION :`vec(OA).vec(OB)=-3`
`RARR 2.3costheta=-3`
`rArr costheta=-1/2 rArr costheta=(2pi)/(3)`
`=4|veca||vecb|^(2)sin^(2)theta+6veca.vecb=4.3/4-61/2=0`
Angle C is `90^(@)`.


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