Below is my working out for some of your questions - Sorry if I made any errors.
1) Obviously in this question, the magnitude of the vector is simple, it is just 1. Therefore, you are concerned mainly with the direction.
Well, you know that a property of a rhombus, is that the diagonals bisect the angles inside it (someone posted this earlier). Imagine these vectors making two sides of a parallelogram (well 4 sides because two pairs of the same vectors). The diagonal will simply be the addition of the two vectors. However, the parallelogram's diagonals do not necessarily bisect the angles. Therefore, you want a rhombus. For a rhombus, you need equal magnitudes, which is what you desire. So, you need to first find the unit vector of a and b (or equate the magnitudes in some other way). (The addition of the unit vectors will form the diagonal of a rhombus, meaning it will bisect the two vectors).
i.e.
 )
 )
 )
 )
Now, to form the diagonal of the rhombus, you would need to add the two vectors together:
+ \frac{1}{5}(3i + 4k) )
let
Now, you need the unit vector:
So:
^2 + (\frac{2}{3})^2 + (\frac{7}{15})^2}}(\frac{19}{15}i - \frac{2}{3}j + \frac{7}{15}k) )
 )
 )
 )
 )
This is my first time doing such a question. Could someone confirm if this is correct?
2)
,
and
For PR
 + (2i + pj + qk) )
j + (q + 1)k )
Now, do the same for PQ:
 + (i - 2j + k) )
If they are collinear, then that ones one will be a scalar multiple of the other (as they have the same origin).
i.e.
, where k is a constant.
By looking at the i components, you know that the scalar multiple is 2.
i.e
j + (q + 1)k] )
j + 2(q +1)k )
Equate the j and k components:
 = -3 )
 = 2 )
3) I don't really understand the question.....
4) let
} + i \sin{(2\theta)} )
})^2 + (\sin{(2\theta)})^2} )
} + \cos^2{(2\theta)} + \sin^2{(2\theta)}} )
Recall that
} + \sin^2{(\theta)} = 1 )
} + 1} )
}} )
} - \sin^2{(\theta)})} )
} - 1)} )
})} )
} )
5) EDIT: Look at Brightsky's post below haha -> I've never used half angle formulas before.
Hope I helped.
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