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Any vector in an arbitrary direction can be replaced by two or three vectors
If we slide a vector parallel to its position it will be the same as before. Any vector in an arbitrary direction can always be replaced by two (or three) arbitrary vectors which have the original vector as their resultant.
The position vector of midpoint of joining the points (2, – 1, 3) and (4, 3, –5) is :
The position vector of point P = 2i  j + 3k
Position Vector of point Q = 4i + 3j  5k
The position vector of R which divides PQ in half is given by:
r = (2i  j + 3k + 4i + 3j  5k)/2
r = (6i + 2j  2k)/2
r = 3i + j  k
The Position vector of a point (12,n) is such that = 13 then n =
Position vector of a =12i+nj
a=√144+n^2=13
squaring
144+n^2=169
n^2=25
n=±5
The vector joining the points A(2, – 3, 1) and B(1, – 2, – 5) directed from B to A is:
Initial coordinates = i  2j 5k
Final coordinates = 2i  3j + k
Final  Initial = [ (21)i + (3+2) j + (1+5)k ]
= i  j + 6k.
ABCD is a parallelogram. If coordinates of A,B,C are (2,3), (1,4) and (0, 2). Coordinates of D =
As ABCD is a parallelogram, to find the fourth coordinate add the adjacent coordinate and then subtract opposite coordinate.
like D = A + C  B
= (2 + 0  1, 3 + (2) 4)
= (1,3)
The position vectors of the end points of diameter of a circle are and , then the position vector of the centre of the circle is:
{(1+5)î +(13)j + (11)k} / 2
= {6i  2j + 0k}/2
= 3i  j
a= ij+2k b= 2i+3j+k
2b = 4i + 6j + 2k
a  2b = (i  j + 2k)  (4i + 6j + 2k)
= 3i 7j + 0k
a  2b = [(3)^{2} + (7)^{2} + (0)^{2}]^{½}
= [9 + 49]^{½}
⇒ [58]^{1/2}
The points with position vectors are collinear vectors, Value of a =
Position vector A = 60i+3j
Position vector B = 40i8j
Position vector C = aj52j
Now, find vector AB and BC
AB = 20i11j
BC= (a40)i44j
To be collinear, angle between the vector AB and BC made by the given position vectors should be 0 or 180 degree.
That’s why the cross product of the vectors should be zero
ABXBC=(20i11j)X(a40)i44j
0i+0j+(880+11(a40))=0
a40= 80
a=40
Therefore, a should be 40 to be the given positions vectors collinear.
The distance between the point (2, 3, 1) and (–1, 2, – 3) is:
is correct because the direction between the A&C is opposite, thats why negative sign is in between the BC and CA.
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