Test: Properties Of Vectors - JEE MCQ

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 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

Position vector of a =12i+nj
|a|=√144+n^2=13
squaring
144+n^2=169
n^2=25
n=±5

Initial coordinates = i - 2j -5k
Final coordinates = 2i - 3j + k
Final - Initial = [ (2-1)i + (-3+2) j + (1+5)k ] 
= i - j + 6k.

 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)

{(1+5)î +(1-3)j + (1-1)k} / 2
= {6i - 2j + 0k}/2
= 3i - j

a= i-j+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

Position vector A = 60i+3j
Position vector B = 40i-8j
Position vector C = aj-52j
Now, find vector AB and BC
AB = -20i-11j
BC= (a-40)i-44j
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=(-20i-11j)X(a-40)i-44j
0i+0j+(880+11(a-40))=0
a-40= -80
a=-40
Therefore, a should be -40 to be the given positions vectors collinear.

is correct because the direction between the A&C is opposite, thats why negative sign is in between the BC and CA.