If A,B,C and D are any four points so that,vector AB.CD + vector BC.AD + vector CA.BD=0.

Given, ABCD is a parallelogram

P is point of intersection of the diagonals

O is any point

Let OA,OB,OC,OD and OP be drawn.

Then we have,

Now,

( For a parallelogram P bisects the two diagonal.)

Solution

Given vectors are

a= (3,-1,-4), b=(-2,4,-3) and c=(-5,7,-1)

a-2b+c = (3,-1,-4) -2(-2,4,-3)+(-5,7,-1)

= (3,-1,-4) -(-4,8,-6)+(-5,7,-1)

={3-(-4)+(-5) , -1-8+7, -4-(-6)+(-1)}

= (2,-2, 1)

I) For unit vector

| a-2b+c |= =

II) For direction cosines of the line represented by a-2b+c

=

Given vectors are

a= 3i+j-k

b= xi-4j+4k

since they are collinear, the relation of their coordinates is given as:

Taking 1^st and 2^nd ratio we get,