HOW TO FIND UNIT VECTOR PARALLEL TO GIVEN VECTOR

Question 1 :

Find the unit vector parallel to 3a − 2b + 4c if a = 3i − j − 4k, b = −2i + 4j − 3k, and c = i + 2 j − k

Solution :

Let n vector  =  3a − 2b + 4c

Unit normal vector =  n vector/|n vector|

3a - 2b + 4c 

  =  3(3i − j − 4k) - 2(−2i + 4j − 3k) + 4(i + 2j − k)

  =  (9+4+4)i+(-3-8+8)j+(-12+6-4)k

n vector  =  17i-3j-10k

|n vector|  =  √17² + (-3)² + (-10)²

=  √289 + 9 + 100)

=  √398

Unit normal vector  =  (17i-3j-10k)/√398

Question 2 :

The position vectors a vector, b vector, c vector of three points satisfy the relation 2a vector - 7b vector + 5c vector. Are these points collinear?

Solution :

Three distinct points A, B and C with position vectors a vector , b vector and c vector are collinear if and only if there exist real numbers x, y, z, none of them is zero, such that x + y + z = 0 and xa vector + yb vector + zc vector  = 0.

In order to prove a, b and c are collinear, we have to find the sum of coefficient of a, b and c and prove it equal to 0.

x = 2, y = -7, z = 5

x + y +  z  =  2 + (-7) + 5  =  0

Hence the points a, b and c are collinear points.

Question 3 :

The position vectors of the points P, Q, R, S are i + j + k, 2i+ 5j, 3i + 2j − 3k, and i − 6j − k respectively. Prove that the line PQ and RS are parallel

Solution :  

OP vector  =  i + j + k

OQ vector  =  2i+ 5j

OR vector  =  3i + 2j − 3k

OS vector  =  i − 6j − k

PQ  =  OQ - OP

  =  (2i+ 5j) - (i + j + k)

PQ  =  (i + 4j - k)  vector

RS  =  OS - OR

  =  (i − 6j − k) - (3i + 2j − 3k)

  =  (-2i - 8j + 2k)  vector

RS  =  -2(i + 4j - k) vector

PQ  =  -2 RS

Hence PQ and RS are parallel.

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