How to Find Unit Vector Parallel to Given Vector :
Here we are going to see 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| = √17^{2} + (-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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