**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.

After having gone through the stuff given above, we hope that the students would have understood,"How to Find Unit Vector Parallel to Given Vector"

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