Correct Answer - Option 1 :
\(\sqrt{3}\)
Concept:
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Vector: A quantity in which both magnitude and direction is considered is called vector.
- Vector follows its own algebra. The calculations are different from normal algebra.
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Resultant or addition of Two Vectors: The resultant of two vectors are given as
\(\overrightarrow{R} =\overrightarrow{A} + \overrightarrow{B}\)
- The Magnitude of the vector is R given as
\(R = \sqrt{A^2 + B^2+ 2ABCos θ }\)
θ is the angle between angles.
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Subtraction of two Vectors: The subtraction of two vectors are given as
\(\overrightarrow{S} =\overrightarrow{A} - \overrightarrow{B}\)
The magnitude of this subtraction vector S is given as
\(S = \sqrt{A^2 + B^2- 2ABCos θ }\)
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Unit Vector: The vector with magnitude one is called the unit vector
Calculation:
Given the sum of two unit vector is a unit vector.
Let \(\overrightarrow{A}\) and \(\overrightarrow{B}\) are unit vectors and θ is the angle between them. Let \(\overrightarrow{R}\) be a there sum which is also a unit vector.
So, A = 1, B = 1, R = 1
\(1 = \sqrt{1^2 + 1^2+2(1)(1)Cos θ }\)
\(\implies 1 = \sqrt{2+ 2Cos θ } \)
Squaring
⇒ 1 = 2 + 2 cosθ
⇒ 2 Cos θ = -1
\(\implies Cos \theta =- \frac{1}{2}\) -- (1)
Subtraction of \(\overrightarrow{A}\) and \(\overrightarrow{B}\) is \(\overrightarrow{S}\) .
\(S = \sqrt{1^2 + 1^2- 2(1)(1)Cos θ }\) -- (2)
Using (1) in (2) we get
\(\implies S = \sqrt{1^2 + 1^2- 2(1)(1)(-\frac{1}{2}) }\)
\(\implies S = \sqrt{1+ 1 + 1 }\)
\(\implies S = \sqrt{3}\)
So, the magnitude of subtraction of vector is \(\sqrt{3}\)
- Usually, unit vector along is direction is represented by using small cap \(\widehat{n}\)
