
The rectangular components of a vector are $\left( {2,2} \right)$. The corresponding rectangular components of another vector are $\left( {1,\sqrt 3 } \right)$. Find the angle between the two vectors.
A. $\dfrac{{5\pi }}{{12}}$
B. $\dfrac{\pi }{{12}}$
C. $\dfrac{{3\pi }}{{12}}$
D. $\dfrac{\pi }{{24}}$
Answer
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Hint: Here we have to use the formula of dot product between two vectors to get the answer.
The vector sections of the vertical and horizontal vector are rectangular elements. Rectangular elements are perpendicular to one another.
Complete step by step answer:
We can resolve a vector into various parts. For the most part there are three segments of vector viz. Here we will examine just two parts X-segment and Y-segment which are opposite to one another. These segments are called rectangular segments of vectors.
A vector unit along any vector is a vector whose direction is the same as the direction of the vector, but whose magnitude is unity is known as a rectangular vector unit.
Any part of a two-dimensional vector is known as a component. The vector components represent the vector’s effect in a given direction. The cumulative influence of both components is equal to the influence of a single two-dimensional vector.
The method of splitting the vector into its components is called resolving of the components. In reality, it is more useful to describe a vector in components that are at right angles to each other, typically horizontal and vertical.
Let vector $\overrightarrow A = 2i + 2j$
And vector $\overrightarrow B = i + \sqrt 3 j$
According to dot product of two vectors we get:
$
\overrightarrow A. \overrightarrow B = AB\cos \theta \\
\Rightarrow \left( {2i + 2j} \right)\left( {i + \sqrt 3 j} \right) = 2\sqrt 2 \times 2 \times \cos \theta \\
\Rightarrow 2 + 2\sqrt 3 = 4\sqrt 2 \cos \theta \\
\Rightarrow \dfrac{{1 + \sqrt 3 }}
{{2\sqrt 2 }} = \cos \theta \\
\Rightarrow \dfrac{1}
{2} \times \dfrac{1}
{{\sqrt 2 }} + \dfrac{1}
{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}
{2} = \cos \theta \\
\Rightarrow \cos \left( {60 - 45} \right) = \cos \theta \\
\Rightarrow \cos {15^ \circ } = \cos \theta \\
\therefore \theta = {15^ \circ } = \dfrac{\pi }
{{12}} \\
$
So, the correct answer is “Option B”.
Note:
Here we have to remember the trigonometric rules and also how to do dot products. Here we have taken the complex form of the vectors to find the answer.
The corresponding vector sum is the addition of two or more vectors. It is the result of two or more vectors being added together. If two or more velocity vectors are applied, the result is the corresponding velocity.
The vector sections of the vertical and horizontal vector are rectangular elements. Rectangular elements are perpendicular to one another.
Complete step by step answer:
We can resolve a vector into various parts. For the most part there are three segments of vector viz. Here we will examine just two parts X-segment and Y-segment which are opposite to one another. These segments are called rectangular segments of vectors.
A vector unit along any vector is a vector whose direction is the same as the direction of the vector, but whose magnitude is unity is known as a rectangular vector unit.
Any part of a two-dimensional vector is known as a component. The vector components represent the vector’s effect in a given direction. The cumulative influence of both components is equal to the influence of a single two-dimensional vector.
The method of splitting the vector into its components is called resolving of the components. In reality, it is more useful to describe a vector in components that are at right angles to each other, typically horizontal and vertical.
Let vector $\overrightarrow A = 2i + 2j$
And vector $\overrightarrow B = i + \sqrt 3 j$
According to dot product of two vectors we get:
$
\overrightarrow A. \overrightarrow B = AB\cos \theta \\
\Rightarrow \left( {2i + 2j} \right)\left( {i + \sqrt 3 j} \right) = 2\sqrt 2 \times 2 \times \cos \theta \\
\Rightarrow 2 + 2\sqrt 3 = 4\sqrt 2 \cos \theta \\
\Rightarrow \dfrac{{1 + \sqrt 3 }}
{{2\sqrt 2 }} = \cos \theta \\
\Rightarrow \dfrac{1}
{2} \times \dfrac{1}
{{\sqrt 2 }} + \dfrac{1}
{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}
{2} = \cos \theta \\
\Rightarrow \cos \left( {60 - 45} \right) = \cos \theta \\
\Rightarrow \cos {15^ \circ } = \cos \theta \\
\therefore \theta = {15^ \circ } = \dfrac{\pi }
{{12}} \\
$
So, the correct answer is “Option B”.
Note:
Here we have to remember the trigonometric rules and also how to do dot products. Here we have taken the complex form of the vectors to find the answer.
The corresponding vector sum is the addition of two or more vectors. It is the result of two or more vectors being added together. If two or more velocity vectors are applied, the result is the corresponding velocity.
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