Question

The magnitude of two vectors are 3 and 4 units and their dot product is 6 units. The angle
between the vectors is what

Answer 1

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✦ Required Answer:

❇ GiveN:

• Magnitude of two vectors are 3 and 4 units.
• Dot product of these vectors is 6 units.

❇ To FinD:

• The angle between the vectors....?

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✦ How to Solve?

The only concept we need to know for solving this question is the Dot product of two vectors. Let's know it's definition and meaning.

$\Large{ \rm{ \underline{ \underline{ \purple{Dot \: product \: of \: two \: vectors}}}}}$

The dot product of two vectors $\large{ \rm{ \overrightarrow{A}}}$ and $\large{ \rm{ \overrightarrow{B}}}$ is equal to the product of magnitudes of the two vectors and the cosine of the smaller angle between them. That is,

$\large{ \boxed{ \rm{ \red{ \overrightarrow{A}. \overrightarrow{B} = AB \cos( \theta) }}}}$

✏The resultant of dot product is a scalar quantity, hence dot product is also known as Scalar product.

By applying this formula, let's find the answer of this Q.

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✦ Solution:

Here,

• A = 3
• B = 4
• $\large{ \rm{ \overrightarrow{ A}}}. \large{ \rm{ \overrightarrow{ B}}}$=6 units

⏺Let the angle between the vectors be θ.

By using formula,

$\large{ \rm{ \longrightarrow \: \large{ \rm{ \overrightarrow{ A}}}. \large{ \rm{ \overrightarrow{ B}}} = AB \cos( \theta) }} \\ \\ \large{ \rm{ \longrightarrow \: 6 = 3 \times 4 \cos( \theta) }} \\ \\ \large{ \rm{ \longrightarrow \: \cos( \theta) = \cancel{\frac{6}{3 \times 4} }}} \\ \\ \large{ \rm{ \longrightarrow \: \cos( \theta) = \frac{1}{2} }} \\ \\ \large{ \rm{ \longrightarrow \: \theta = \cos {}^{ - 1} ( \frac{1}{2} ) }} \\ \\ \large{ \rm{ \longrightarrow \: \theta = \boxed{ \rm{ \purple{60 \degree}}}}} \\ \\ \large{ \therefore{ \underline{ \underline{ \green{ \rm{Hence, \: solved \: \dag}}}}}}$

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