#### Question

Find the angle between two vectors \[\vec{a} \text{ and } \vec{b}\] if

\[\left| \vec{a} \right| = \sqrt{3}, \left| \vec{b} \right| = 2 \text{ and } \vec{a} \cdot \vec{b} = \sqrt{6}\]

#### Solution

\[\text{ Given that }\]

\[\left| \vec{a} \right| = 2, \left| \vec{b} \right| = 3 \text{ and } \vec{a} . \vec{b} = 4 . . . \left( 1 \right)\]

\[\text{ We know that }\]

\[ \left| \vec{a} - \vec{b} \right|^2 = \left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 - 2 \vec{a} . \vec{b} \]

\[ = 2^2 + 3^2 - 2 \left( 4 \right) \left[ \text{ Using } \left( 1 \right) \right]\]

\[ = 4 + 9 - 8\]

\[ = 5\]

\[ \therefore \left| \vec{a} - \vec{b} \right| = \sqrt{5}\]

Is there an error in this question or solution?

Solution Find the Angle Between Two Vectors → a and → B If | → a | = √ 3 , ∣ ∣ → B ∣ ∣ = 2 and → a ⋅ → B = √ 6 Concept: Multiplication of a Vector by a Scalar.