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Using the Scalar product

The scalar product a.b is defined as a . b = \left| a \right|\;\left| b \right|\;\cos \theta where \theta is angle between a and b. From this definition it can also be shown that a . b = a_x b_x + a_y b_y + a_z b_z. You'll find both these versions on the formula sheet.

The main use of the scalar product is to calculate the angle \theta.

\left| a \right|\;\left| b \right|\;\cos \theta = a . b

\cos \theta = {{a . b} \over {\left| a \right|\;\left| b \right|}} where a . b = a_x b_x + a_y b_y + a_z b_z

Here's a worked example. Calculate the angle \theta on the diagram below.

Angle between two lines

Angle between two lines

 

State the rule you are using for this question

\cos \theta = {{p . q} \over {\left| p \right|\;\left| q \right|}}

Calculate {p . q}

\eqalign{ && p_x q_x + p_y q_y + p_z q_z = \cr && 3 \times 2 + ( - 1) \times 4 + 4 \times 2 \cr && = 10} \cr

Calculate {\left| p \right|} and {\left| q \right|}

\eqalign{ & \left| p \right| = \sqrt {9 + 1 + 16} = \sqrt {26} \cr & \left| q \right| = \sqrt {4 + 16 + 4} = \sqrt {24} \cr}

Substitute

\cos \theta = {{10} \over {\sqrt {26} \sqrt {24} }} = 0.400

Evaluate \theta

\theta = 66.4^\circ

If your answer at the substitution stage works out negative then the angle lies between 90° and 180°. For example, if {\rm{cos}}\;\theta\;{\rm{ = - 0.362}} then \theta\;{\rm{ = 111}}^\circ

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