Dot Product — Question 8

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Question 8

Prove the identity ∥u→+v→∥2=∥u→∥2+2u→⋅v→+∥v→∥2\|\vec u+\vec v\|^2=\|\vec u\|^2+2\vec u\cdot\vec v+\|\vec v\|^2.

Original worksheet page 1: question and worked solution for 5-3-008
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Question 8 – Solution

Compute the dot product carefully, then choose the formula that matches the question. A zero, positive, or negative dot product also gives immediate geometric information.

See the diagram in the original worksheet below.

Start with the squared norm: ∥u→+v→∥2=(u→+v→)⋅(u→+v→)\|\vec u+\vec v\|^2=(\vec u+\vec v)\cdot(\vec u+\vec v).

Distribute the dot product to obtain u→⋅u→+u→⋅v→+v→⋅u→+v→⋅v→\vec u\cdot\vec u+\vec u\cdot\vec v+\vec v\cdot\vec u+\vec v\cdot\vec v.

Use symmetry and u→⋅u→=∥u→∥2\vec u\cdot\vec u=\|\vec u\|^2 to get the stated identity. Proved\boxed{\text{Proved}}

The result follows from the defining vector formulas used above, and each component, magnitude, or scalar condition has been checked against the information in the question.

The dot product converts two vectors into a scalar measuring directional agreement: positive means generally aligned, zero means perpendicular, and negative means generally opposed.

Original worksheet page 2: question and worked solution for 5-3-008

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