Dot Product — Question 10

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Question 10

A spotlight beam points along d→=⟨2,−1,2⟩\vec d=\langle2,-1,2\rangle. A surface has unit normal n→=13⟨1,2,2⟩\vec n=\frac13\langle1,2,2\rangle. Find the magnitude of the component of d→\vec d normal to the surface.

Original worksheet page 1: question and worked solution for 5-3-010
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Question 10 – 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.

Because n→\vec n is a unit vector, the normal component magnitude is |d→⋅n→||\vec d\cdot\vec n|.

Compute d→⋅n→=(2−2+4)/3=4/3\vec d\cdot\vec n=(2-2+4)/3=4/3.

Therefore the requested magnitude is 4/3\boxed{4/3}. The absolute value makes the result independent of the chosen normal orientation.

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 5-3-010

Original worksheet layout. Use Enlarge or open the PDF for a closer view.