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Worked solution

Explain why an acute-angle projection has positive sign

Relative to O, A,B have position vectors a,b and c is the position vector of C on AB dividing AB in the ratio 3:1. Given angle AOB is acute, show that the length d of the projection of OC onto OB is d=34∣b∣+a⋅b4∣b∣d=\frac34|\mathbf b|+\frac{\mathbf a\cdot\mathbf b}{4|\mathbf b|}.

CJC · Review problems · Question 13

Answer

d=34∣b∣+a⋅b4∣b∣d=\frac34|b|+\frac{a\cdot b}{4|b|}.

Apply the section formula

c=(a+3b)/(3+1)=14(a+3b)\mathbf c=(\mathbf a+3\mathbf b)/(3+1)=\frac14(\mathbf a+3\mathbf b).

How did this part go?

Justify removing the absolute value

The projection length is d=∣c⋅b^∣d=|\mathbf c\cdot\widehat{\mathbf b}|. Since AOB is acute and OC lies inside that angle, COB is also acute. Equivalently a⋅b>0\mathbf a\cdot\mathbf b>0 and b⋅b>0\mathbf b\cdot\mathbf b>0 show c⋅b>0\mathbf c\cdot\mathbf b>0. Thus d=c⋅b^d=\mathbf c\cdot\widehat{\mathbf b}.

How did this part go?

Expand and simplify

d=14(a+3b)⋅b∣b∣=a⋅b4∣b∣+3b⋅b4∣b∣=a⋅b4∣b∣+3∣b∣24∣b∣=34∣b∣+a⋅b4∣b∣.\begin{aligned}&d\\ & =\frac14(\mathbf a+3\mathbf b)\cdot\frac{\mathbf b}{|\mathbf b|}\\ & =\frac{\mathbf a\cdot\mathbf b}{4|\mathbf b|}+\frac{3\mathbf b\cdot\mathbf b}{4|\mathbf b|}\\ & =\frac{\mathbf a\cdot\mathbf b}{4|\mathbf b|}+\frac{3|\mathbf b|^2}{4|\mathbf b|}\\ & =\frac34|\mathbf b|+\frac{\mathbf a\cdot\mathbf b}{4|\mathbf b|}.\end{aligned}

How did this part go?

Compare your sketch

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Acute angle AOB, with O lower left,B lower right and A above. C lies three quarters along AB from A, so AC:CB=3:1. A dashed perpendicular from C meets OB at F, and OF is the projection length d. Arrows OA and OB show a and b. This is an illustrative acute triangle, not a coordinate specification.
Acute angle AOB, with O lower left,B lower right and A above. C lies three quarters along AB from A, so AC:CB=3:1. A dashed perpendicular from C meets OB at F, and OF is the projection length d. Arrows OA and OB show a and b. This is an illustrative acute triangle, not a coordinate specification.
Acute angle AOB, with O lower left,B lower right and A above. C lies three quarters along AB from A, so AC:CB=3:1. A dashed perpendicular from C meets OB at F, and OF is the projection length d. Arrows OA and OB show a and b. This is an illustrative acute triangle, not a coordinate specification.

Common mistakes

A projection length is nonnegative; a signed scalar projection is not automatically the requested length.

Solution notes

Complete source proof and its ratio/projection diagram retained. No marks are printed.

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