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junior college 1 | H2 Maths
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Mason
Mason

junior college 1 chevron_right H2 Maths chevron_right Singapore

I’m having trouble doing Part 2 of the question. I have been trying different ways but cannot break through.

Date Posted: 10 months ago
Views: 150

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J
J's answer
1022 answers (A Helpful Person)
1st
More explanation in the comments
J
J
10 months ago
length of line segment OE = the length of projection of vector a onto b
= |a|cosθ

Length of line segment BE

= length of line segment OB - length of line segment OE

= Magnitude of vector b - |a|cosθ

= |b|- |a|cosθ


The length of projection of vector c onto b is actually the length of line segment OF.

But triangles OCF and BAE are congruent (you can use RH congruency to prove)

OF and BE are congruent sides and equal in length. So OF = BE = |b|- |a|cosθ

(You could rotate the parallelogram about its centre and OF will map onto BE exactly)
J
J
10 months ago
As for why there is a modulus sign flanking |b| - |a|cosθ,

We need to consider the case whereby OB is the shorter diagonal of the parallelogram and is shorter than the length of projection of vector a onto b.

i.e |a|cosθ > |b|
Which means |b|- a|cosθ| is negative.

(This also means ∠OBA is obtuse and OA is longer than OB)

The length of projection of vector c onto b would fall outside the triangle BCO, and would be congruent to the difference between the length of projection of vector a onto b and OB

(Since triangles OAB and BCO are congruent)

Length of projection of vector c onto b
= a|cosθ| - b

So, the modulus sign is there to ensure the absolute (non-negative) value is obtainedand take care of the possible negative sign of |b|- a|cosθ|
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J
J's answer
1022 answers (A Helpful Person)
The other possibility