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junior college 1 | H2 Maths
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vectors
Date Posted: 1 year ago
Views: 217
(i)
Given the vectors are perpendicular
a.(2a + b) = 0
2a.a + a.b = 0
2(2)² + 2|b|cos(3π/4) = 0
|b| = 4√2
(ii)
Length of projection
|a.b|/|b| = 8/(4√2) = √2 units
(iii)
OP = 3/5 OB
Area of ΔAPB
= Area of ΔOAB - Area of ΔOAP
= 1/2 |a × b| - 1/2 |a × p|
= 1/2 |a × b| - 1/2 × 3/5 |a × b|
= 1/2 (1 - 3/5) |a × b|
= 1/5 (2)(4√2)sin(3π/4)
= 8/5 units²
Given the vectors are perpendicular
a.(2a + b) = 0
2a.a + a.b = 0
2(2)² + 2|b|cos(3π/4) = 0
|b| = 4√2
(ii)
Length of projection
|a.b|/|b| = 8/(4√2) = √2 units
(iii)
OP = 3/5 OB
Area of ΔAPB
= Area of ΔOAB - Area of ΔOAP
= 1/2 |a × b| - 1/2 |a × p|
= 1/2 |a × b| - 1/2 × 3/5 |a × b|
= 1/2 (1 - 3/5) |a × b|
= 1/5 (2)(4√2)sin(3π/4)
= 8/5 units²
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Date Posted:
1 year ago
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