Try application problems involving the dot product.
Cosine of an Angle in a Triangle
Example ①
For A(1, 1), B(4, 2), and C(2, 5), find the cosine of angle A.
Solution: →AB=(3, 1), →AC=(1, 4). Dot product=7, |→AB|=√10, |→AC|=√17. cos A=7/√170
For A(1, 1), B(4, 2), and C(2, 5), find the cosine of angle A.
Solution: →AB=(3, 1), →AC=(1, 4). Dot product=7, |→AB|=√10, |→AC|=√17. cos A=7/√170
Parameter t and the Dot Product
Example ②
For →a=(2, 1) and →b=(1, 3), find t such that →a + t→b is perpendicular to →a.
Solution: |→a|²+t(→a・→b)=5+5t=0 → t=−1
For →a=(2, 1) and →b=(1, 3), find t such that →a + t→b is perpendicular to →a.
Solution: |→a|²+t(→a・→b)=5+5t=0 → t=−1
Projection
The length of the projection of →a in the direction of →b is (→a・→b)/|→b|.
Example ③
Find the length of the projection of →a=(4, 2) in the direction of →b=(3, 1).
Solution: →a・→b=14, |→b|=√10. Projection=7√10/5
Find the length of the projection of →a=(4, 2) in the direction of →b=(3, 1).
Solution: →a・→b=14, |→b|=√10. Projection=7√10/5
Key Points
- Angle in a triangle: cos = (→AB・→AC)/(|→AB||→AC|)
- Perpendicularity condition → an equation for t
- Projection length: (→a・→b)/|→b|
Practice Problems
- For A(0,0), B(5,0), and C(3,4), find the cosine of angle B.
- For →a=(3, 1) and →b=(1, 2), find t such that →a+t→b is perpendicular to →b.
- For →a=(6, 8) and →b=(1, 0), find the length of the projection of →a in the direction of →b.
Answers and Explanations
- Answer: √5/5
Explanation: →BA=(−5,0), →BC=(−2,4). Dot product 10, |→BA|=5, |→BC|=2√5. cos B=1/√5=√5/5. - Answer: t=−1
Explanation: →a・→b=5, |→b|²=5. 5+5t=0. - Answer: 6
Explanation: →a・→b=6, |→b|=1.