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2.2 Operations on Algebraic Vectors

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1 2.2 Operations on Algebraic Vectors
Consider the two vectors, 𝑒 = π‘Ž,𝑏,𝑐 and 𝑣 =[π‘₯,𝑦,𝑧] and the scalar π‘˜βˆˆβ„. Vector Addition 𝑒 + 𝑣 = π‘Ž+π‘₯,𝑏+𝑦,𝑐+𝑧 Proof of Vector Addition: 𝑒 + 𝑣 =π‘Ž 𝑖 +𝑏 𝑗 +𝑐 π‘˜ +π‘₯ 𝑖 +𝑦 𝑗 +𝑧 π‘˜ = π‘Ž+π‘₯ 𝑖 + 𝑏+𝑦 𝑗 +(𝑐+𝑧) π‘˜ = π‘Ž+π‘₯,𝑏+𝑦,𝑐+𝑧 Scalar Multiplication π‘˜ 𝑒 = π‘˜π‘Ž,π‘˜π‘,π‘˜π‘

2 Example 1: If 𝑒 = 2,3,βˆ’1 and 𝑣 =[0,1,4] find 𝑀 =5 𝑒 βˆ’2 𝑣 .
Solution: 𝑀 =5 𝑒 βˆ’2 𝑣 =5 2,3,βˆ’1 βˆ’2[0,1,4] = 10,15,βˆ’5 βˆ’ 0,2,8 = 10,13,βˆ’13 *So basically just like β€œregular” algebra! Example 2: Given π‘Ž =3 𝑖 βˆ’3 𝑗 + π‘˜ and 𝑏 = 𝑖 +2 π‘˜ , calculate βˆ’2 π‘Ž + 𝑏 . Solution: βˆ’2 π‘Ž + 𝑏 =βˆ’2 3 𝑖 βˆ’3 𝑗 + π‘˜ + 𝑖 +2 π‘˜ =βˆ’6 𝑖 +6 𝑗 βˆ’2 π‘˜ + 𝑖 +2 π‘˜ =βˆ’5 𝑖 +6 𝑗 ∴ βˆ’2 π‘Ž + 𝑏 = βˆ’ =61

3 Example 3: Given points 𝐴(1,βˆ’4) and 𝐡 3,βˆ’1 , find 𝐴𝐡 .
Solution: We can form the two position vectors 𝑂𝐴 =[1,βˆ’4] and 𝑂𝐡 =[3,βˆ’1] then, 𝐴𝐡 = 𝐴𝑂 + 𝑂𝐡 =βˆ’ 𝑂𝐴 + 𝑂𝐡 = βˆ’1,4 +[3,βˆ’1] =[2,3] * [3-1, -1-(-4)] 𝑂 𝐡 𝐴 In general when given two points 𝐴 π‘Ž π‘₯ , π‘Ž 𝑦 and B 𝑏 π‘₯ , 𝑏 𝑦 then 𝐴𝐡 = 𝑏 𝑦 βˆ’ 𝑏 π‘₯ , π‘Ž 𝑦 βˆ’ π‘Ž π‘₯ i.e. the distance between the two points.

4 Example 4: Using vectors, show that the points 𝑃 βˆ’1,βˆ’3 , 𝑄 0,2 and 𝑅(3,17) are collinear. Solution: 𝑃𝑄 = 0βˆ’ βˆ’1 ,2βˆ’ βˆ’3 =[1,5] and 𝑄𝑅 = 3βˆ’0,17βˆ’2 =[3,15] Recall that two vectors, π‘Ž and 𝑏 are parallel iff there exists a π‘˜βˆˆβ„ such that π‘Ž =π‘˜ 𝑏 . 𝑃𝑄 =3 𝑄𝑅 β‡’ 𝑃𝑄 𝑄𝑅 βˆ΄π‘ƒ, 𝑄, π‘Žπ‘›π‘‘ 𝑅 are collinear.

5 Example 5: If 𝐴 1,βˆ’5,2 and 𝐡(βˆ’3,4,4) are opposite vertices of a parallelogram 𝑂𝐴𝑃𝐡, with 𝑂 at the origin, find the coordinates of 𝑃. Solution: Let 𝑃 π‘Ž,𝑏,𝑐 . 𝐡𝑃 = 𝑂𝐴 = [1,βˆ’5,2] and 𝐴𝑃 = 𝑂𝐡 = βˆ’3,4,4 𝐴𝑃 = π‘Žβˆ’1, 𝑏+5, π‘βˆ’2 =[βˆ’3,4,4] π‘Žβˆ’1=βˆ’3 𝑏+5=4 π‘βˆ’2=4 π‘Ž=βˆ’ 𝑏=βˆ’ 𝑐=6 βˆ΄π‘ƒ(βˆ’2,βˆ’1,6).

6 Example 6: Find the point on the π‘§βˆ’axis that is equidistant from the points 𝐴(βˆ’5,2,1) and 𝐡(2,βˆ’6,3). Solution: Let 𝑃 be the point on the π‘§βˆ’axis that is equidistant form the points 𝐴(βˆ’5,2,1) and 𝐡(2,βˆ’6,3), then the π‘₯ and 𝑦 coordinates are 0 so 𝑃 is 0,0,𝑧 . 𝑃𝐴 =(βˆ’5,2,1βˆ’π‘§) and 𝑃𝐡 =(2,βˆ’6,3βˆ’π‘§)


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