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2.2 Operations on Algebraic Vectors
Consider the two vectors, π’ = π,π,π and π£ =[π₯,π¦,π§] and the scalar πββ. Vector Addition π’ + π£ = π+π₯,π+π¦,π+π§ Proof of Vector Addition: π’ + π£ =π π +π π +π π +π₯ π +π¦ π +π§ π = π+π₯ π + π+π¦ π +(π+π§) π = π+π₯,π+π¦,π+π§ Scalar Multiplication π π’ = ππ,ππ,ππ
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Example 1: If π’ = 2,3,β1 and π£ =[0,1,4] find π€ =5 π’ β2 π£ .
Solution: π€ =5 π’ β2 π£ =5 2,3,β1 β2[0,1,4] = 10,15,β5 β 0,2,8 = 10,13,β13 *So basically just like βregularβ algebra! Example 2: Given π =3 π β3 π + π and π = π +2 π , calculate β2 π + π . Solution: β2 π + π =β2 3 π β3 π + π + π +2 π =β6 π +6 π β2 π + π +2 π =β5 π +6 π β΄ β2 π + π = β =61
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Example 3: Given points π΄(1,β4) and π΅ 3,β1 , find π΄π΅ .
Solution: We can form the two position vectors ππ΄ =[1,β4] and ππ΅ =[3,β1] then, π΄π΅ = π΄π + ππ΅ =β ππ΄ + ππ΅ = β1,4 +[3,β1] =[2,3] * [3-1, -1-(-4)] π π΅ π΄ In general when given two points π΄ π π₯ , π π¦ and B π π₯ , π π¦ then π΄π΅ = π π¦ β π π₯ , π π¦ β π π₯ i.e. the distance between the two points.
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Example 4: Using vectors, show that the points π β1,β3 , π 0,2 and π
(3,17) are collinear. Solution: ππ = 0β β1 ,2β β3 =[1,5] and ππ
= 3β0,17β2 =[3,15] Recall that two vectors, π and π are parallel iff there exists a πββ such that π =π π . ππ =3 ππ
β ππ ππ
β΄π, π, πππ π
are collinear.
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Example 5: If π΄ 1,β5,2 and π΅(β3,4,4) are opposite vertices of a parallelogram ππ΄ππ΅, with π at the origin, find the coordinates of π. Solution: Let π π,π,π . π΅π = ππ΄ = [1,β5,2] and π΄π = ππ΅ = β3,4,4 π΄π = πβ1, π+5, πβ2 =[β3,4,4] πβ1=β3 π+5=4 πβ2=4 π=β π=β π=6 β΄π(β2,β1,6).
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Example 6: Find the point on the π§βaxis that is equidistant from the points π΄(β5,2,1) and π΅(2,β6,3). Solution: Let π be the point on the π§βaxis that is equidistant form the points π΄(β5,2,1) and π΅(2,β6,3), then the π₯ and π¦ coordinates are 0 so π is 0,0,π§ . ππ΄ =(β5,2,1βπ§) and ππ΅ =(2,β6,3βπ§)
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