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EXAMPLE 6 Find the scalar projection and vector projection of b = (4, 2, 1) onto a = (-2, 4, 3). SOLUTION Since $|a| = \sqrt{(-2)^2 + 4^2 + 3^2} = \sqrt{29}$, the scalar projection of b onto a is $comp_a b = \frac{a \cdot b}{|a|} = \frac{(-2)(4) + 4(2) + 3(1)}{\sqrt{29}}$ $= \frac{3\sqrt{29}}{29}$ The vector projection is this scalar projection times the unit vector in the direction of a, below. $proj_a b = \frac{3\sqrt{29}}{29} \cdot \frac{a}{|a|} = \frac{3\sqrt{29}}{29} \cdot a$

          EXAMPLE 6 Find the scalar projection and vector projection of b = (4, 2, 1) onto a = (-2, 4, 3).
SOLUTION Since $|a| = \sqrt{(-2)^2 + 4^2 + 3^2} = \sqrt{29}$, the scalar projection of b onto a is
$comp_a b = \frac{a \cdot b}{|a|} = \frac{(-2)(4) + 4(2) + 3(1)}{\sqrt{29}}$
$= \frac{3\sqrt{29}}{29}$
The vector projection is this scalar projection times the unit vector in the direction of a, below.
$proj_a b = \frac{3\sqrt{29}}{29} \cdot \frac{a}{|a|} = \frac{3\sqrt{29}}{29} \cdot a$
        
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EXAMPLE 6 Find the scalar projection and vector projection of b = (4, 2, 1) onto a = (-2, 4, 3).
SOLUTION Since |a| = √((-2)^2 + 4^2 + 3^2) = √(29), the scalar projection of b onto a is
compa b = (a · b)/(|a|) = ((-2)(4) + 4(2) + 3(1))/(√(29))
= (3√(29))/(29)
The vector projection is this scalar projection times the unit vector in the direction of a, below.
proja b = (3√(29))/(29)·(a)/(|a|) = (3√(29))/(29)· a

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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EXAMPLE 6: Find the scalar projection and vector projection of b = 4,2,1 onto a = -2,4,3. SOLUTION: Since |a| = √((-2)^2 + 4^2 + 3^2) = √(4 + 16 + 9) = √29, the scalar projection of b onto a is: proj.b = (b · a) / |a| = (4*(-2) + 2*4 + 1*3) / √29 = (-8 + 8 + 3) / √29 = 3 / √29. The vector projection is this scalar projection times the unit vector in the direction of a, below: proj.b = (3 / √29) * (a / |a|) = (3 / √29) * (-2/√29, 4/√29, 3/√29) = (-6/29, 12/29, 9/29).
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Transcript

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00:01 Hi, in this question, govern that b equals 3 ,1 ,1 on to a equals minus 4 ,2 ,3.
00:14 We have to find the scalar projection.
00:17 So, first we have to find modulus of a which is equal to square root of minus 4 the whole square plus 2 the whole square plus 3 square which is equal to square root of 16 plus 4 plus 9 which is equal to square root of 29.
00:37 Here component of ab equals a dot b divided by modulus a.
00:41 On multiplying we get 3 into minus 4 minus 12, 1 into 2, 2, 1 into 3, 3 divided by square root of 29 which is equal to minus 7 divided by square root of 29 which can be written as minus 7 square root of 29 divided by 29.
01:02 Next we have to find the projection of ab which is equal to component ab a divided by modulus a which can be written as here minus 7 square root of 29 divided by 29 into a divided by modulus a...
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