What is the projection of ## onto ##?

Question

What is the projection of #<-2,4,2 ># onto #<1,8,-3 >#?

Adrian Ayala · Accepted Answer

The projection is #=12/37<1,8,-3>#

Allow the vectors to

#vecu=<-2,4,2>#

additionally

#vecv=<1,8,-3>#

The projection of #vecu# onto #vecv# is

#proj_(vecv)vecu=(vecu.vecv)/(||vecv||^2)vecv#

#=(<-2,4,2>.<1,8,-3>)/(||<1,8,-3>||^2)<1,8,-3>#

#=(-2+32-6)/(1+64+9)<1,8,-3>#

#=24/74<1,8,-3>#

#=12/37<1,8,-3>#

Sebastian Acosta · Answer

undefined To find the projection of vector $ \mathbf{a} $ onto vector $ \mathbf{b} $, we use the formula: $$ \text{proj}_{\mathbf{b}}(\mathbf{a}) = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{b} \|^2} \mathbf{b} $$ Given vectors $ \mathbf{a} = <-2, 4, 2> $ and $ \mathbf{b} = <1, 8, -3> $, we can calculate their dot product as $ \mathbf{a} \cdot \mathbf{b} = (-2)(1) + (4)(8) + (2)(-3) = 30 $. The magnitude of vector $ \mathbf{b} $ is $ \| \mathbf{b} \| = \sqrt{1^2 + 8^2 + (-3)^2} = \sqrt{74} $. Using the formula, the projection of $ \mathbf{a} $ onto $ \mathbf{b} $ is: $$ \text{proj}_{\mathbf{b}}(\mathbf{a}) = \frac{30}{74} <1, 8, -3> \approx <0.4054, 3.243, -1.216> $$

Charles Aeschliman · Answer

undefined To find the projection of one vector onto another, you can use the formula: $$ \text{proj}_{\mathbf{v}}(\mathbf{u}) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} \mathbf{v} $$ Given vectors $ \mathbf{u} = <-2, 4, 2> $ and $ \mathbf{v} = <1, 8, -3> $, first calculate the dot product of $ \mathbf{u} $ and $ \mathbf{v} $, and then the magnitude of $ \mathbf{v} $. Finally, use the formula to find the projection: $$ \mathbf{u} \cdot \mathbf{v} = (-2)(1) + (4)(8) + (2)(-3) $$ $$ \| \mathbf{v} \| = \sqrt{1^2 + 8^2 + (-3)^2} $$ $$ \text{proj}_{\mathbf{v}}(\mathbf{u}) = \frac{(-2)(1) + (4)(8) + (2)(-3)}{\sqrt{1^2 + 8^2 + (-3)^2}^2} <1, 8, -3> $$ Calculate these values to find the projection vector.

What is the projection of #&lt;-2,4,2 &gt;# onto #&lt;1,8,-3 &gt;#?

What is the projection of #<-2,4,2 ># onto #<1,8,-3 >#?