What is the projection of #<-5,3,7 ># onto #<0,8,-2 >#?

Let #veca=〈0,8,-2〉#

and #vecb=〈-5,3,7〉#

The projection of #vecb# onto #veca# is

#=(veca.vecb)/(∥veca∥^2)veca#

#veca.vecb=〈0,8,-2〉.〈-5,3,7〉=0*-5+8*3*-2*7=0+24-14=10#

Then, we calculate the modulus of #veca#

#∥veca∥=∥〈0,8,-2〉∥=sqrt(0+64+4)=sqrt68#

The ptojection is #=10/68〈0,8,-2〉=5/34〈0,8,-2〉#

#=〈0,40/34,-10/34〉#

#=〈0,20/17,-5/17〉#

To find the projection of vector \(<-5, 3, 7>\) onto vector \(<0, 8, -2>\), you can use the formula for the projection of one vector onto another: \[\text{proj}_\text{v}(u) = \left(\frac{u \cdot v}{\|v\|^2}\right) v\] Where: \(u\) is the vector being projected (\(<-5, 3, 7>\)), \(v\) is the vector onto which \(u\) is projected (\(<0, 8, -2>\)), \(\cdot\) denotes the dot product of vectors, and \(\|v\|\) denotes the magnitude (or length) of vector \(v\). 1. Calculate the dot product of \(u\) and \(v\): \[<-5, 3, 7> \cdot <0, 8, -2> = (-5)(0) + (3)(8) + (7)(-2) = 24\] 2. Calculate the magnitude of vector \(v\): \[ \|<0, 8, -2>\| = \sqrt{0^2 + 8^2 + (-2)^2} = \sqrt{68} = 2\sqrt{17}\] 3. Plug the values into the formula: \[\text{proj}_\text{v}(u) = \left(\frac{24}{(2\sqrt{17})^2}\right) <0, 8, -2>\] \[= \left(\frac{24}{68}\right) <0, 8, -2>\] \[= \left(\frac{6}{17}\right) <0, 8, -2>\] \[= <0, \frac{48}{17}, -\frac{12}{17}>\] So, the projection of \(<-5, 3, 7>\) onto \(<0, 8, -2>\) is \(<0, \frac{48}{17}, -\frac{12}{17}>\).