Answer:
[tex]\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\[/tex]
Step-by-step explanation:
Given A = 5i + 11j – 2k and B = 4i + 7k, the vector projection of B unto a is expressed as [tex]proj_ab = \dfrac{b.a}{||a||^2} * a[/tex]
b.a = (5i + 11j – 2k)*( 4i + 0j + 7k)
note that i.i = j.j = k.k =1
b.a = 5(4)+11(0)-2(7)
b.a = 20-14
b.a = 6
||a|| = √5²+11²+(-2)²
||a|| = √25+121+4
||a|| = √130
square both sides
||a||² = (√130)
||a||² = 130
[tex]proj_ab = \dfrac{6}{130} * (5i+11j-2k)\\\\proj_ab = \frac{30}{130} i+\frac{11}{130} j-\frac{12}{130} k\\\\proj_ab = \frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\\\[/tex]
Hence the projection of b unto a is expressed as [tex]\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\[/tex]
