Answer:
Part 1) Projection of A along B is 19 units
Part 2) Projection of B' along A' is 35 units.
Step-by-step explanation:
The projection of any vector A on another vector B is given by the dot product between the two vectors
Given vector A [tex]\overrightarrow{v_A}=a\widehat{i}+b\widehat{j}+c\widehat{k}[/tex]
And Vector B
[tex]\overrightarrow{v_B}=l\widehat{i}+m\widehat{j}+n\widehat{k}[/tex]
The dot product between them is given by
[tex]\overrightarrow{v_A}\cdot \overrightarrow{v_B}=al+bm+cn[/tex]
Comparing with the given vectors we have
for vector A
a = 1 ,b = -2, c =1
For vector B
l = 4, m = -4, n = 7
Thus the projection of A along B is
[tex]\overrightarrow{v_A}\cdot \overrightarrow{v_B}=4+8+7=19[/tex]
Part b)
For the second case
[tex]\overrightarrow{v_A}=2\widehat{i}+3\widehat{j}+6\widehat{k}[/tex]
and
[tex]\overrightarrow{v_B}=1\widehat{i}+5\widehat{j}+3\widehat{k}[/tex]
Thus the projection of B along A is
[tex]\overrightarrow{v_B}\cdot \overrightarrow{v_A}=2+15+18=35[/tex]
