A=10ax−4ay+6az and B=2ax+ay
a) If A=Axax+Ayay+Azaz , then the component of A along ay is Ay .
For A=10ax−4ay+6az we have that Ay=−4 is the component of A along ay.
b) 3A−B=3(10ax−4ay+6az)−(2ax+ay)=(3×10−2)ax+(3×(−4)−1)ay+3×6az
3A−B=28ax−13ay+18az
∣3A−B∣=282+(−13)2+182=1277=35.74
c) A+2B=10ax−4ay+6az+2(2ax+ay)=(10+2×2)ax+(−4+2)ay+6az
A+2B=14ax−2ay+6az
A unit vector along A+2B is vector 142+(−2)2+6214ax−2ay+6az=23614ax−2ay+6az≈0.91ax−0.13ay+0.39az
Answer: a) −4 ; b) 35.74 ; c) 0.91ax−0.13ay+0.39az
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