Question

Solve the following
r1=3i−2j+kr2=2i−4j−3kr3=−i+2j+2kFindr2r3

-16
16
15
-15

Answer 1

Answer on Question #46940 – Math – Vector Calculus

Question.

Solve the following r1=3i-2j+k, r2=2i-4j-3k, r3=-i+2j+2k. Find r2r3

-16

16

15

-15

Solution.

$$r_1 = (3, -2, 1)$$ $$r_2 = (2, -4, -3)$$ $$r_3 = (-1, 2, 2)$$ $$r_2 r_3 = 2 \cdot (-1) + (-4) \cdot 2 + (-3) \cdot 2 = -2 -8 -6 = -16$$

Answer: -16

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