Answer:
Step-by-step explanation:
First, let's find A x (B+C):
A x (B+C) = A x B + A x C (distributive property of cross product over vector addition)
Now, we need to find A x B and A x C separately:
A x B = (3i + 3j - 2k) x (i - 4j + 2k) (using the cross product formula)
= (31 + 32) i - (34 - 21) j + (3*(-2) - 3*1) k
= 9i + 10j - 9k
Similarly,
A x C = (3i + 3j - 2k) x (4i - j + 2k)
= (34 - 2(-1)) i - (32 + 31) j + (31 - 34) k
= 14i - 9j - 9k
Now, we can substitute these values into the original equation:
A x (B+C) = A x B + A x C
(3i + 3j - 2k) x [(i - 4j + 2k) + (4i - j + 2k)] = (9i + 10j - 9k) + (14i - 9j - 9k)
Simplifying the right side:
(9i + 10j - 9k) + (14i - 9j - 9k) = 23i + 1j - 18k
Now, let's calculate the left side:
(3i + 3j - 2k) x [(i - 4j + 2k) + (4i - j + 2k)]
= (3i + 3j - 2k) x (5i - 5j + 4k)
= (35 + 24) i - (3*(-5) + 34) j + (35 - 2*(-5)) k
= 23i + 1j + 15k
Therefore, we have:
23i + 1j - 18k = 23i + 1j + 15k
Since the left and right sides are equal, we have shown that:
A x (B+C) = A x B + A x C.