What is A(6i-2j-3k)\times B(2i-4j+4k) ?

The solution to A(6i-2j-3k)\times B(2i-4j+4k) is AB(8j^2+4jk-12k^2-12)+AB(-28j+18k)i

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simplify A(6i-2j-3k)\times B(2i-4j+4k)

simplify

A (6 i - 2 j - 3 k) \cdot B (2 i - 4 j + 4 k)

A B (8 j^{2} + 4 jk - 12 k^{2} - 12) + A B (- 28 j + 18 k) i

Solution steps

A (6 i - 2 j - 3 k) B (2 i - 4 j + 4 k)

Rewrite

6 i - 2 j - 3 k

in standard complex form:

(- 2 j - 3 k) + 6 i

Rewrite

2 i - 4 j + 4 k

in standard complex form:

(- 4 j + 4 k) + 2 i

= A ((- 2 j - 3 k) + 6 i) B ((- 4 j + 4 k) + 2 i)

Apply complex arithmetic rule:

(a + bi) (c + d i) = (a c - b d) + (a d + b c) i

= A (((- 2 j - 3 k) (- 4 j + 4 k) - 6 \cdot 2) + ((- 2 j - 3 k) \cdot 2 + 6 (- 4 j + 4 k)) i) B

Simplify

((- 2 j - 3 k) (- 4 j + 4 k) - 6 \cdot 2) + ((- 2 j - 3 k) \cdot 2 + 6 (- 4 j + 4 k)) i : (8 j^{2} + 4 jk - 12 k^{2} - 12) + (- 28 j + 18 k) i

= A ((8 j^{2} + 4 jk - 12 k^{2} - 12) + (- 28 j + 18 k) i) B

Distribute parentheses

= A B (8 j^{2} + 4 jk - 12 k^{2} - 12) + A B (- 28 j + 18 k) i

\frac{( t ^{4} + 3 t ^{2} ) ^{2}}{9 ( t ^{2} + 1 ) ^{4}}

\frac{( e ^{x} + e ^{- x} ) ^{2}}{( e ^{x} - e ^{- x} ) ^{2}} - 1

(u cos (v) + u sin (v)) (u cos (v) - u sin (v))

F (- 2, - 8) d i rec t r i z - 10

5 (53 - 9)

What is A(6i-2j-3k)\times B(2i-4j+4k) ?
The solution to A(6i-2j-3k)\times B(2i-4j+4k) is AB(8j^2+4jk-12k^2-12)+AB(-28j+18k)i