Question

Calculate the cross product assuming that u × w = 0, 1, 3 . (2u + 3w) × w

Answer 1

Answer- (2u + 3w) × w =  < 0, 2, 6 >

Solution -

Given in the problem

(u × w) = < 0, 1, 3 >

Then,

(2u + 3w) × w = (2u × w) + (3w × w)

(2u × w) = 2(u × w) = < 0×2, 1×2, 3×2 > = < 0, 2, 6 >  ( ∵ Multiplying it by 3 with all components)

(3w × w) = 3(w × w) = 0                                               ( ∵ The cross product of any vector with itself is 0)

∴ (2u + 3w) × w =  < 0, 2, 6 >

Answer 2

The cross product $(2u \,+ \,3w) \,\times \,w$ is $(0,2,6)$.

Let be $u\times w = (0, 1, 3)$, we proceed to simplify $(2u \,+ \,3w) \,\times \,w$ by algebraic means:

1) $u\times w = (0, 1, 3)$  Given.

2) $(2u \,+ \,3w) \,\times \,w$  Given.

3) $2u \,\times \,w + 3w\,\times \,w$ Distributive property.

4) $2u\,\times\,w + O$ Nilpotency property.

5) $2\cdot (u\,\times \,w) + O$ Scalar multiplication.

6) $2\cdot (0, 1, 3) + (0,0,0)$ Definition of zero vector, 1)

7) $(0, 2, 6) + (0,0,0)$ Scalar multiplication.

8) $(0,2,6)$ Vector sum/Result.

The cross product $(2u \,+ \,3w) \,\times \,w$ is $(0,2,6)$.

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