Se d1=3i-2j+4k e d2=-5i+2j-k, determine (d1+d2)x(d1xd2) .

d1 = 3i-2j+4k, v1 = (3, -2, 4)

d2 = -5i+2j-k, v2 = (-5, 2, -1)

(d1+d2)x(d1xd2)

= ((3, -2, 4)+(-5, 2, -1)) x ((3, -2, 4)x(-5, 2, -1))

= (-2, 0, 3) x(-15, -4, -4)

= (30, 0,-12)

Para encontrarmos o resultado da expressão acima, realizaremos os cálculos abaixo:

\(\begin{align} & {{d}_{1}}=3i-2j+4k \\ & {{v}_{1}}=(3,-2,4) \\ & {{d}_{2}}=-5i+2j-k \\ & {{v}_{2}}=(-5,2,-1) \\ & \backslash \backslash \\ & T=({{d}_{1}}+{{d}_{2}})({{d}_{1}}{{d}_{2}}) \\ & T=\left( \left( 3,\text{ }-2,\text{ }4 \right)+\left( -5,\text{ }2,\text{ }-1 \right) \right)\cdot \text{ }\left( \left( 3,\text{ }-2,\text{ }4 \right)x\left( -5,\text{ }2,\text{ }-1 \right) \right) \\ & T=~\left( -2,\text{ }0,\text{ }3 \right)\text{ }\cdot \left( -15,\text{ }-4,\text{ }-4 \right) \\ & T=(30,0,-12) \\ \end{align}\ \)

Portanto, obtemos que \(\boxed{T = \left( {30,0, - 12} \right)}\).