Respuesta :
Answer:
Step-by-step explanation:
we know
[tex]\vec{i}\times \vec{j}=\vec{k}[/tex]
[tex]\vec{j}\times \vec{k}=\vec{i}[/tex]
[tex]\vec{k}\times \vec{i}=\vec{j}[/tex]
(a)[tex]\left [ \left ( \hat{i}+\hat{j}\right )\times \hat{i}\right ]\times \hat{j}[/tex]
[tex]=\left [ \hat{i}\times \hat{i}+\hat{j}\times \hat{i}\right ]\times \hat{j}[/tex]
[tex]=\left [ 0-\hat{k}\right ]\times \hat{j}[/tex]
[tex]=\hat{i}[/tex]
(b)[tex]\left ( \hat{j}+\hat{k}\right )\times \left ( \hat{j}\times \hat{k}\right )[/tex]
[tex]=\left ( \hat{j}+\hat{k}\right )\times \left ( \hat{i}\right )[/tex]
[tex]=\hat{k}+\hat{j}[/tex]
(c)[tex]4\hat{i}\times \left ( \hat{i}+\hat{j}\right )[/tex]
[tex]=4\hat{i}\times \hat{i}+4\hat{i}\times \hat{j}[/tex]
[tex]=0+4\hat{k}[/tex]
(d)[tex]\left ( \hat{k}+\hat{j}\right )\times \left ( \hat{k}-\hat{j}\right )[/tex]
[tex]=\hat{k}\times \hat{k}-\hat{k}\times \hat{j}+\hat{j}\times \hat{k}-\hat{j}\times \hat{j}[/tex]
[tex]=0+\hat{i}+\hat{i}-0[/tex]
[tex]=2\hat{i}[/tex]
The direction of the cross product depends on the right-hand rule and the resulting cross-product is located on the plane that is perpendicular to the vectors undergoing the cross product.
Taking i, j, k as the unit vector along x, y, z-direction.
Since [tex]i \times i = 0[/tex] it implies the angle∠ between them is 0;
Then: sin(0) = 0
Also;
[tex]j \times j = 0 \\ \\ k\times k = 0[/tex]
Similarly, [tex]i \times j = k[/tex] which implies that the angle ∠ between them = 90°;
Then: sin (90°) = 1
Also;
[tex]j \times k= i \\ \\ k \times i = j[/tex]
[tex]j \times i = - k[/tex] which implies that the angle ∠ between them = -90° or 270°
Then; sin (-90) or Sin ( 270) = -1
Also;
[tex]i \times k = -j \\ \\ j \times i = -k[/tex]
As such i, j, k are unit vectors for x, y, and z-axis.
To determine the following calculations, we have;
(a)
[tex]\Big ( ( i^{\to }+ j^{\to })\times i^{\to }\Big)\times j^{\to }[/tex]
[tex]= ( i^{\to } \times i^{\to } + j^{\to }\times i^{\to }) \times j^{\to }[/tex]
[tex]=(0 - k^{\to })\times j^{\to }[/tex]
[tex]= - k^{\to } \times j^{\to }\\\\= -(-i^{\to }) \\ \\ \mathbf{= i^{\to }}[/tex]
(b)
[tex](j^{\to }+ k^{\to }) \times (j^{\to } + k^{\to })[/tex]
[tex]= j^{\to } \times (j^{\to } \times k ) + k^{\to } \times (j \times k)\\ \\ = j^{\to } \times i^{\to } + k^{\to } \times i^{\to } \\ \\ \mathbf{ = -k^{\to } + j^{\to }}[/tex]
(c)
[tex]4i^{\to } \times (j \times k^{\to }) \\ \\ = 4 (i^{\to } \times i^{\to }) \\ \\ \mathbf{= 0}[/tex]
(d)
[tex](k^{\to} + j^{\to}) \times (k^{\to} - j^{\to})) \\ \\ =(k^{\to} \times k^{\to}) - (k^{\to} \times j^{\to})+(j^{\to} + k^{\to}) - (j^{\to} \times j^{\to}) \\ \\ = 0 + i^{\to} + i^{\to} -0 \\ \\ \mathbf{ = 2 i^{\to}}[/tex]
Therefore, we can conclude that the calculations of the cross-product are well defined from the above explanations.
Learn more about cross-product here:
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