Given:
The two vectors are:
To find:
The value of 
.
Solution:
We have,
The cross product of these two vectors is:
![\overrightarrow{a}\times \overrightarrow{b}=\hat{i}[(-1)(5)-(1)(-3)]-\hat{j}[(2)(5)-(1)(1)]+\hat{k}[(2)(-3)-(-1)(1)]](https://tex.z-dn.net/?f=%5Coverrightarrow%7Ba%7D%5Ctimes%20%5Coverrightarrow%7Bb%7D%3D%5Chat%7Bi%7D%5B%28-1%29%285%29-%281%29%28-3%29%5D-%5Chat%7Bj%7D%5B%282%29%285%29-%281%29%281%29%5D%2B%5Chat%7Bk%7D%5B%282%29%28-3%29-%28-1%29%281%29%5D)
![\overrightarrow{a}\times \overrightarrow{b}=\hat{i}[-5+3]-\hat{j}[10-1]+\hat{k}[-6+1]](https://tex.z-dn.net/?f=%5Coverrightarrow%7Ba%7D%5Ctimes%20%5Coverrightarrow%7Bb%7D%3D%5Chat%7Bi%7D%5B-5%2B3%5D-%5Chat%7Bj%7D%5B10-1%5D%2B%5Chat%7Bk%7D%5B-6%2B1%5D)
Now the magnitude of the cross product is:
Therefore, the value of is 
.
64 and 28 have a GCF of 4, since 64 = 4 x 16, 28 = 4 x 7, and 7 and 16 have no factors other than 1 in common. Knowing that, we can rewrite 64 + 28 as
4 x 16 + 4 x 7
and then use the distributive property to rewrite it again as
4 x (16 + 7)
Answer:
X = {8n – 7n – 1 : n ϵ N}
Y = {49 (n –1) : n ϵ N} — (1) [all terms divisible by 49]
X = {8n – 7n – 1}
= (1 + 7)n – 7n – 1
= 1 + nC1 7 + nC2 72 + ….. + nCn 7n – 7n – 1
= nC2 72 + …… + nCn 7n
= 49 [nC2 + nC3 7 + …. nCn 7n-2] — (2)
From (1) and (2),
X is divisible by 49.
Y has all multiples of 49.
X ⊂ Y
88=2(4b+6)-4
88=8b+12-4
88=8b+8
-8 -8
80=8b
80/8=10b
3 is positive because it is more than 0. Anything more than zero is positive.
