Find the cross product (7,9,6) x (-4,1,5). Is the resulting vector perpendicular to the given vectors

Expand each vector as linear combinations of the standard basis vectors:

(7, 9, 6) = 7(1, 0, 0) + 9(0, 1, 0) + 6(0, 0, 1)

(-4, 1, 5) = -4(1, 0, 0) + (0, 1, 0) + 5(0, 0, 1)

For brevity, write

i = (1, 0, 0)

j = (0, 1, 0)

k = (0, 0, 1)

Then by definition of the cross product,

i x i = j x j = k x k = (0, 0, 0)

i x j = k

j x k = i

k x i = j

and for any two vectors a and b, we have a x b = - b x a.

Now compute the product:

(7i + 9j + 6k) x (-4i + j + 5k)

= -28 (i x i) - 36 (j x i) - 24 (k x i)

...   + 7 (i x j) + 9 (j x j) + 6 (k x j)

...   + 35 (i x k) + 45 (j x k) + 30 (k x k)

= -36 (-k) - 24 j + 7 k + 6 (-i) + 35 (-j) + 45 i

= 39 i - 59 j + 43 k

which is the same as the vector

(39, -59, 43)

And yes, this vector is perpendicular to both given vectors.

the aswere is the one with all 8's and the second one is with the 10,8 and 6

step-by-step explanation:

the terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. a recursive formula for a geometric sequence with common ratio r is given by an=ran−1 a n = r a n − 1 for n≥2 n ≥ 2 .

step-by-step explanation: