A) If ā= 2i + 3j + 4k and 5 = i- j + k, Find
(b) à x 5
(ii) Sine of the angle between these vectors.
(iii) Unit vector perpendicular to each vector

Looks like you're given

a = 2 i + 3 j + 4 k

b = i - j + k

Recall the properties of the cross product:

i x i = j x j = k x k = 0

i x j = k

j x k = i

k x i = j

and for any two vectors u and v,

u x v = -(v x u)

(i) Compute the cross product:

a x b = (2 i + 3 j + 4 k) x (i - j + k)

a x b = 2 (i x i) + 3 (j x i) + 4 (k x i)

  - 2 (i x j) - 3 (j x j) - 4 (k x j)

  + 2 (i x k) + 3 (j x k) + 4 (k x k)

a x b = (3 + 4) (j x k) + (4 - 2) (k x i) + (-2 - 3) (i x j)

a x b = 7 i + 2 j - 5 k

(ii) Recall the formula,

||a x b|| = ||a|| ||b|| sin(θ)

where θ is the angle between the vectors a and b.

We have

||a x b|| = √(7^2 + 2^2 + (-5)^2) = √78

||a|| = √(2^2 + 3^2 + 4^2) = √29

||b|| = √(1^2 + (-1)^2 + 1^2) = √3

so that

sin(θ) = √(78/87)

(iii) The cross product a x b is already perpendicular to both a and b, so you just need to normalize this vector. This is done by dividing it by its magnitude:

(a x b)/||a x b|| = 7/√78 i + 2/√78 j - 5/√78 k

The slope of AB is different from the slope of BC is NOT TRUE.

Step-by-step explanation:

AB and BC are on the same straight line, so their slopes are the same.