Answer:
b = 0, -6 and 6
Explanation:
Given the vectors ⟨−39,b,3⟩ and ⟨b,b²,b⟩, for the vectors to be orthogonal, the dot product of both vectors must be equal to zero.
Taking the dot product of both vectors and equating it t zero we will have;
⟨−39,b,3⟩*⟨b,b²,b⟩ = -39(b)+ b(b²)+3(b)
(Note that when multiplying, we multiply component wise and take the sum of the product if each component)
Since the vectors are orthogonal, then -39(b)+ b(b²)+3(b) = 0
open the parenthesis
-39b+b³+3b = 0
on rearranging
b³+3b-39b= 0
b³-36b= 0
factor out b from the equation
b(b²-36) = 0
b = 0 and b²-36 = 0
from b²-36 = 0
add 36 to both sides of the equation
b²-36+36= 0+36
b² = 36
take the square root of both sides
√b² = ±√36
b = ±6
Hence the values of b for the vectors to be orthogonal are 0, -6 and 6.