For what values of b are the vectors ⟨−39,b,3⟩ and ⟨b, b2,b⟩ orthogonal?

Step-by-step 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.