Answer:
Therefore the given vectors are orthogonal for b = 0,±3.
Step-by-step explanation:
If [tex]\vec a[/tex] and [tex]\vec b[/tex] are two vectors orthogonal, then the dot product of [tex]\vec a[/tex] and [tex]\vec b[/tex] will be zero.
i.e [tex]\vec a. \vec b =0[/tex]
If [tex]\vec a = x_1\hat i+y_1\hat j +z_1\hat k[/tex] and [tex]\vec b = x_2\hat i+y_2\hat j +z_2\hat k[/tex]
[tex]\vec a. \vec b=( x_1\hat i+y_1\hat j +z_1\hat k).( x_2\hat i+y_2\hat j +z_2\hat k)[/tex]
[tex]=x_1 x_2+y_1y_2+z_1z_2[/tex]
Given two vectors are (-18,b,9) and (b,b²,b)
Let
[tex]\vec P= -18 \hat i+b\hat j +9 \hat k[/tex]
and
[tex]\vec Q = b \hat i+b^2 \hat j +b\hat k[/tex]
Therefore,
[tex]\vec P.\vec Q[/tex]
[tex]=( -18 \hat i+b\hat j +9 \hat k).( b \hat i+b^2 \hat j +b\hat k)[/tex]
=(-18).b+b.b²+9.b
= -18b+b³+9b
= b³-9b
Since [tex]\vec P[/tex] and [tex]\vec Q[/tex] are orthogonal. Then [tex]\vec P.\vec Q[/tex] = 0.
Therefore,
b³-9b= 0
⇒b(b²-9)=0
⇒b =0 or b²=9
⇒b=0 or b =±3
Therefore the given vectors are orthogonal for b = 0,±3.