Respuesta :

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.

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