How will you prove that three vectors are mutually perpendicular?

The scalar triple product for vectors a, b, c is

a • (b x c)

First, let

e = b x c (Eq 0)

then, recall from the definition of the cross product

|e| = |b x c| = |b||c| sinβ (Eq 1)

where: β = angle between b & c

Next, recall from the definition of the dot product

a • e = |a||e| cosα (Eq 2)

where: α = angle between a & e

Equating Eq0, Eq1 & Eq2

a • (b x c) = |a||b||c|cosα sinβ (Eq3)

Our goal is to show

a • (b x c) = |a||b||c|

and from Eq3, it's obvious that the only way this is possible is if

cosα sinβ = 1

and THIS is only possible if

cosα =1

sinβ =1

which implies

α = 0

β = π/2

So, the angle between b & c = π/2, thus they are perpendicular. By definition, e = (b x c) is perpendicular to both b & c. And the angle between a & e is 0, so a is perpendicular to b & c. Thus, a, b, c are mutually perpendicular when a•(bxc) = |a||b||c|.

Are you confident you typed the question wisely? If OB and OC are seen to be vectors then OC won't equivalent a million/3 OB. For one vector to be a relentless diverse of yet another they might desire to be parallel yet they have been defined to be perpendicular. If OB and OC are line segments from the beginning to the factors B and C respectively, then there is not any longer something proscribing the size of those line segments. the only advice given is they are perpendicular. you're able to make the line segments as long (or short) as you like and that they're going to nevertheless be perpendicular. So it should not be provable. OA isn't even used for something. Why is it there?