is (-1, sec X, tan x) a unit vector?

Question

if yes why and if not why?

if it's not how do u know it's not a unit vector?

how to prove it's not a unit vector?

Answer 1

A unit vector satisfies the equation

|| (x, y, z) || = 1

To prove that it's a unit vector, all we have to do is prove that

|| (-1, sec(x), tan(x) ) || is equal to 1.
If it's NOT equal to 1, then it is not a unit vector.

By definition,

|| (x, y, z) || = √(x² + y² + z²)

|| (-1, sec(x), tan(x) || = √( [-1]² + sec²(x) + tan²(x) )

= √( 1 + sec²(x) + tan²(x) )

By the identity tan²(x) + 1 = sec²(x), we get

= √( sec²(x) + sec²(x) )
= √(2sec²(x) )
= √(2) sec(x)

And as you can see, this is not equal to 1.
Therefore, it is not a unit vector.

To make it into a unit vector, we divide each element by
√(2) sec(x).

(-1, sec(x), tan(x) ) is not a unit vector, BUT

(-1/[√(2)sec(x)] , sec(x)/[√(2)sec(x)], tan(x)/[√(2)sec(x)]) IS a unit vector.

Just for fun, let's reduce that. Skipping some details, the above should reduce to

( (-1/√(2))cos(x) , 1/√(2) , (1/√(2))sin(x) )

Answer 2

Doesn't appear to be one.

A unit vector must have a norm of 1, that is, x^2+y^2+z^2=1.

Here, your norm is

1^2+sec^2(X)+tan^2(X)

and using the trig identity 1+tan^2X=sec^2X,

1^2+sec^2(X)+tan^2(X) = 2sec^2X,

which is not a unit vector for any real X.

Good luck, work hard, and stay away from drugs.

Answer 3

magnitude = √[(-1)² + sec²x + tan²x]
= √[ 1 + tan²x + sec²x
= √ [ 2.sec²x ]
= √2.sec x
magnitude does not = 0
Is not a unit vector.

Answer 4

No. This is because its modulus is
sqrt((-1)^2 + (secx)^2 + (tanx)^2)
= sqrt(1 + (secx)^2 + (tanx)^2)
= sqrt(2(secx)^2)
= secx*sqrt2 > 1 for all x.

Answer 5

no and thear is no reason y not