Question

For what values of t, if any, is given vector parallel to vector u = (4,-1): (8t, 2t)? and (1,t^2)?

Answer 1

the vector would have to be a scalar of (4,-1) in order to be parallel to it
so
 4 = 8t   -1 = 2t

t= 4/8 = 1/2
-1=2t
t= -1/2

Answer 2

For t=±1/2 given vectors will be parallel to vector u.

Given vectors are:

(4,-1)

(8t, 2t)

(1, t²)

What are parallel vectors?

If two vectors are parallel, the angle between them is zero.

If the vector (4, -1) is parallel to vector (8t, 2t), the angle between these vectors should be zero.

Similarly, If the given vector (4, -1) is parallel to vector (1, t²), the angle between these vectors also should be zero.

The angle between vector u and vector (8t, 2t) = Angle between vector u and vector (1, t²).

So, $\frac{32t-2t}{\sqrt{17} \sqrt{68} t} =\frac{4-t^{2} }{\sqrt{17} \sqrt{1+t^4} }$

$\frac{30}{\sqrt{68} } =\frac{4-t^{2} }{\sqrt{1+t^4} }$

$t=+-\frac{1}{2}$

Hence, for t=±1/2 given vectors will be parallel to vector u.

To get more about vectors visit:

https://brainly.com/question/25705666