3. Determine that the vectors u=[t, 4, 2t+1] and v=[t+2, 1-t, -1] are perpendicular, find the possible values of the contant, t.

Same as previous question,

http://www.jiskha.com/display.cgi?id=1337914331

Except that the dot-product equated to zero results in a quadratic equation.

So solve for possible values of t by solving the quadratic.

t(t+2) + 4(1-t) - (2t+1) = 0

t = 1,3

like that?

Correct!

thanks

To determine if two vectors are perpendicular, their dot product must be zero. In this case, we need to find the dot product of vectors u and v and set it equal to zero:

u • v = (t)(t+2) + (4)(1-t) + (2t+1)(-1) = t^2 + 2t + 4 - 4t - 2 + 2t + 1 = t^2 - t + 3 = 0

Now, we need to solve the quadratic equation t^2 - t + 3 = 0 to find the values of t that make the two vectors perpendicular.

Using the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:

x = (-b ± √(b^2 - 4ac))/(2a)

For our equation t^2 - t + 3 = 0, a = 1, b = -1, and c = 3.

Using the quadratic formula:

t = (-(-1) ± √((-1)^2 - 4(1)(3)))/(2(1))
t = (1 ± √(1 - 12))/2
t = (1 ± √(-11))/2

Since the square root of a negative number is not real, there are no real values of t that make vectors u and v perpendicular.