For what values of c are the vectors < 6, c, 2 > and <c, c2,c > orthogonal ?
Recall that if u⊥v then u•v = 0
So, if "c2" means c^2, then just solve
6c + c^3 + 2c = 0
To determine the values of c for which the given vectors are orthogonal, we need to find when their dot product is equal to zero.
The dot product of two vectors can be found by multiplying their corresponding components and summing the results.
Let's calculate the dot product of the given vectors:
< 6, c, 2 > · < c, c^2, c >
= (6 * c) + (c * c^2) + (2 * c)
= 6c + c^3 + 2c
To find when the dot product is equal to zero, we set the equation equal to zero and solve for c:
6c + c^3 + 2c = 0
Combining like terms:
c^3 + 8c = 0
Factoring out c:
c(c^2 + 8) = 0
Setting each factor equal to zero:
c = 0 or c^2 + 8 = 0
For the first factor, c = 0.
For the second factor, we solve for c^2:
c^2 = -8
Taking the square root of both sides (considering both positive and negative square roots):
c = ± √(-8)
Since the square root of a negative number is not defined in the real number system, there are no real solutions for c in this case.
Therefore, the values of c for which the given vectors are orthogonal are:
c = 0
To determine the values of c for which the given vectors are orthogonal, we need to check if their dot product is zero.
The dot product of two vectors < a, b, c > and < d, e, f > is given by the formula:
a * d + b * e + c * f
In this case, the dot product of the two vectors < 6, c, 2 > and < c, c^2, c > would be:
(6 * c) + (c * c^2) + (2 * c)
Simplifying this expression, we have:
6c + c^3 + 2c
Combining like terms, we get:
c^3 + 8c
To find the values of c for which the dot product is zero (i.e., the vectors are orthogonal), we need to solve the equation:
c^3 + 8c = 0
Factoring out a common factor of c, we have:
c(c^2 + 8) = 0
Setting each factor equal to zero, we get:
c = 0 or c^2 + 8 = 0
For the first case, c = 0 satisfies the equation. Therefore, one value of c for which the vectors are orthogonal is c = 0.
For the second case, we need to solve the quadratic equation:
c^2 + 8 = 0
Subtracting 8 from both sides, we have:
c^2 = -8
Taking the square root of both sides, we get:
c = ±√(-8)
Since the square root of a negative number is not real, there are no real solutions for c when c^2 + 8 = 0.
Therefore, the only value of c for which the vectors < 6, c, 2 > and < c, c^2, c > are orthogonal is c = 0.