Question

1. find two integer values of c such that \\(\langle c^2, -7, c + 3\

angle\\) is perpendicular to \\(\langle 1, c, 2\
angle\\).

Explanation:

Step1: Use dot product for perpendicular vectors

The dot product of two perpendicular vectors is 0. For vectors $\langle c^2, -7, c+3
angle$ and $\langle 1, c, 2
angle$, their dot product is:
$$(c^2)(1) + (-7)(c) + (c+3)(2) = 0$$

Step2: Simplify the equation

Expand and combine like terms:
$$c^2 - 7c + 2c + 6 = 0$$
$$c^2 - 5c + 6 = 0$$

Step3: Factor the quadratic equation

Factor the polynomial to solve for $c$:
$$(c-2)(c-3) = 0$$

Step4: Solve for integer values of c

Set each factor equal to 0 and solve:
$c-2=0 \implies c=2$; $c-3=0 \implies c=3$

Answer:

The two integer values are $c=2$ and $c=3$.