The angle between vectors $A = 2i - j + 2k$ and $B = 6i - 3j + 6k$ is:

Question

The integer value printed by the ANSI-C program given below is ________.


int funcp(){
    static int x = 1;
    x++;
    return x;
}

int main(){
    int x,y;
    x = funcp();
    y = funcp() + x;
    printf("%d\n", (x+y));
    return 0;
}

Answer 1

Step 1: A quick way to test whether two vectors are parallel is to check whether their cross product is the zero vector, since the cross product magnitude is $|A||B|\sin\theta$, which vanishes only when $\theta = 0$ or $180$ degrees.
Step 2: Compute the cross product for $A = (2,-1,2)$ and $B = (6,-3,6)$:\[A \times B = \begin{vmatrix} i & j & k \\ 2 & -1 & 2 \\ 6 & -3 & 6 \end{vmatrix}\]
Step 3: Expand along the top row. The i-component is $(-1)(6) - (2)(-3) = -6+6 = 0$. The j-component is $-[(2)(6)-(2)(6)] = 0$. The k-component is $(2)(-3)-(-1)(6) = -6+6 = 0$. So $A \times B = (0,0,0)$.
Step 4: A zero cross product confirms the vectors are parallel. Since $B = 3A$ with a positive multiplier, both vectors point in the same direction, ruling out 180 degrees.
Step 5: Therefore the angle between A and B is exactly 0 degrees, ruling out the nonzero options 30, 45 and 60.\[\boxed{\theta = 0^\circ}\]

The angle between vectors \(A = 2i - j + 2k\) and \(B = 6i - 3j + 6k\) is:

Show Hint

The Correct Option is A

Solution and Explanation

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