Vector and Plane Geometry Problems: Find q if A and B are perpendicular, find the normal vector to a plane, and simplify an expression involving arctan.

Let $\mathbf{A} = 2\mathbf{i} - q\mathbf{j} - \mathbf{k}$ and $\mathbf{B} = -2\mathbf{i} - 2\mathbf{j} + q\mathbf{k}$.

If two vectors are perpendicular, their dot product is zero ($\mathbf{A} \cdot \mathbf{B} = 0$).

The correct option is B.
The scalar $q$ is $\boxed{\text{B. } 4}$.

Q34.

For a plane given by the equation $Ax + By + Cz + D = 0$, the vector perpendicular to the plane (normal vector) is $\mathbf{N} = A\mathbf{i} + B\mathbf{j} + C\mathbf{k}$.

The equation of the plane is $3x - 5y + z + 7 = 0$.
Here, $A=3$, $B=-5$, and $C=1$.

The vector perpendicular to the plane is $3\mathbf{i} - 5\mathbf{j} + 1\mathbf{k}$.
The correct option is B.
The vector perpendicular to the plane is $\boxed{\text{B. } 3\mathbf{i} - 5\mathbf{j} + \mathbf{k}}$.

Q35.

The formula for $\tan(x+y)$ is $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$.
Let $x = \tan^{-1} 5$ and $y = \tan^{-1} (-3)$. Then $\tan x = 5$ and $\tan y = -3$.

The correct option is C.
Then $\tan A = \boxed{\text{C. } \frac{1}{8}}$.

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