Chapter 11 - Introduction to Three Dimensional Geometry

The distance of a point $(x, y, z)$ from the yz-plane is given by $|x|$. Here, the point is $(3, 4, 5)$, so the distance is $|3| = 3$ units.

The length of the perpendicular from $P(x, y, z)$ to the y-axis is $\sqrt{x^2 + z^2}$. Here, it is $\sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$. The options provided (41, 34, 5) seem to represent the square of the distance. Since $34$ is the square of the actual length, option (B) is the intended answer in this context.

The distance from the origin is $\sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$. Similar to the previous question, the options represent the square of the distance. Option (A) 50 is the square of $\sqrt{50}$.

Let the distance be $\sqrt{27}$ (assuming the square root symbol was omitted in the question text, consistent with the pattern of squared values). Then distance$^2 = 27$. Using the distance formula: $(a-0)^2 + (0-1)^2 + (1-2)^2 = 27 \Rightarrow a^2 + 1 + 1 = 27 \Rightarrow a^2 = 25 \Rightarrow a = \pm 5$.

The x-axis is the line where $y=0$ and $z=0$. The xy-plane is $z=0$ and the xz-plane is $y=0$. Thus, their intersection is the x-axis.

On the y-axis, the coordinates are of the form $(0, y, 0)$. Thus, both $x$ and $z$ coordinates are zero. The equations are $x=0$ and $z=0$.

The signs of the coordinates are $(-,-,-)$. This corresponds to the 7th octant (VII).

The yz-plane is perpendicular to the x-axis. Any plane parallel to the yz-plane will also be perpendicular to the x-axis.

The condition $y=0$ and $z=0$ defines the x-axis.

The equation $x=0$ represents the yz-plane.

The length of the diagonal is the distance between the two points: $\sqrt{(5-3)^2 + (8-6)^2 + (10-8)^2} = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{4+4+4} = \sqrt{12} = 2\sqrt{3}$.

The foot of the perpendicular from $(x, y, z)$ on the xy-plane is $(x, y, 0)$. For $P(3, 4, 5)$, it is $(3, 4, 0)$. This is not in the options (A), (B), or (C).

The foot of the perpendicular from $(x, y, z)$ on the x-axis is $(x, 0, 0)$. For $(3, 4, 5)$, it is $(3, 0, 0)$.

Diagonal = $\sqrt{10^2 + 13^2 + 8^2} = \sqrt{100 + 169 + 64} = \sqrt{333}$.

$\sqrt{(a-1)^2 + (2-(-1))^2 + (1-1)^2} = 5 \Rightarrow (a-1)^2 + 9 = 25 \Rightarrow (a-1)^2 = 16 \Rightarrow a-1 = \pm 4$. So $a=5$ or $a=-3$.

The centroid of $\triangle ABC$ is the same as the centroid of the triangle formed by the midpoints of its sides. $G = (\frac{1+3-1}{3}, \frac{2+0+1}{3}, \frac{-3+1-4}{3}) = (1, 1, -2)$.