# Introduction to Three Dimensional Geometry - Class 11 - Mathematics

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## Extra Questions - Introduction to Three Dimensional Geometry | NCERT | Mathematics | Class 11

The distance of the plane passing through $(1,1,1)$ and perpendicular to the line $\frac{x-1}{3}=\frac{y-1}{0}=\frac{z-1}{4}$ from the origin is

A. $\frac{3}{4}$

B. $\frac{4}{5}$

C. $\frac{7}{5}$

D. 0

The correct answer is **Option C: $\frac{7}{5}$**.

To find the equation of a plane that passes through a given point $(1, 1, 1)$ and is perpendicular to the line defined by $\frac{x-1}{3}=\frac{y-1}{0}=\frac{z-1}{4}$, we first need to identify the normal vector to the plane. The direction ratios of the line give us the components of the normal vector directly, which are $(3, 0, 4)$.

The general equation for a plane with a normal vector $(A, B, C)$ passing through the point $(x_1, y_1, z_1)$ is: $$ A(x-x_1) + B(y-y_1) + C(z-z_1) = 0 $$

Substituting the normal vector and the point through which the plane passes, we get: $$ 3(x-1) + 0(y-1) + 4(z-1) = 0 $$

Expanding and simplifying this equation, we derive the equation of the plane: $$ 3x + 4z - 7 = 0 $$

To find the distance (*d*) of this plane from the origin (0,0,0) using the formula:
$$
d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}
$$
where $(x_0, y_0, z_0)$ is the point (0,0,0), and $Ax + By + Cz + D = 0$ is the equation of the plane, we substitute $A = 3$, $B = 0$, $C = 4$, and $D = -7$:
$$
d = \frac{|3 \cdot 0 + 0 \cdot 0 + 4 \cdot 0 - 7|}{\sqrt{3^2 + 0^2 + 4^2}}
= \frac{|-7|}{\sqrt{9 + 16}}
= \frac{7}{5}
$$

Thus, the distance of the plane from the origin is $\frac{7}{5}$, making **Option C** correct.