Intercepts Calculation Using Vector Equation

The plane equation is given as $\vec{r} \cdot \vec{n} = d$, where $d \neq 0$. $\vec{r}$ is the position vector ($x\hat{i} + y\hat{j} + z\hat{k}$) and $\vec{n}$ is the normal vector.

Cartesian Form Conversion

Substituting $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ into the plane equation:

$ \left( x\hat{i} + y\hat{j} + z\hat{k} \right) \cdot \vec{n} = d $

Using the distributive property of the dot product:

$ x(\hat{i} \cdot \vec{n}) + y(\hat{j} \cdot \vec{n}) + z(\hat{k} \cdot \vec{n}) = d $

Note: To align with the provided options, we assume $\vec{n}$ represents the unit normal vector, $\hat{n}$. Thus, the equation is treated as:

$ x(\hat{i} \cdot \hat{n}) + y(\hat{j} \cdot \hat{n}) + z(\hat{k} \cdot \hat{n}) = d $

Coordinate Axis Intercepts Derivation

Intercepts are determined by finding the point where the plane crosses each axis. This is done by setting the other two coordinates to zero.

• x-intercept: Set $y=0$ and $z=0$.

  $ x(\hat{i} \cdot \hat{n}) = d $ $ x = \frac{d}{\hat{i} \cdot \hat{n}} $

• y-intercept: Set $x=0$ and $z=0$.

  $ y(\hat{j} \cdot \hat{n}) = d $ $ y = \frac{d}{\hat{j} \cdot \hat{n}} $

• z-intercept: Set $x=0$ and $y=0$.

  $ z(\hat{k} \cdot \hat{n}) = d $ $ z = \frac{d}{\hat{k} \cdot \hat{n}} $

Intercepts Match Option C

The calculated intercepts are $\frac{d}{\hat{i} \cdot \hat{n}}$, $\frac{d}{\hat{j} \cdot \hat{n}}$, and $\frac{d}{\hat{k} \cdot \hat{n}}$. This corresponds exactly to Option C.