Question

Intercepts of the plane \(\vec{r}\cdot\vec{n}=d \ (\ne0)\) on the coordinate axes respectively are:

• A. $\frac{\hat{i}\cdot\vec{n}}{d}, \frac{\hat{j}\cdot\vec{n}}{d}, \frac{\hat{k}\cdot\vec{n}}{d}$
• B. $\left|\frac{\hat{i}\cdot\hat{n}}{d}\right|, \left|\frac{\hat{j}\cdot\vec{n}}{d}\right|, \left|\frac{\hat{k}\cdot\vec{n}}{d}\right|$
• C. $\frac{d}{\hat{i}\cdot\hat{n}}, \frac{d}{\hat{j}\cdot\hat{n}}, \frac{d}{\hat{k}\cdot\hat{n}}$
• D. $\frac{d}{\hat{i}\cdot\vec{n}}, \frac{d}{\hat{j}\cdot\vec{n}}, \frac{d}{\hat{k}\cdot\vec{n}}$

Hint:

To find the $x$-intercept of any vector plane equation, simply set the other coordinates to zero by evaluating $\vec{r} = x\hat{i}$. Substituting this into the plane equation gives $(x\hat{i})\cdot\vec{n} = d \implies x(\hat{i}\cdot\vec{n}) = d \implies x = \frac{d}{\hat{i}\cdot\vec{n}}$. This method allows you to verify the answer choice in a single step!

Answer 1

The Correct Option is D

Concept: To find where a plane intersects the coordinate axes, we convert its equation into the standard Cartesian intercept form: $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$ where $a, b,$ and $c$ represent the explicit intercept lengths along the $x, y,$ and $z$ axes respectively. Step 1: Convert the vector equation into Cartesian form.
Let the normal vector be expressed in terms of its Cartesian components, $\vec{n} = n_x\hat{i} + n_y\hat{j} + n_z\hat{k}$, and the position vector be $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Computing their dot product: $$\vec{r}\cdot\vec{n} = d \quad \Rightarrow \quad n_x x + n_y y + n_z z = d$$

Step 2: Rearrange into standard intercept form.
Divide the entire scalar equation by the constant $d$ to make the right side equal to 1: $$\frac{n_x x}{d} + \frac{n_y y}{d} + \frac{n_z z}{d} = 1$$ Rearrange the fractions to move the coefficients into the denominators: $$\frac{x}{\left(\frac{d}{n_x}\right)} + \frac{y}{\left(\frac{d}{n_y}\right)} + \frac{z}{\left(\frac{d}{n_z}\right)} = 1$$ This shows that the intercepts along the axes are exactly $a = \frac{d}{n_x}, \ b = \frac{d}{n_y},$ and $c = \frac{d}{n_z}$.

Step 3: Express individual normal components using dot products.
We can isolate the individual scalar components of the normal vector $\vec{n}$ by taking the dot product with the respective coordinate unit vectors: $$n_x = \hat{i}\cdot\vec{n}, \quad n_y = \hat{j}\cdot\vec{n}, \quad n_z = \hat{k}\cdot\vec{n}$$

Step 4: Assemble the final vector intercept expressions.
Substitute these dot product expressions back into our denominator equations: $$\text{Intercepts} = \left( \frac{d}{\hat{i}\cdot\vec{n}}, \ \frac{d}{\hat{j}\cdot\vec{n}}, \ \frac{d}{\hat{k}\cdot\vec{n}} \right)$$ This matches option (D) perfectly.