Derive r⃗·n̂ = d and convert to Cartesian form ax + by + cz = d.

Equation of Plane in Normal Form

Question

A plane is at a perpendicular distance of 5 units from the origin. The unit normal to the plane is $\hat{n} = \frac{1}{3}(2\hat{i} - \hat{j} + 2\hat{k})$ . Write the vector equation of the plane and convert it to Cartesian form.

Solution — Step by Step

Any plane can be described by two things: the direction perpendicular to it (the normal) and how far it sits from the origin. If $\hat{n}$ is the unit normal and $d$ is that perpendicular distance, then for any point $\vec{r}$ on the plane, the component of $\vec{r}$ along $\hat{n}$ must equal $d$ .

This gives us the vector equation directly: $\vec{r} \cdot \hat{n} = d$ .

We’re given $\hat{n} = \frac{1}{3}(2\hat{i} - \hat{j} + 2\hat{k})$ . Let’s confirm it’s actually a unit vector:

|\hat{n}| = \frac{1}{3}\sqrt{2^2 + (-1)^2 + 2^2} = \frac{1}{3}\sqrt{4 + 1 + 4} = \frac{1}{3} \times 3 = 1 \checkmark

Always verify this — if $\hat{n}$ is not a unit vector, the " $d$ " you get won’t be the perpendicular distance.

We have $\hat{n} = \frac{1}{3}(2\hat{i} - \hat{j} + 2\hat{k})$ and $d = 5$ .

\vec{r} \cdot \frac{1}{3}(2\hat{i} - \hat{j} + 2\hat{k}) = 5

This is the vector equation of the plane.

Substitute $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and carry out the dot product:

\frac{1}{3}(2x - y + 2z) = 5

Multiply both sides by 3:

\boxed{2x - y + 2z = 15}

Why This Works

The key insight is that $\vec{r} \cdot \hat{n}$ gives the scalar projection of $\vec{r}$ onto $\hat{n}$ . For any point on the plane, this projection is always the same value — the perpendicular distance $d$ from the origin. Points off the plane have a different projection value.

This is why we need $\hat{n}$ specifically as a unit vector. If we used a general normal $\vec{n}$ instead, the equation becomes $\vec{r} \cdot \vec{n} = d|\vec{n}|$ , which is the more general form. The normal form is the clean, normalized version.

The Cartesian conversion is mechanical — just expand the dot product with $\vec{r} = (x, y, z)$ . The coefficient of $x$ is the $\hat{i}$ -component of $\vec{n}$ , and so on.

Alternative Method

If you’re given the Cartesian equation first and need to find $d$ (the perpendicular distance from origin), use the formula directly:

For the plane $ax + by + cz = p$ , the perpendicular distance from origin is:

d = \frac{p}{\sqrt{a^2 + b^2 + c^2}}

Here, $2x - y + 2z = 15$ gives $d = \frac{15}{\sqrt{4+1+4}} = \frac{15}{3} = 5$ . ✓

This cross-check is useful in exams — if your $d$ doesn’t match the given value, you’ve made an error somewhere in the normal computation.

In JEE Main, this formula is often tested in reverse: you’re given a plane like $4x - 3y + 5z = 10$ and asked for the perpendicular distance from origin. Just divide the RHS by $\sqrt{a^2+b^2+c^2}$ . No need to fully derive the unit normal.

Common Mistake

The classic error: students write the equation as $\vec{r} \cdot \vec{n} = d$ using the non-unit normal vector, then take $d = 5$ directly. This is wrong.

If the given normal is $\vec{n} = 2\hat{i} - \hat{j} + 2\hat{k}$ (without the $\frac{1}{3}$ ), the correct equation would be $\vec{r} \cdot \vec{n} = 5 \times |\vec{n}| = 5 \times 3 = 15$ , giving $2x - y + 2z = 15$ .

The final Cartesian answer is the same, but the vector equation $\vec{r} \cdot \vec{n} = d$ is only valid when $\vec{n}$ is a unit vector. Mixing up the two forms — especially in MCQ derivation questions from JEE Main 2023 pattern — costs marks.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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