The foot of the perpendicular from a point on the circle x² + y² = 1, z = 0 to the plane 2x + 3y + z

Q: Wait, why is the line $PQ$ parallel to the normal vector $(2, 3, 1)$?

Great question! By definition, the foot of the perpendicular from a point to a plane is the point on the plane closest to the original point. The vector connecting the point to its projection must be perpendicular to the plane itself. Since the normal vector $(2, 3, 1)$ is perpendicular to the plane, any line segment perpendicular to the plane must be parallel to this normal vector.

Q: Why do we use the parameter $k$ in the line equation?

The parameter $k$ is a powerful tool in 3D geometry. It allows us to express the coordinates of any point on the line $PQ$ in terms of a single variable. By setting the symmetric ratios equal to $k$, we create a bridge between the coordinates of the point $P$ on the circle and the coordinates of the foot of the perpendicular $Q$ on the plane.

Q: Do I really need to substitute $z = 6 - 2x - 3y$ at the end?

Yes, absolutely. The problem asks for the curve on which the foot of the perpendicular lies. Since the foot $Q$ is on the plane, its coordinates must satisfy the plane's equation. By substituting $z$ in terms of $x$ and $y$, we project the 3D locus onto the $xy$-plane, which is exactly what the options represent.

Question

The foot of the perpendicular from a point on the circle $x^2 + y^2 = 1, z = 0$ to the plane $2x + 3y + z = 6$ lies on which one of the following curves ?

Options:

(A) $(6x + 5y - 12)^2 + 4(3x + 7y - 8)^2 = 1, z = 6 - 2x - 3y$

(B) $(5x + 6y - 12)^2 + 4(3x + 5y - 9)^2 = 1, z = 6 - 2x - 3y$

(C) $(6x + 5y - 14)^2 + 9(3x + 5y - 7)^2 = 1, z = 6 - 2x - 3y$

(D) $(5x + 6y - 14)^2 + 9(3x + 7y - 8)^2 = 1, z = 6 - 2x - 3y$

Sigma Percentile · Accepted Answer

### The Geometry of Projection: A Journey into 3D Space

Imagine you are standing in a vast, empty room. On the floor, there is a perfect unit circle, $x^2 + y^2 = 1$, resting silently in the $xy$-plane where $z = 0$.

Now, imagine a tilted plane hovering above this circle, defined by the equation $2x + 3y + z = 6$. You are tasked with finding the path traced by the foot of the perpendicular dropped from every point on that circle onto the plane.

This is not just a math problem; it is a study of how shapes transform when projected into different dimensions.

### Phase 1: The Parametric Dance

To begin, we must capture the essence of our circle. A point $P$ moving along the unit circle can be described elegantly using trigonometry.

We define $P$ as $(\cos 	heta, \sin 	heta, 0)$. As $	heta$ sweeps from $0$ to $2\pi$, $P$ traces the entire circle. This parametric representation is our starting point, our anchor in the $xy$-plane.

### Phase 2: The Normal Vector Bridge

Now, consider the plane $2x + 3y + z = 6$. The coefficients of $x, y,$ and $z$ give us the normal vector to the plane: $\vec{n} = (2, 3, 1)$.

When we drop a perpendicular from $P$ to the plane, the resulting line segment $PQ$ is inherently parallel to this normal vector. This is the geometric secret that unlocks the problem.

The line $PQ$ passing through $P(\cos 	heta, \sin 	heta, 0)$ with direction ratios $(2, 3, 1)$ can be written in symmetric form:

$$\frac{x - \cos 	heta}{2} = \frac{y - \sin 	heta}{3} = \frac{z - 0}{1} = k$$

Here, $k$ is our constant of proportionality. It acts as a scalar that scales the normal vector to reach from $P$ to $Q$.

### Phase 3: The Algebraic Bridge

Look closely at the third ratio: $\frac{z}{1} = k$. This is a gift! It tells us that $k = z$.

Now, we can rewrite the first two ratios using $z$ instead of $k$:

$$\frac{x - \cos 	heta}{2} = z \implies \cos 	heta = x - 2z$$

$$\frac{y - \sin 	heta}{3} = z \implies \sin 	heta = y - 3z$$

We have successfully expressed the circle's parameter $	heta$ in terms of the coordinates of the foot of the perpendicular $Q(x, y, z)$.

### Phase 4: The Final Constraint

We know that $\cos^2 	heta + \sin^2 	heta = 1$. Substituting our expressions, we get:

$$(x - 2z)^2 + (y - 3z)^2 = 1$$

This equation describes a cylinder in 3D space. But our point $Q$ is not just anywhere; it is constrained to the plane $2x + 3y + z = 6$.

We can solve for $z$: $z = 6 - 2x - 3y$. Now, we substitute this into our cylinder equation.

Let's handle the terms one by one. First, $(x - 2z)$ becomes $x - 2(6 - 2x - 3y) = 5x + 6y - 12$.

Next, $(y - 3z)$ becomes $y - 3(6 - 2x - 3y) = 6x + 10y - 18$, which simplifies to $2(3x + 5y - 9)$.

### The Elegant Conclusion

Putting it all together, our final locus equation is:

$$(5x + 6y - 12)^2 + [2(3x + 5y - 9)]^2 = 1$$

$$(5x + 6y - 12)^2 + 4(3x + 5y - 9)^2 = 1$$

This is the curve traced by the foot of the perpendicular. It is a beautiful, precise result that perfectly matches the second option.

You have just navigated the projection of a circle onto a tilted plane, turning a complex 3D visualization into a concrete algebraic reality. Keep this intuition, and no geometry problem will ever be too daunting again!

Animated Solution for Mathematics - Three Dimensional Geometry: The foot of the perpendicular from a point on the circle $x^{2} + y^{2} = 1, z = 0$ to the plane $2 x + 3 y + z = 6$ lies on which one of the following curves ?

Select Answer:

Visualized Solution

The Setup: Circle and Point $P$

The Plane and Foot of Perpendicular

Foot of Perpendicular Formula

Substituting the Coordinates

Finding the Constant $k$

Isolating $cos θ$

Isolating $sin θ$

Eliminating $θ$

The Role of the Plane

Substituting $z$ into the First Term

Substituting $z$ into the Second Term

Final Locus Equation

The Sigma Insight: Intersection of a Line and a Plane

The Geometry of Projection

Phase 1

Phase 2

Phase 3

Phase 4

The Elegant Conclusion

Similar Questions

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Let the plane $P : 8 x + α_{1} y + α_{2} z + 12 = 0$ be parallel to the line $L : \frac{x + 2}{2} = \frac{y - 3}{3} = \frac{z + 4}{5}$ . If the intercept of $P$ on the $y$ -axis is 1, then the distance between $P$ and $L$ is :