A perpendicular is drawn from a point on the line (x-1)/2 = (y+1)/(-1) = z/1 to the plane x + y + z

Q: Wait, why did we subtract the two plane equations? Is that always the first step?

It is a strategic choice! When you see two planes, their intersection is a line. Subtracting or adding equations is the fastest way to find the relationship between variables on that line. Here, it isolated $y$ instantly, which saved us from a massive algebraic headache. Always look for these 'geometric shortcuts' before diving into heavy calculations.

Q: Is it mathematically sound to use only the $y$-component of the foot of perpendicular formula?

Absolutely! The formula $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = k$ implies that each component is equal to the same constant $k$. If you know one coordinate of the foot, you can solve for the parameter without needing the others. It is a powerful technique to reduce the complexity of the problem.

Q: I got confused with the signs in the foot of perpendicular formula. How do I remember it?

The formula is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = -\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$. Think of the right side as the 'signed distance' scaled by the normal vector's magnitude. The negative sign is crucial because it points *towards* the plane. If the point is on the positive side of the plane, the expression $ax_1+by_1+cz_1+d$ is positive, so the negative sign ensures we move in the direction of the normal vector to reach the plane.

Question

A perpendicular is drawn from a point on the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z}{1}$ to the plane $x + y + z = 3$ such that the foot of the perpendicular Q also lies on the plane $x - y + z = 3$. Then the co-ordinates of Q are :

Options:

(A) (2, 0, 1)

(B) (4, 0, -1)

(C) (-1, 0, 4)

(D) (1, 0, 2)

Sigma Percentile · Accepted Answer

### The Dance of Geometry: Finding the Foot of the Perpendicular

Imagine you are standing in a vast, three-dimensional space. You have a line $L$ stretching out into infinity, and two planes, $\pi_1$ and $\pi_2$, slicing through the void.

This is not just an algebra problem; it is a story of intersection and alignment. We are tasked with finding a specific point $Q$, the foot of a perpendicular dropped from a point $P$ on line $L$ to plane $\pi_1$.

But there is a twist: $Q$ is not just any point; it is a prisoner of two worlds, lying on both $\pi_1$ and $\pi_2$. Let us embark on this journey to uncover its coordinates.

### Phase 1: The Parametric Leap

Our first challenge is to tame the line $L$. The equation is given as $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z}{1}$.

In the world of JEE Advanced, we never treat a line as a static object. We treat it as a path. By setting this entire expression equal to a scalar parameter $\lambda$, we unlock the ability to describe any point $P$ on this line.

$$P = (2\lambda+1, -\lambda-1, \lambda)$$

This is our 'moving point.' As $\lambda$ changes, $P$ glides along the line. Our mission is to find the specific value of $\lambda$ that forces the perpendicular from $P$ to land exactly on the intersection of our two planes.

### Phase 2: The Intersection Insight

Before we touch the perpendicular formula, let us look at the planes: $\pi_1: x + y + z = 3$ and $\pi_2: x - y + z = 3$. If point $Q$ lies on both, it must satisfy both equations simultaneously.

This is where the magic happens. Instead of solving for $x, y,$ and $z$ individually, let us look at the system as a whole. Subtracting the second equation from the first:

$$(x + y + z) - (x - y + z) = 3 - 3$$

$$2y = 0 \Rightarrow y = 0$$

This is a massive breakthrough! We have just discovered that the $y$-coordinate of our target point $Q$ is zero. We have reduced the dimensionality of our problem with a single stroke of subtraction.

### Phase 3: The Bridge of the Perpendicular

Now, we need to connect $P$ to $Q$. We use the standard formula for the foot of the perpendicular from a point $(x_1, y_1, z_1)$ to a plane $ax+by+cz+d=0$:

$$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = -\frac{ax_1+by_1+cz_1+d}{a^2+b^2+c^2}$$

For our plane $\pi_1$ ($x+y+z-3=0$), the coefficients are $a=1, b=1, c=1$, and $d=-3$. The denominator becomes $1^2+1^2+1^2 = 3$.

We know $y_Q = 0$. Let us focus solely on the $y$-component of the formula:

$$y_Q - y_P = -\frac{x_P + y_P + z_P - 3}{3}$$

### Phase 4: The Final Calculation

Substitute our parametric coordinates of $P$ into the equation. We know $y_Q = 0$ and $y_P = -\lambda - 1$.

$$0 - (-\lambda - 1) = -\frac{(2\lambda + 1) + (-\lambda - 1) + \lambda - 3}{3}$$

Simplifying the left side gives $\lambda + 1$. Simplifying the numerator on the right: $(2\lambda - \lambda + \lambda) + (1 - 1 - 3) = 2\lambda - 3$.

$$\lambda + 1 = -\frac{2\lambda - 3}{3}$$

Multiply both sides by $3$:

$$3\lambda + 3 = -(2\lambda - 3)$$

$$3\lambda + 3 = -2\lambda + 3$$

$$5\lambda = 0 \Rightarrow \lambda = 0$$

The math has collapsed beautifully into $\lambda = 0$. This means our point $P$ is located at $(1, -1, 0)$.

### Conclusion: The Coordinates of $Q$

With $\lambda = 0$, we can find the coordinates of $Q$. The common ratio in our formula was $-\frac{2(0)-3}{3} = 1$. Using $x_Q - x_P = 1(a)$, $y_Q - y_P = 1(b)$, and $z_Q - z_P = 1(c)$:

$$x_Q - 1 = 1(1) \Rightarrow x_Q = 2$$

$$y_Q - (-1) = 1(1) \Rightarrow y_Q = 0$$

$$z_Q - 0 = 1(1) \Rightarrow z_Q = 1$$

The coordinates of $Q$ are **$(2, 0, 1)$**. We have traversed the geometry, used the parametric form, exploited the intersection, and arrived at the solution. Remember, in JEE Advanced, the most complex problems often yield to the simplest geometric insights.

Select Answer:

Visualized Solution

Visualizing the Geometry

Parametric Form of Point $P$

The Intersection Logic

Foot of the Perpendicular Formula

Focusing on the $y$ -coordinate

Simplifying the Equation

Solving for $λ$

Finding Coordinates of $Q$

Final Verification

The Sigma Insight: Intersection of a Line and a Plane

The Dance of Geometry

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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