Let the foot of perpendicular from a point P(1, 2, -1) to the straight line L : x/1 = y/0 = z/-1 be

Q: Wait, why do we use two different parameters, $\lambda$ and $\mu$, for points N and Q? Can't we just use one?

That is a fantastic question! Think of it this way: N and Q are two distinct points on the line L. If you used the same parameter for both, you would be forcing N and Q to be the exact same point. By using $\lambda$ for N and $\mu$ for Q, you grant them the freedom to exist at different locations on the line, which is exactly what the problem requires.

Q: Why is the line PQ being parallel to the plane equivalent to PQ being perpendicular to the normal vector?

Imagine the plane as a flat sheet of paper. The normal vector is a pencil sticking straight out of that paper at a 90-degree angle. If a line is parallel to the paper, it will never touch it. Consequently, that line must also be perpendicular to the pencil (the normal vector). It is a beautiful geometric necessity: parallel to the plane means perpendicular to the normal!

Q: Is there a faster way to find the foot of the perpendicular N without using the dot product?

You could use the projection formula, but the dot product method is the 'gold standard' for JEE Advanced. It is robust, less prone to calculation errors, and builds the intuition you need for more complex 3D problems. Stick with the dot product—it is the most reliable tool in your arsenal.

Question

Let the foot of perpendicular from a point $P(1, 2, -1)$ to the straight line $L : \frac{x}{1} = \frac{y}{0} = \frac{z}{-1}$ be $N$. Let a line be drawn from $P$ parallel to the plane $x + y + 2z = 0$ which meets $L$ at point $Q$. If $\alpha$ is the acute angle between the lines $PN$ and $PQ$, then $\cos \alpha$ is equal to

Options:

(A) $\frac{1}{\sqrt{5}}$

(B) $\frac{\sqrt{3}}{2}$

(C) $\frac{1}{\sqrt{3}}$

(D) $\frac{1}{2\sqrt{3}}$

Sigma Percentile · Accepted Answer

### The Geometry of 3D Space: A Journey to the Foot of the Perpendicular

Welcome, fellow explorer of the mathematical universe. Today, we are not just solving a problem; we are navigating the elegant landscape of 3D coordinate geometry.

Imagine yourself standing at point $P(1, 2, -1)$, looking out at a line $L$ that stretches infinitely through space. Our mission is to find the angle between two specific vectors, $\vec{PN}$ and $\vec{PQ}$, where $N$ is the foot of the perpendicular from $P$ to $L$, and $Q$ is a point on $L$ such that $PQ$ is parallel to a given plane.

### Phase 1: Taming the Line with Parameters

In 3D geometry, a line is often a slippery beast. To catch it, we use the power of parametric representation.

The line $L$ is given by $\frac{x}{1} = \frac{y}{0} = \frac{z}{-1}$. By setting this equal to a parameter, say $\lambda$, we can express any point on this line as $(\lambda, 0, -\lambda)$.

This is our anchor. Whether we are looking for point $N$ or point $Q$, they must both obey this rule. This simple act of parameterization transforms a vague geometric object into a concrete algebraic coordinate.

### Phase 2: The Perpendicularity Condition

Now, let us find $N$. We know that $N$ is the foot of the perpendicular from $P$ to $L$, which means the vector $\vec{PN}$ must be perfectly perpendicular to the line $L$. The direction vector of $L$, which we can call $\vec{d}$, is simply $(1, 0, -1)$.

We define $\vec{PN}$ by subtracting the coordinates of $P$ from $N$: 
$$\vec{PN} = (\lambda - 1, 0 - 2, -\lambda - (-1)) = (\lambda - 1, -2, -\lambda + 1)$$

For $\vec{PN}$ to be perpendicular to $\vec{d}$, their dot product must vanish: $\vec{PN} \cdot \vec{d} = 0$. 
$$(\lambda - 1)(1) + (-2)(0) + (-\lambda + 1)(-1) = 0$$

Solving this, we find $(\lambda - 1) + (\lambda - 1) = 0$, which leads us to $2\lambda = 2$, or $\lambda = 1$. With $\lambda = 1$, our point $N$ is $(1, 0, -1)$, and our vector $\vec{PN}$ becomes $(0, -2, 0)$.

### Phase 3: The Parallelism Constraint

Next, we turn our attention to $Q$. We know $Q$ lies on $L$, so its coordinates are $(\mu, 0, -\mu)$. The vector $\vec{PQ}$ is $(\mu - 1, -2, -\mu + 1)$.

We are told that $PQ$ is parallel to the plane $x + y + 2z = 0$. If a line is parallel to a plane, it must be perpendicular to the plane's normal vector. The normal vector $\vec{n}$ of our plane is $(1, 1, 2)$.

Thus, $\vec{PQ} \cdot \vec{n} = 0$:
$$(\mu - 1)(1) + (-2)(1) + (-\mu + 1)(2) = 0$$

Expanding this, we get $\mu - 1 - 2 - 2\mu + 2 = 0$, which simplifies to $-\mu - 1 = 0$, giving us $\mu = -1$. Substituting this back, our vector $\vec{PQ}$ becomes $(-2, -2, 2)$.

### Phase 4: The Final Convergence

We have arrived at the final stage. We have our two vectors: $\vec{PN} = (0, -2, 0)$ and $\vec{PQ} = (-2, -2, 2)$. The angle $\alpha$ between them is governed by the dot product formula:
$$\cos \alpha = \frac{|\vec{PN} \cdot \vec{PQ}|}{|\vec{PN}| |\vec{PQ}|}$$

Calculating the dot product: $(0)(-2) + (-2)(-2) + (0)(2) = 4$. The magnitudes are $|\vec{PN}| = 2$ and $|\vec{PQ}| = \sqrt{(-2)^2 + (-2)^2 + 2^2} = \sqrt{12} = 2\sqrt{3}$.

Plugging these into our formula:
$$\cos \alpha = \frac{4}{2 \cdot 2\sqrt{3}} = \frac{4}{4\sqrt{3}} = \frac{1}{\sqrt{3}}$$

The final value for the cosine of the angle is **$\frac{1}{\sqrt{3}}$**. You have mastered the logic—now go forth and conquer the next problem!

Select Answer:

Visualized Solution

Visualizing the 3D Setup

Parametric Form of Point $N$

Defining Vector $P N$

Applying Perpendicularity Condition

Solving for $λ$

Final Vector $P N$

Parametric Form of Point $Q$

Defining Vector $P Q$

Parallel to Plane Condition

Solving for $μ$

Final Vector $P Q$

The Angle Formula

Calculating Dot Product and Magnitudes

Final Calculation of $cos α$

Conclusion and Takeaway

The Sigma Insight: Angle Between Two Lines

The Geometry of 3D Space

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the 3D Setup

Parametric Form of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

Defining Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PN

Applying Perpendicularity Condition

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Final Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PN

Parametric Form of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Defining Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ​