From a point P(λ, λ, λ), perpendicular PQ and PR are drawn respectively on the lines y = x, z = 1 an

Q: Wait, why do we reject $\lambda = 1$?

This is a classic trap! If you substitute $\lambda = 1$ into the coordinates of $P$ and $Q$, you get $P(1, 1, 1)$ and $Q(1, 1, 1)$. They are the same point. A vector $\vec{PQ}$ of zero length has no direction, and therefore, the angle $\angle QPR$ cannot be defined. We must reject any value that makes the vectors degenerate.

Q: Why can't we just use the distance formula to solve this?

You could, but it would be significantly more complex. Using the distance formula would require you to equate $PQ^2 + PR^2 = QR^2$ (Pythagoras theorem). While valid, it involves calculating the square of the distance between points, which leads to much more tedious algebra compared to the elegant dot product approach.

Q: How do I know if the lines are skew or intersecting?

In this specific problem, you don't actually need to check if the lines are skew. The problem defines $Q$ and $R$ as the feet of the perpendiculars from $P$ to the lines. The existence of these points is guaranteed by the geometry of lines in 3D space. Focus on the perpendicularity condition $\vec{PQ} \cdot \vec{d} = 0$ for each line.

Question

From a point $P(\lambda, \lambda, \lambda)$, perpendicular $PQ$ and $PR$ are drawn respectively on the lines $y = x, z = 1$ and $y = -x, z = -1$. If $P$ is such that $\angle QPR$ is a right angle, then the possible value(s) of $\lambda$ is/(are)

Options:

(A) $\sqrt{2}$

(B) 1

(C) -1

(D) $-\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional coordinate system. You have a point $P$ with coordinates $(\lambda, \lambda, \lambda)$, which is essentially a point moving along the line $y=x, z=x$.

Below and above this point, there are two distinct lines, $L_1$ and $L_2$. We are looking for the specific position of $P$ such that the lines connecting it to its projections, $Q$ and $R$, form a perfect right angle.

### Parametrizing the Lines

To find the foot of the perpendicular $Q$ on line $L_1$, we define $L_1$ mathematically. Given $L_1: y = x, z = 1$, we introduce a parameter $t$.

If we set $x = t$, then $y = t$, and $z$ is fixed at $1$. Thus, any point on $L_1$ can be represented as $(t, t, 1)$. The direction vector of this line is $\vec{d_1} = (1, 1, 0)$.

Similarly, for $L_2: y = -x, z = -1$, we use a parameter $s$. A general point on $L_2$ is $(s, -s, -1)$, with a direction vector $\vec{d_2} = (1, -1, 0)$.

### The Power of Orthogonality

Now, let's find $Q$. The vector $\vec{PQ}$ is the difference between the coordinates of $Q$ and $P$:
$$\vec{PQ} = (t-\lambda, t-\lambda, 1-\lambda)$$

Because $Q$ is the foot of the perpendicular, $\vec{PQ}$ must be orthogonal to the line $L_1$. This means the dot product $\vec{PQ} \cdot \vec{d_1} = 0$.

Performing the calculation:
$$(t-\lambda)(1) + (t-\lambda)(1) + (1-\lambda)(0) = 0$$
$$2(t-\lambda) = 0 \implies t = \lambda$$

This result tells us that $Q$ is $(\lambda, \lambda, 1)$. We repeat this process for $R$ on $L_2$:
$$\vec{PR} = (s-\lambda, -s-\lambda, -1-\lambda)$$
$$\vec{PR} \cdot \vec{d_2} = (s-\lambda)(1) + (-s-\lambda)(-1) + (-1-\lambda)(0) = 0$$
$$s - \lambda + s + \lambda = 0 \implies 2s = 0 \implies s = 0$$

Thus, $R$ is fixed at $(0, 0, -1)$.

### The Climax: The Angle Condition

We are given that $\angle QPR = 90^\circ$. This implies the vectors $\vec{PQ}$ and $\vec{PR}$ are perpendicular, so their dot product must be zero.

Substituting our coordinates:
$$\vec{PQ} = (0, 0, 1-\lambda)$$
$$\vec{PR} = (-\lambda, -\lambda, -1-\lambda)$$

Now, calculate the dot product:
$$(0)(-\lambda) + (0)(-\lambda) + (1-\lambda)(-1-\lambda) = 0$$
$$-(1-\lambda)(1+\lambda) = 0$$
$$\lambda^2 - 1 = 0$$

### The Final Trap

We arrive at $\lambda^2 = 1$, which gives $\lambda = 1$ or $\lambda = -1$. However, we must check for degeneracy.

If $\lambda = 1$, then $P = (1, 1, 1)$ and $Q = (1, 1, 1)$. The vector $\vec{PQ}$ becomes the zero vector, which cannot form a $90^\circ$ angle.

Therefore, we must reject $\lambda = 1$. The only valid solution is **$\lambda = -1$**.

Animated Solution for Mathematics - Three Dimensional Geometry: From a point $P (λ, λ, λ)$ , perpendicular $P Q$ and $P R$ are drawn respectively on the lines $y = x, z = 1$ and $y = - x, z = - 1$ . If $P$ is such that $∠ QP R$ is a right angle, then the possible value(s) of $λ$ is/(are)

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Visualized Solution

The 3D Setup

Parametrizing Line $L_{1}$

Foot of Perpendicular $Q$

Coordinates of $Q$

Parametrizing Line $L_{2}$

Foot of Perpendicular $R$

Coordinates of $R$

Applying $∠ QP R = 9 0^{\circ}$

Calculating the Dot Product

Checking the Roots

Rejecting the Invalid Root

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Parametrizing the Lines

The Power of Orthogonality

The Climax

The Final Trap

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Visualized Solution

The 3D Setup

Parametrizing Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

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

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

Parametrizing Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Foot of Perpendicular .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∠QPR=90∘