If the length of the perpendicular from the point (β, 0, β) (β ≠ 0) to the line (x-1)/1 = (y-0)/0 =

Q: Wait, why can we just ignore the y-coordinate? Isn't this a 3D geometry problem?

That is a fantastic question! In 3D geometry, if the $y$-coordinate of the point is $0$ and the $y$-coordinate of every point on the line is also $0$ (because $\frac{y-0}{0}$ implies $y=0$), then the entire interaction between the point and the line happens on the plane $y=0$. By restricting our view to this $XZ$-plane, we simplify the problem significantly without losing any information.

Q: Why is the foot of the perpendicular the origin $(0,0)$?

Great observation. In the $XZ$-plane, the line $L$ simplifies to $z = -x$, which has a slope of $-1$. The point $P$ is at $(\beta, \beta)$, so the line connecting the origin $(0,0)$ to $P$ has a slope of $\frac{\beta-0}{\beta-0} = 1$. Since the product of the slopes is $(-1) \times 1 = -1$, these two lines are perfectly perpendicular. Because the line $z = -x$ passes through the origin, the origin is the intersection point, and thus, the foot of the perpendicular.

Q: Why do we use the absolute value $|\beta|$ when simplifying the distance?

When we simplify $\sqrt{2\beta^2}$, we get $\sqrt{2} \cdot \sqrt{\beta^2}$. The square root of a squared variable is always its absolute value, $|\beta|$. This is crucial because $\beta$ could be negative, and distance must always be positive. Keeping the absolute value ensures our algebraic steps remain rigorous.

Question

If the length of the perpendicular from the point $(\beta, 0, \beta) (\beta 
eq 0)$ to the line $\frac{x-1}{1} = \frac{y-0}{0} = \frac{z+1}{-1}$ is $\sqrt{3/2}$, then $\beta$ is equal to :

Options:

(A) -1

(B) 2

(C) -2

(D) 1

Sigma Percentile · Accepted Answer

### The Illusion of 3D Complexity

Welcome, future engineer! Today, we are going to dismantle a problem that, at first glance, looks like a daunting 3D geometry nightmare.

You see a point $(\beta, 0, \beta)$ and a line in 3D space, and your brain immediately starts reaching for complex vector projection formulas. But stop! Take a deep breath.

The secret to mastering JEE Advanced is not just knowing formulas; it is knowing when to look for the hidden simplicity. Let us embark on this journey together.

### Phase 1: Dimensional Reduction

Look closely at the point $P(\beta, 0, \beta)$. Its $y$-coordinate is $0$.

Now, look at the line equation:

$$ \frac{x-1}{1} = \frac{y-0}{0} = \frac{z+1}{-1} $$

The presence of the $0$ in the denominator for the $y$-term tells us that for any point on this line, $y$ must be $0$. This is our 'Aha!' moment.

The point $P$ and the entire line $L$ exist exclusively in the $XZ$-plane. We have just collapsed a 3D problem into a 2D one. We are no longer lost in space; we are just drawing on a flat sheet of paper.

### Phase 2: The Geometry of the Line

Now that we are in the $XZ$-plane, let us define our line $L$. We take the $x$ and $z$ parts of the equation:

$$ \frac{x-1}{1} = \frac{z+1}{-1} $$

Cross-multiplying gives us $-1(x - 1) = 1(z + 1)$, which simplifies to $-x + 1 = z + 1$.

Rearranging this, we get the beautiful, elegant equation: $x + z = 0$, or simply $z = -x$. This line passes perfectly through the origin $(0,0)$ with a slope of $-1$.

### Phase 3: The Perpendicularity Catch

Where is our point $P$? In our $XZ$-plane, it is at $(\beta, \beta)$.

Now, consider the line segment connecting the origin $(0,0)$ to $P(\beta, \beta)$. The slope of this segment is:

$$ m = \frac{\beta - 0}{\beta - 0} = 1 $$

Look at that! The slope of our line $L$ is $-1$, and the slope of the segment from the origin to $P$ is $1$.

Since $(-1) 	imes 1 = -1$, these two lines are perfectly perpendicular. This means the foot of the perpendicular from $P$ to $L$ is simply the origin $(0,0)$. The geometry has aligned perfectly for us.

### Phase 4: The Final Calculation

We are now at the finish line. We need the perpendicular distance from $P(\beta, \beta)$ to the line $L$.

Since the foot of the perpendicular is the origin, this is just the distance from $(0,0)$ to $(\beta, \beta)$. Using the distance formula:

$$ d = \sqrt{(\beta - 0)^2 + (\beta - 0)^2} = \sqrt{2\beta^2} = \sqrt{2}|\beta| $$

The problem states this distance is $\sqrt{\frac{3}{2}}$. So, we set up our final equation:

$$ \sqrt{2}|\beta| = \sqrt{\frac{3}{2}} $$

Squaring both sides, we get $2\beta^2 = \frac{3}{2}$, which leads to $\beta^2 = \frac{3}{4}$.

Thus, **$\beta = \pm \frac{\sqrt{3}}{2}$**. You have navigated the 3D trap and found the truth through 2D clarity. Keep this mindset, and no problem will ever be too complex for you!

Animated Solution for Mathematics - Three Dimensional Geometry: If the length of the perpendicular from the point $(β, 0, β) (β \neq = 0)$ to the line $\frac{x - 1}{1} = \frac{y - 0}{0} = \frac{z + 1}{- 1}$ is $3/2$ , then $β$ is equal to :

Select Answer:

Visualized Solution

Visual Anchor: Dimensional Reduction

Logic Bridge: Line Equation in 2D

Visual Anchor: Plotting the Point

Logic Bridge: The Perpendicularity Catch

Raw Setup: Distance Formula

Atomic Compute: Simplifying Distance

Raw Setup: Equating to Given Value

Atomic Compute: Squaring Both Sides

The Way Forward: Final Answer

The Sigma Insight: Equation of a Line in Space

The Illusion of 3D Complexity

Phase 1

Phase 2

Phase 3

Phase 4

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