If the length of the perpendicular drawn from the point P(a, 4, 2), a 0 on the line (x+1)/2 = (y-3

Q: Wait, why do we reject the negative root for $\lambda$?

Great catch! The problem explicitly states that $a > 0$. When we solve the quadratic equation for $\lambda$, we get two possible values. We must substitute both back into our expression for $a$ (which we found to be $a = 7\lambda - 2$). Only the $\lambda$ that results in a positive $a$ is physically valid for this problem.

Q: Why can't we just use the projection formula directly?

You absolutely can! The projection formula is a powerful tool. However, using the parametric form $M = (2\lambda-1, 3\lambda+3, -\lambda+1)$ is often more intuitive for JEE problems because it naturally leads to a single-variable equation, reducing the chance of algebraic errors in 3D coordinate geometry.

Q: How do we know $M$ is the midpoint of $PQ$?

This is a fundamental property of reflection. When we find the image $Q$ of a point $P$ in a line, the line acts as a mirror. The perpendicular segment $PM$ is equal in length to $MQ$, and $M$ lies exactly on the line. Therefore, $M$ is the midpoint of the segment $PQ$.

Question

If the length of the perpendicular drawn from the point $P(a, 4, 2), a > 0$ on the line $\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1}$ is $2\sqrt{6}$ units and $Q(\alpha_1, \alpha_2, \alpha_3)$ is the image of the point $P$ in this line, then $a + \sum_{i=1}^3 \alpha_i$ is equal to :

Options:

(A) 7

(B) 8

(C) 12

(D) 14

Sigma Percentile · Accepted Answer

### The Geometry of Reflection: A 3D Journey

Imagine you are standing in a vast, three-dimensional room. You see a straight wire stretched across the space, and a point $P$ hovering in the air, not on the wire.

You are asked to find the image of this point $P$ if the wire were a mirror. This is exactly what this JEE problem asks us to do. It is not just about crunching numbers; it is about understanding the spatial relationship between a point and a line.

### Phase 1: The Parametric Dance

To find the image $Q$, we first need the foot of the perpendicular, $M$. How do we pin down a point on a line in 3D? We use a parameter, $\lambda$.

By setting the symmetric equations of the line equal to $\lambda$:

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

We can express any point on the line as $M = (2\lambda-1, 3\lambda+3, -\lambda+1)$. This is our anchor, where every point on that wire is described by a single value of $\lambda$.

### Phase 2: The Orthogonality Constraint

Now, we need to find the specific $\lambda$ that makes $PM$ perpendicular to the line. We construct the vector $\vec{PM} = M - P$.

Since $P = (a, 4, 2)$, we get $\vec{PM} = (2\lambda-1-a, 3\lambda-1, -\lambda-1)$. The line's direction vector is $\vec{d} = (2, 3, -1)$.

Because $\vec{PM}$ is perpendicular to the line, their dot product must be zero: $\vec{PM} \cdot \vec{d} = 0$. Expanding this gives us a linear relationship:

$$ 2(2\lambda-1-a) + 3(3\lambda-1) - 1(-\lambda-1) = 0 $$

Simplifying this, we find $a = 7\lambda - 2$. This is our bridge between the unknown coordinate $a$ and the parameter $\lambda$.

### Phase 3: The Distance Constraint

We are given that the perpendicular distance $PM = 2\sqrt{6}$, so $PM^2 = 24$. Using the distance formula, we write:

$$ (2\lambda-1-a)^2 + (3\lambda-1)^2 + (-\lambda-1)^2 = 24 $$

Substituting our expression for $a$, the equation becomes a quadratic in $\lambda$. After careful expansion and collection of terms, we arrive at:

$$ 35\lambda^2 - 14\lambda - 21 = 0 \Rightarrow 5\lambda^2 - 2\lambda - 3 = 0 $$

Factoring this, we find $\lambda = 1$ or $\lambda = -\frac{3}{5}$.

### Phase 4: The Final Reflection

We check our condition $a > 0$. If $\lambda = 1$, then $a = 7(1) - 2 = 5$, which is valid. If $\lambda = -\frac{3}{5}$, then $a$ is negative, which we reject.

With $\lambda = 1$, the foot $M$ is $(1, 6, 0)$. Since $M$ is the midpoint of $PQ$, we use the midpoint formula $\frac{P+Q}{2} = M$ to find $Q = (-3, 8, -2)$.

Finally, calculating $a + \sum \alpha_i$ gives us $5 + (-3 + 8 - 2) = 8$. The elegance of this result is the reward for our patience. You have mastered the reflection! The final answer is **8**.

Select Answer:

Visualized Solution

Visualize Point $P$ and Line $L$

Parametrize the Foot $M$

Construct Vector $P M$

Apply $P M \cdot d = 0$

Express $a$ in terms of $λ$

Apply Distance Formula $P M^{2} = 24$

Substitute $a = 7 λ - 2$

Expand to Quadratic in $λ$

Solve for $λ$

Validate $a > 0$

Find Coordinates of $M$

Use Midpoint Formula for $Q$

Calculate $a + \sum α_{i}$

The Sigma Insight: Equation of a Line in Space

The Geometry of Reflection

Phase 1

Phase 2

Phase 3

Phase 4

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Select Answer:

Visualized Solution

Visualize Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

Parametrize the Foot .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

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

Apply .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PM⋅d=0

Express .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Apply Distance Formula .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PM2=24

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=7λ−2

Expand to Quadratic in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

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

Validate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a>0

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

Use Midpoint Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+∑αi​