The foot of perpendicular of the point (2, 0, 5) on the line (x+1)/2 = (y-1)/5 = (z+1)/-1 is (α, β,

Q: Why do we set the line equation equal to \lambda? Is there another way?

Setting the line equation to \lambda is the most powerful tool in 3D geometry. It transforms a line, which is a set of infinite points, into a single-variable function. While you could use the projection formula, the parametric method is more robust and less prone to memory errors during high-pressure exams.

Q: Why is the dot product of \vec{AP} and the line's direction vector zero?

The dot product of two vectors is defined as |a||b|cos(\theta). If the vectors are perpendicular, the angle \theta is 90 degrees. Since cos(90) is zero, the entire dot product must be zero. This is the mathematical definition of orthogonality.

Q: Why did we have to check all the options? Can't we just find the point and stop?

The question asks for the statement that is NOT correct. This is a classic JEE trap. You must verify every option because the error might be hidden in the very last one you check. Always read the question carefully—it's testing your attention to detail as much as your math skills.

Question

The foot of perpendicular of the point $(2, 0, 5)$ on the line $\frac{x+1}{2} = \frac{y-1}{5} = \frac{z+1}{-1}$ is $(\alpha, \beta, \gamma)$. Then Which of the following is NOT correct?

Options:

(A) $\frac{\alpha\beta}{\gamma} = \frac{4}{15}$

(B) $\frac{\alpha}{\beta} = -8$

(C) $\frac{\beta}{\gamma} = -5$

(D) $\frac{\gamma}{\alpha} = \frac{5}{8}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are floating in the vast, silent expanse of three-dimensional space. You have a fixed point, $A$, located at coordinates $(2, 0, 5)$.

Nearby, there is a straight line defined by the equation:

$$
\frac{x+1}{2} = \frac{y-1}{5} = \frac{z+1}{-1}
$$

Your mission is to drop a perfectly straight perpendicular line from point $A$ onto this path. The exact point of impact, where the perpendicular meets the line, is the foot of the perpendicular, $P(\alpha, \beta, \gamma)$.

### The Parametric Bridge

To find a point on a line, we use a parameter $\lambda$. By setting the equation of the line equal to $\lambda$, we express any point on the line as a function of this single variable:

$$
\frac{x+1}{2} = \frac{y-1}{5} = \frac{z+1}{-1} = \lambda
$$

This yields the coordinates of any point $P$ on the line as:

$$
x = 2\lambda - 1, \quad y = 5\lambda + 1, \quad z = -\lambda - 1
$$

Think of $\lambda$ as a GPS coordinate along the line. Our goal is to find the specific $\lambda$ that makes the vector $\vec{AP}$ perpendicular to the line itself.

### The Orthogonality Condition

We define the vector $\vec{AP}$ by subtracting the coordinates of $A$ from the coordinates of $P$:

$$
\vec{AP} = (2\lambda - 1 - 2, \, 5\lambda + 1 - 0, \, -\lambda - 1 - 5) = (2\lambda - 3, \, 5\lambda + 1, \, -\lambda - 6)
$$

For this path to be the perpendicular, it must be orthogonal to the line's direction vector, which is $(2, 5, -1)$. The condition for perpendicularity is that the dot product of these two vectors must be zero:

$$
2(2\lambda - 3) + 5(5\lambda + 1) + (-1)(-\lambda - 6) = 0
$$

### Solving the Equation

Expanding the terms, we obtain:

$$
4\lambda - 6 + 25\lambda + 5 + \lambda + 6 = 0
$$

Combining the $\lambda$ terms and the constants, we get:

$$
30\lambda + 5 = 0 \quad \Rightarrow \quad \lambda = -\frac{1}{6}
$$

Substituting this value of $\lambda$ back into our parametric coordinates, we find the foot of the perpendicular $P(\alpha, \beta, \gamma)$:

$$
P = \left( 2\left(-\frac{1}{6}\right) - 1, \, 5\left(-\frac{1}{6}\right) + 1, \, -\left(-\frac{1}{6}\right) - 1 \right) = \left( -\frac{4}{3}, \, \frac{1}{6}, \, -\frac{5}{6} \right)
$$

### The Final Verification

We now evaluate the given options to identify the incorrect statement:

1. $\frac{\alpha\beta}{\gamma} = \frac{(-4/3)(1/6)}{-5/6} = \frac{-4/18}{-5/6} = \frac{4}{15}$ (Correct)
2. $\frac{\alpha}{\beta} = \frac{-4/3}{1/6} = -8$ (Correct)
3. $\frac{\beta}{\gamma} = \frac{1/6}{-5/6} = -\frac{1}{5}$ (The option claims $-5$, which is **incorrect**)
4. $\frac{\gamma}{\alpha} = \frac{-5/6}{-4/3} = \frac{5}{8}$ (Correct)

Through systematic calculation, we have identified that **Option 3 is the incorrect statement**. In JEE Advanced, precision is your greatest ally.

Animated Solution for Mathematics - Three Dimensional Geometry: The foot of perpendicular of the point $(2, 0, 5)$ on the line $\frac{x + 1}{2} = \frac{y - 1}{5} = \frac{z + 1}{- 1}$ is $(α, β, γ)$ . Then Which of the following is NOT correct?

Select Answer:

Visualized Solution

Visualizing the Point and the Line

Dropping the Perpendicular

Parametric Coordinates of Point $P$

Direction Ratios of Vector $A P$

Applying the Orthogonality Condition

Solving for the Scalar $λ$

Finding the Coordinates $(α, β, γ)$

Verifying Option 1

Verifying Option 2

Verifying Option 3

Verifying Option 4

Final Conclusion

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Parametric Bridge

The Orthogonality Condition

Solving the Equation

The Final Verification

Similar Questions

If the foot of the perpendicular drawn from $(1, 9, 7)$ to the line passing through the point $(3, 2, 1)$ and parallel to the planes $x + 2 y + z = 0$ and $3 y - z = 3$ is $(α, β, γ)$ , then $α + β + γ$ is equal to

If the foot of perpendicular drawn from the point $(1, 0, 3)$ on a line passing through $(α, 7, 1)$ is $(\frac{5}{3}, \frac{7}{3}, \frac{17}{3})$ , then $α$ is equal to ________.

If the foot of the perpendicular drawn from the point $(1, 0, 3)$ on a line passing through $(α, 7, 1)$ is $(5/3, 7/3, 17/3)$ , then $α$ is equal to .....

Let P be the foot of the perpendicular from the point (1, 2, 2) on the line $L : \frac{x - 1}{1} = \frac{y + 1}{- 1} = \frac{z - 2}{2}$ Let the line $r = (- \hat{i} + \hat{j} - 2 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$ , $λ \in R,$ intersect the line L at Q. Then $2 (P Q)^{2}$ is equal to:

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 :

Let $P$ be the point $(10, - 2, - 1)$ and $Q$ be the foot of the perpendicular drawn from the point $R (1, 7, 6)$ on the line passing through the points $(2, - 5, 11)$ and $(- 6, 7, - 5)$ . Then the length of the line segment $P Q$ is equal to ________

Let P be the foot of the perpendicular from the point $Q (10, - 3, - 1)$ on the line $\frac{x - 3}{7} = \frac{y - 2}{- 1} = \frac{z + 1}{- 2}$ Then the area of the right angled triangle PQR, where R is the point $(3, - 2, 1)$ , is

Let $L$ be the line of intersection of planes $r \cdot (\hat{i} - \hat{j} + 2 \hat{k}) = 2$ and $r \cdot (2 \hat{i} + \hat{j} - \hat{k}) = 2$ . If $P (α, β, γ)$ is the foot of perpendicular on $L$ from the point $(1, 2, 0)$ , then the value of $35 (α + β + γ)$ is equal to :

The equation of the line through the point $(0, 1, 2)$ and perpendicular to the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{- 2}$ is :