If the foot of the perpendicular from the point A(-1, 4, 3) on the plane P : 2x + my + nz = 4, is (-

Q: Wait, why do we equate the direction ratios of $\vec{AF}$ and $\vec{n}$? Can't we just use the dot product?

You absolutely can! Since $\vec{AF}$ is parallel to the normal vector $\vec{n}$, you could also say $\vec{AF} = k\vec{n}$ for some scalar $k$. Equating the components gives you the same proportionality. Using ratios is just a faster way to see that $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.

Q: Why can't we just use the perpendicular distance formula for a point to a plane?

That is a classic trap! The formula $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$ gives the *shortest* distance, which is the perpendicular distance. The question specifically asks for the distance measured *parallel* to a given line, which is a different path entirely. You must find the intersection point $Q$ along that specific direction.

Q: How do I know if my parameter $\lambda$ is correct?

Great question! After finding $\lambda$, substitute your coordinates of $Q$ back into the plane equation $2x + y + 3z = 4$. If the result is $4$, your intersection point is correct. It is the ultimate self-check for any 3D geometry problem.

Question

If the foot of the perpendicular from the point $A(-1, 4, 3)$ on the plane $P : 2x + my + nz = 4$, is $\left(-2, \frac{7}{2}, \frac{3}{2}\right)$, then the distance of the point $A$ from the plane $P$, measured parallel to a line with direction ratios $3, -1, -4$, is equal to :

Options:

(A) 1

(B) $\sqrt{26}$

(C) $2\sqrt{2}$

(D) $\sqrt{14}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given point $A(-1, 4, 3)$ and the foot of the perpendicular $F(-2, \frac{7}{2}, \frac{3}{2})$ on the plane $P: 2x + my + nz = 4$.

The vector $\vec{AF}$ represents the normal to the plane, calculated as:
$$\vec{AF} = F - A = (-2 - (-1), \frac{7}{2} - 4, \frac{3}{2} - 3) = (-1, -\frac{1}{2}, -\frac{3}{2})$$

### Decoding the Plane

Since $\vec{AF}$ is parallel to the normal vector $\vec{n} = (2, m, n)$, their components must be proportional:
$$\frac{-1}{2} = \frac{-1/2}{m} = \frac{-3/2}{n}$$

Solving these proportions, we find:
$$m = 1, \quad n = 3$$

Thus, the equation of the plane $P$ is fully defined as:
$$2x + y + 3z = 4$$

### The Path Less Traveled

We now travel from $A$ along the direction vector $\vec{d} = (3, -1, -4)$. Any point $Q$ on this path can be expressed using a parameter $\lambda$:
$$x = -1 + 3\lambda, \quad y = 4 - \lambda, \quad z = 3 - 4\lambda$$

### The Intersection

The point $Q$ lies on the plane $P$, so its coordinates must satisfy the plane equation $2x + y + 3z = 4$. Substituting the parametric expressions:
$$2(-1 + 3\lambda) + (4 - \lambda) + 3(3 - 4\lambda) = 4$$

Expanding the equation:
$$-2 + 6\lambda + 4 - \lambda + 9 - 12\lambda = 4$$

Simplifying the terms involving $\lambda$:
$$-7\lambda + 11 = 4 \Rightarrow -7\lambda = -7 \Rightarrow \lambda = 1$$

Substituting $\lambda = 1$ back into our parametric equations, we find the intersection point $Q$:
$$Q = (-1 + 3(1), 4 - 1, 3 - 4(1)) = (2, 3, -1)$$

### Final Calculation

The distance $AQ$ represents the length of the path traveled from $A(-1, 4, 3)$ to $Q(2, 3, -1)$. Using the distance formula:
$$AQ = \sqrt{(2 - (-1))^2 + (3 - 4)^2 + (-1 - 3)^2}$$

$$AQ = \sqrt{3^2 + (-1)^2 + (-4)^2} = \sqrt{9 + 1 + 16} = \sqrt{26}$$

The length of the path is **$\sqrt{26}$**.

Select Answer:

Visualized Solution

Visualizing the Setup

Vector $A F$ Calculation

Parallelism with Normal Vector

Solving for $m$ and $n$

Equation of Plane $P$

Line Parallel to Given Direction

General Point $Q$ on Line

Intersection with Plane $P$

Solving for $λ$

Coordinates of Point $Q$

Distance $A Q$ Calculation

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

Decoding the Plane

The Path Less Traveled

The Intersection

Final Calculation

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A perpendicular is drawn from a point on the line $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z}{1}$ to the plane $x + y + z = 3$ such that the foot of the perpendicular Q also lies on the plane $x - y + z = 3$ . Then the co-ordinates of Q are :

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Select Answer:

Visualized Solution

Visualizing the Setup

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

Parallelism with Normal Vector

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

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

Line Parallel to Given Direction

General Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q on Line

Intersection with Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

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

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

Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AQ Calculation