If for a 0, the feet of perpendiculars from the points A(a, -2a, 3) and B(0, 4, 5) on the plane lx

Q: Why is \vec{AC} parallel to the normal vector of the plane?

By definition, the foot of the perpendicular from a point to a plane is the point where the line passing through the point and perpendicular to the plane intersects the plane. Since the normal vector is perpendicular to the plane, any line perpendicular to the plane must be parallel to the normal vector.

Q: Can we just use the distance formula between C and D directly?

Unfortunately, no. We don't know the coordinates of D initially. We only know D is the foot of the perpendicular from B. Finding D directly would require finding the intersection of a line and a plane, which is much more tedious. Using the right-angled triangle \triangle BDC is the elegant shortcut.

Q: Why did we assume m \neq 0 when solving for a?

Great catch! If m were zero, then n would also be zero (from n = -ma). If l, m, n were all zero, the plane equation would be invalid. Since we are dealing with a valid plane, m cannot be zero. It is a safe assumption in this context.

Question

If for $a > 0$, the feet of perpendiculars from the points $A(a, -2a, 3)$ and $B(0, 4, 5)$ on the plane $lx + my + nz = 0$ are points $C(0, -a, -1)$ and $D$ respectively, then the length of line segment $CD$ is equal to :

Options:

(A) $\sqrt{31}$

(B) $\sqrt{41}$

(C) $\sqrt{55}$

(D) $\sqrt{66}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a plane defined by the equation $lx + my + nz = 0$. Two points, $A(a, -2a, 3)$ and $B(0, 4, 5)$, have perpendiculars dropped onto this plane, landing at points $C(0, -a, -1)$ and $D$, respectively.

Our objective is to determine the length of the segment $CD$.

### Phase 1: The Plane and the Point

Since point $C(0, -a, -1)$ lies on the plane $lx + my + nz = 0$, its coordinates must satisfy the plane equation. Substituting these coordinates, we obtain:

$$l(0) + m(-a) + n(-1) = 0$$

This simplifies to the vital relationship:

$$n = -ma$$

### Phase 2: The Vector Bridge

The vector $\vec{AC}$ represents the perpendicular dropped from $A$ to the plane. Therefore, $\vec{AC}$ must be parallel to the normal vector of the plane, $\vec{n} = (l, m, n)$.

Calculating the vector $\vec{AC}$ by subtracting the coordinates of $A$ from $C$:

$$\vec{AC} = (0 - a, -a - (-2a), -1 - 3) = (-a, a, -4)$$

Because $\vec{AC}$ is parallel to $\vec{n}$, their components are proportional:

$$\frac{l}{-a} = \frac{m}{a} = \frac{n}{-4}$$

From these ratios, we find $n = \frac{-4m}{a}$. Equating this to our earlier result $n = -ma$, we get:

$$-ma = \frac{-4m}{a}$$

Assuming $m 
eq 0$, we solve for $a$:

$$a^2 = 4$$

Given the constraint $a > 0$, we conclude that **$a = 2$**.

### Phase 3: The Right-Angled Triangle

With $a = 2$, the coordinates of $C$ are $(0, -2, -1)$. Substituting $a=2$ into our ratio $\frac{l}{-2} = \frac{m}{2} = \frac{n}{-4}$, we can choose $l=1, m=-1, n=2$. Thus, the plane equation is $x - y + 2z = 0$.

Consider the triangle $	riangle BDC$. Since $BD$ is the perpendicular from $B$ to the plane, $	riangle BDC$ is a right-angled triangle with the right angle at $D$. By the Pythagorean theorem:

$$CD^2 = BC^2 - BD^2$$

First, we calculate the distance $BC^2$ between $B(0, 4, 5)$ and $C(0, -2, -1)$:

$$BC^2 = (0 - 0)^2 + (4 - (-2))^2 + (5 - (-1))^2 = 0 + 36 + 36 = 72$$

Next, we calculate the perpendicular distance $BD$ from $B(0, 4, 5)$ to the plane $x - y + 2z = 0$:

$$BD = \frac{|0 - 4 + 2(5)|}{\sqrt{1^2 + (-1)^2 + 2^2}} = \frac{6}{\sqrt{6}} = \sqrt{6}$$

Thus, $BD^2 = 6$. Substituting these values into our equation for $CD^2$:

$$CD^2 = 72 - 6 = 66$$

The final length of the segment $CD$ is **$\sqrt{66}$**.

Animated Solution for Mathematics - Three Dimensional Geometry: If for $a > 0$ , the feet of perpendiculars from the points $A (a, - 2 a, 3)$ and $B (0, 4, 5)$ on the plane $l x + m y + n z = 0$ are points $C (0, - a, - 1)$ and $D$ respectively, then the length of line segment $C D$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Geometry

Point $C$ lies on the Plane

Normal Vector and $A C$

Proportionality of Direction Ratios

Solving for $a$

Finding the Plane Equation

The Geometry of Triangle $B D C$

Calculating Distance $B C$

Calculating Perpendicular Distance $B D$

Final Calculation for $C D$

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

Visualizing the Geometry

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C lies on the Plane

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

Proportionality of Direction Ratios

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

Finding the Plane Equation

The Geometry of Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BDC

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

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

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