Consider a line L passing through the points P(1, 2, 1) and Q(2, 1, -1). If the mirror image of the

Q: Why can't I just use the distance formula to find the image?

The distance formula only gives you the magnitude, not the direction. To find the mirror image, you need the exact coordinates, which requires the vector projection or the foot of the perpendicular to define the line of reflection.

Q: How do I know if the dot product should be zero?

The vector from the point $A$ to the foot of the perpendicular $C$ must be perpendicular to the line $L$ itself. Orthogonality is the geometric definition of the shortest distance, which is exactly what the foot of the perpendicular represents.

Q: Is there a shortcut formula for the mirror image of a point in a line?

While there are formulas, deriving it via the foot of the perpendicular is safer and builds deeper intuition. Don't just memorize; understand the process of projecting the point onto the line.

Question

Consider a line $L$ passing through the points $P(1, 2, 1)$ and $Q(2, 1, -1)$. If the mirror image of the point $A(2, 2, 2)$ in the line $L$ is $(\alpha, \beta, \gamma)$, then $\alpha + \beta + 6\gamma$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the mirror image of point $A(2, 2, 2)$ across the line $L$ passing through $P(1, 2, 1)$ and $Q(2, 1, -1)$, we first define the line's geometry. The direction vector of the line is given by:

$$\vec{v} = Q - P = (2-1, 1-2, -1-1) = (1, -1, -2)$$

Any point $C$ on the line $L$ can be expressed in terms of a parameter $k$ using the symmetric form:

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

This yields the general coordinates for a point on the line: $C = (k+1, -k+2, -2k+1)$.

### The Perpendicular Drop

The foot of the perpendicular $C$ is the point on $L$ such that the vector $\vec{AC}$ is orthogonal to the direction vector $\vec{v}$. We calculate $\vec{AC}$ as:

$$\vec{AC} = C - A = (k+1-2, -k+2-2, -2k+1-2) = (k-1, -k, -2k-1)$$

For orthogonality, the dot product $\vec{AC} \cdot \vec{v}$ must be zero:

$$(k-1)(1) + (-k)(-1) + (-2k-1)(-2) = 0$$

Expanding this equation, we obtain:

$$k - 1 + k + 4k + 2 = 0$$

$$6k + 1 = 0 \Rightarrow k = -\frac{1}{6}$$

### Finding the Image

Substituting $k = -\frac{1}{6}$ into the expression for $C$, we find the coordinates of the foot of the perpendicular:

$$C = \left( -\frac{1}{6} + 1, \frac{1}{6} + 2, \frac{2}{6} + 1 \right) = \left( \frac{5}{6}, \frac{13}{6}, \frac{4}{3} \right)$$

Let $B(\alpha, \beta, \gamma)$ be the mirror image of $A$. Since $C$ is the midpoint of $AB$, we apply the midpoint formula:

$$\frac{\alpha + 2}{2} = \frac{5}{6}, \quad \frac{\beta + 2}{2} = \frac{13}{6}, \quad \frac{\gamma + 2}{2} = \frac{4}{3}$$

Solving these equations, we find:

$$\alpha = \frac{5}{3} - 2 = -\frac{1}{3}, \quad \beta = \frac{13}{3} - 2 = \frac{7}{3}, \quad \gamma = \frac{8}{3} - 2 = \frac{2}{3}$$

### Final Calculation

We are tasked with finding the value of $\alpha + \beta + 6\gamma$:

$$\alpha + \beta + 6\gamma = -\frac{1}{3} + \frac{7}{3} + 6\left( \frac{2}{3} \right)$$

$$\alpha + \beta + 6\gamma = \frac{6}{3} + 4 = 2 + 4 = 6$$

The final result is **6**.

Animated Solution for Mathematics - Three Dimensional Geometry: Consider a line $L$ passing through the points $P (1, 2, 1)$ and $Q (2, 1, - 1)$ . If the mirror image of the point $A (2, 2, 2)$ in the line $L$ is $(α, β, γ)$ , then $α + β + 6 γ$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing Line $L$ through $P$ and $Q$

Direction Ratios of Line $L$

Parametric Equation of $L$

Introducing Point $A$ and Foot $C$

Defining Vector $A C$

The Orthogonality Condition

Solving for Parameter $k$

Coordinates of Foot $C$

Image $B$ as Midpoint

Finding $α, β, γ$

Final Calculation

Key Takeaway

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Perpendicular Drop

Finding the Image

Final Calculation

Similar Questions

Let $(α, β, γ)$ be mirror image of the point $(2, 3, 5)$ in the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ . Then $2 α + 3 β + 4 γ$ is equal to

If the mirror image of the point $P (3, 4, 9)$ in the line $\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 2}{1}$ is $(α, β, γ)$ , then $14 (α + β + γ)$ is :

Let $a, b \in R$ . If the mirror image of the point $P (a, 6, 9)$ with respect to the line $\frac{x - 3}{7} = \frac{y - 2}{5} = \frac{z - 1}{- 9}$ is $(20, b, - a - 9)$ , then $∣ a + b ∣$ is equal to:

If the image of the point $P (a, 2, a)$ in the line $\frac{x}{2} = \frac{y + a}{1} = \frac{z}{1}$ is $Q$ and the image of $Q$ in the line $\frac{x - 2 b}{2} = \frac{y - a}{1} = \frac{z + 2 b}{- 5}$ is $P$ , then $a + b$ is equal to .

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 $26$ units and $Q (α_{1}, α_{2}, α_{3})$ is the image of the point $P$ in this line, then $a + \sum_{i = 1}^{3} α_{i}$ is equal to :

If the image of the point $P (1, 0, 3)$ in the line joining the points $A (4, 7, 1)$ and $B (3, 5, 3)$ is $Q (α, β, γ)$ , then $α + β + γ$ is equal to

If $(a, b, c)$ is the image of the point $(1, 2, - 3)$ in the line $\frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1}$ , then $a + b + c$ is equal to:

Let the image of the point $P (1, 2, 3)$ in the line $L : \frac{x - 6}{3} = \frac{y - 1}{2} = \frac{z - 2}{3}$ be $Q$ . let $R (α, β, γ)$ be a point that divides internally the line segment $P Q$ in the ratio $1 : 3$ . Then the value of $22 (α + β + γ)$ is equal to ____.

If the distances of the point $(1, 2, a)$ from the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ along the lines $L_{1} : \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - a}{b}$ and $L_{2} : \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c}$ are equal, then $a + b + c$ is equal to

If the image of the point $P (1, 2, a)$ in the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{7 - z}{2}$ is $Q (5, b, c)$ , then $a^{2} + b^{2} + c^{2}$ is equal to

Enter Numerical Value:

Visualized Solution

Direction Ratios of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L

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

Introducing Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and Foot .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

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

The Orthogonality Condition

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

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

Image .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B as Midpoint

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α,β,γ

Final Calculation

Key Takeaway

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Perpendicular Drop

Finding the Image

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let (α,β,γ) be mirror image of the point (2,3,5) in the line 2x−1​=3y−2​=4z−3​. Then 2α+3β+4γ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the mirror image of the point P(3,4,9) in the line 3x−1​=2y+1​=1z−2​ is (α,β,γ), then 14(α+β+γ) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,b∈R. If the mirror image of the point P(a,6,9) with respect to the line 7x−3​=5y−2​=−9z−1​ is (20,b,−a−9), then ∣a+b∣ is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the image of the point P(a,2,a) in the line 2x​=1y+a​=1z​ is Q and the image of Q in the line 2x−2b​=1y−a​=−5z+2b​ is P, then a+b is equal to .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the image of the point P(1,0,3) in the line joining the points A(4,7,1) and B(3,5,3) is Q(α,β,γ), then α+β+γ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (a,b,c) is the image of the point (1,2,−3) in the line 2x+1​=−2y−3​=−1z​, then a+b+c is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the image of the point P(1,2,a) in the line 3x−6​=2y−7​=27−z​ is Q(5,b,c), then a2+b2+c2 is equal to

Direction Ratios of Line $L$

Parametric Equation of $L$

Introducing Point $A$ and Foot $C$

Defining Vector $A C$

Solving for Parameter $k$

Coordinates of Foot $C$

Image $B$ as Midpoint

Finding $α, β, γ$

Let $(α, β, γ)$ be mirror image of the point $(2, 3, 5)$ in the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ . Then $2 α + 3 β + 4 γ$ is equal to

If the mirror image of the point $P (3, 4, 9)$ in the line $\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 2}{1}$ is $(α, β, γ)$ , then $14 (α + β + γ)$ is :

Let $a, b \in R$ . If the mirror image of the point $P (a, 6, 9)$ with respect to the line $\frac{x - 3}{7} = \frac{y - 2}{5} = \frac{z - 1}{- 9}$ is $(20, b, - a - 9)$ , then $∣ a + b ∣$ is equal to:

If the image of the point $P (a, 2, a)$ in the line $\frac{x}{2} = \frac{y + a}{1} = \frac{z}{1}$ is $Q$ and the image of $Q$ in the line $\frac{x - 2 b}{2} = \frac{y - a}{1} = \frac{z + 2 b}{- 5}$ is $P$ , then $a + b$ is equal to .

If the image of the point $P (1, 0, 3)$ in the line joining the points $A (4, 7, 1)$ and $B (3, 5, 3)$ is $Q (α, β, γ)$ , then $α + β + γ$ is equal to

If $(a, b, c)$ is the image of the point $(1, 2, - 3)$ in the line $\frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1}$ , then $a + b + c$ is equal to:

If the image of the point $P (1, 2, a)$ in the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{7 - z}{2}$ is $Q (5, b, c)$ , then $a^{2} + b^{2} + c^{2}$ is equal to