Let the mirror image of the point (a, b, c) with respect to the plane 3x - 4y + 12z + 19 = 0 be (a-6

Q: Why is the line segment connecting the point and its image parallel to the normal vector?

Think of the plane as a flat surface. The shortest path from a point to the plane is along the normal. Since the image is just as far behind the plane as the object is in front, the line connecting them must be perpendicular to the plane's surface, which is exactly what the normal vector defines!

Q: Do I really need to find the values of a, b, and c individually?

Not at all! This is a classic JEE trap. Often, the variables are linked in such a way that they cancel out or form a specific expression. Focus on finding the relationship between the variables rather than solving for each one.

Q: How do I know the midpoint lies on the plane?

By definition, the mirror image is symmetric with respect to the plane. The point where the line segment crosses the plane is the midpoint of the segment connecting the object and its image. Since it's on the plane, it must satisfy the plane's equation.

Q: What if I get confused with the signs in the normal vector?

The normal vector $\vec{n} = (A, B, C)$ comes directly from the coefficients of $x, y, z$ in the plane equation $Ax + By + Cz + D = 0$. Just be careful to carry the negative sign with the coefficient of $y$!

Q: Is there a faster way to find the image of a point?

You can use the formula $\frac{x-a}{A} = \frac{y-b}{B} = \frac{z-c}{C} = -2 \frac{Aa+Bb+Cc+D}{A^2+B^2+C^2}$. It's a powerful tool, but understanding the derivation via the midpoint method is what builds your intuition for tougher problems.

Question

Let the mirror image of the point $(a, b, c)$ with respect to the plane $3x - 4y + 12z + 19 = 0$ be $(a-6, \beta, \gamma)$. If $a + b + c = 5$, then $7\beta - 9\gamma$ is equal to ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a room, looking at a mirror. In the world of 3D geometry, reflecting a point $(a, b, c)$ across the plane $3x - 4y + 12z + 19 = 0$ is a precise mathematical relationship.

This process is not just about plugging numbers into a formula; it is about understanding how space itself folds around a plane.

### The Normal: The Compass of the Plane

Every plane has a soul, and that soul is its normal vector, $\vec{n} = (3, -4, 12)$. This vector acts as the compass for our problem.

When we reflect a point $P(a, b, c)$ to its image $P'(a-6, \beta, \gamma)$, the line segment $PP'$ must be perpendicular to the plane. This means the vector $\vec{PP'}$ must be parallel to the normal vector $\vec{n}$.

We can write this relationship as a ratio:

$$\frac{(a-6)-a}{3} = \frac{\beta-b}{-4} = \frac{\gamma-c}{12} = k$$

Look at the first term: $\frac{-6}{3} = -2$. This tells us that our constant of proportionality $k$ is $-2$.

With $k = -2$, we can immediately express $\beta$ and $\gamma$ in terms of $b$ and $c$:

$$\beta = b + 8$$

$$\gamma = c - 24$$

### The Midpoint: The Bridge to the Plane

The midpoint $M$ of the segment $PP'$ must lie exactly on the plane. Let's calculate the coordinates of $M$:

$$M = \left( \frac{a+(a-6)}{2}, \frac{b+\beta}{2}, \frac{c+\gamma}{2} \right)$$

Substituting our expressions for $\beta$ and $\gamma$, we find:

$$M = (a-3, b+4, c-12)$$

This point $M$ is the bridge. Because it lies on the plane $3x - 4y + 12z + 19 = 0$, it must satisfy the equation.

When we substitute these coordinates into the plane equation, we get:

$$3(a-3) - 4(b+4) + 12(c-12) + 19 = 0$$

### The Beauty of Algebraic Cancellation

Expanding this, we get $3a - 9 - 4b - 16 + 12c - 144 + 19 = 0$, which simplifies to $3a - 4b + 12c = 150$. Now, we use the constraint given in the problem: $a + b + c = 5$.

This allows us to replace $a$ with $5 - b - c$. Substituting this into our equation:

$$3(5 - b - c) - 4b + 12c = 150$$

$$15 - 3b - 3c - 4b + 12c = 150$$

$$-7b + 9c = 135$$

This gives us the vital relationship $7b - 9c = -135$.

### The Final Reveal

We are asked to find the value of $7\beta - 9\gamma$. Let's substitute our expressions for $\beta$ and $\gamma$ back in:

$$7(b+8) - 9(c-24) = 7b + 56 - 9c + 216$$

Rearranging the terms, we see the expression we just solved for:

$$(7b - 9c) + 272$$

Substituting $-135$ for $(7b - 9c)$, we get $-135 + 272 = 137$.

The complexity of the variables $a, b,$ and $c$ dissolves, leaving us with a clean, elegant result. The final answer is **137**.

Animated Solution for Mathematics - Three Dimensional Geometry: Let the mirror image of the point $(a, b, c)$ with respect to the plane $3 x - 4 y + 12 z + 19 = 0$ be $(a - 6, β, γ)$ . If $a + b + c = 5$ , then $7 β - 9 γ$ is equal to ____.

Enter Numerical Value:

Visualized Solution

Visualizing the Reflection

The Direction Ratio Property

Finding $β$ and $γ$

Locating the Midpoint $M$

Simplifying Midpoint Coordinates

Substituting $M$ into the Plane

Simplifying the Equation

Using the Constraint $a + b + c = 5$

The Relation Between $b$ and $c$

Calculating $7 β - 9 γ$

Final Result

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Normal

The Midpoint

The Beauty of Algebraic Cancellation

The Final Reveal

Similar Questions

If the mirror image of the point $(2, 4, 7)$ in the plane $3 x - y + 4 z = 2$ is $(a, b, c)$ , the $2 a + b + 2 c$ is equal to :

Let the plane $2 x + 3 y + z + 20 = 0$ be rotated through a right angle about its line of intersection with the plane $x - 3 y + 5 z = 8$ . If the mirror image of the point $(2, - \frac{1}{2}, 2)$ in the rotated plane is $B (a, b, c)$ , then :

Let $α, β, γ, δ$ be real numbers such that $α^{2} + β^{2} + γ^{2} \neq = 0$ and $α + γ = 1$ . Suppose the point $(3, 2, - 1)$ is the mirror image of the point $(1, 0, - 1)$ with respect to the plane $α x + β y + γ z = δ$ . Then which of the following statements is/are TRUE?

Let $S$ be the mirror image of the point $Q (1, 3, 4)$ with respect to the plane $2 x - y + z + 3 = 0$ and let $R (3, 5, γ)$ be a point of this plane. Then the square of the length of the line segment $S R$ is .

Let the image of the point $P (1, 2, 3)$ in the plane $2 x - y + z = 9$ be $Q$ . If the coordinates of the point $R$ are $(6, 10, 7)$ , then the square of the area of the triangle $P QR$ is _______.

Let $P$ be a plane containing the line $\frac{x - 1}{3} = \frac{y + 6}{4} = \frac{z + 5}{2}$ and parallel to the line $\frac{x - 3}{4} = \frac{y - 2}{- 3} = \frac{z + 5}{7}$ . If the point $(1, - 1, α)$ lies on the plane $P$ , then the value of $∣5 α ∣$ is equal to ___

If the equation of the plane passing through the line of intersection of the planes $2 x - y + z = 3, 4 x - 3 y + 5 z + 9 = 0$ and parallel to the line $\frac{x + 1}{- 2} = \frac{y + 3}{4} = \frac{z - 2}{5}$ is $a x + b y + cz + 6 = 0$ , then $a + b + c$ is equal to

The mirror image of the point $(1, 2, 3)$ in a plane is $(- \frac{7}{3}, - \frac{4}{3}, - \frac{1}{3})$ . Which of the following points lies on this plane?

The mirror image of the point $(1, 2, 3)$ in a plane is $(- \frac{7}{3}, - \frac{4}{3}, - \frac{1}{3})$ . Which of the following points lies on this plane?

Enter Numerical Value:

Visualized Solution

Visualizing the Reflection

The Direction Ratio Property

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

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

Simplifying Midpoint Coordinates

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

Simplifying the Equation

Using the Constraint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+b+c=5

The Relation Between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 7β−9γ

Final Result

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Normal

The Midpoint

The Beauty of Algebraic Cancellation

The Final Reveal

Similar Questions

.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 (2,4,7) in the plane 3x−y+4z=2 is (a,b,c), the 2a+b+2c is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the image of the point P(1,2,3) in the plane 2x−y+z=9 be Q. If the coordinates of the point R are (6,10,7), then the square of the area of the triangle PQR is _______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The mirror image of the point (1,2,3) in a plane is (−37​,−34​,−31​). Which of the following points lies on this plane?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The mirror image of the point (1,2,3) in a plane is (−37​,−34​,−31​). Which of the following points lies on this plane?

Finding $β$ and $γ$

Locating the Midpoint $M$

Substituting $M$ into the Plane

Using the Constraint $a + b + c = 5$

The Relation Between $b$ and $c$

Calculating $7 β - 9 γ$

If the mirror image of the point $(2, 4, 7)$ in the plane $3 x - y + 4 z = 2$ is $(a, b, c)$ , the $2 a + b + 2 c$ is equal to :

Let the image of the point $P (1, 2, 3)$ in the plane $2 x - y + z = 9$ be $Q$ . If the coordinates of the point $R$ are $(6, 10, 7)$ , then the square of the area of the triangle $P QR$ is _______.

The mirror image of the point $(1, 2, 3)$ in a plane is $(- \frac{7}{3}, - \frac{4}{3}, - \frac{1}{3})$ . Which of the following points lies on this plane?

The mirror image of the point $(1, 2, 3)$ in a plane is $(- \frac{7}{3}, - \frac{4}{3}, - \frac{1}{3})$ . Which of the following points lies on this plane?