Let the image of the point (1, 0, 7) in the line x/1 = (y-1)/2 = (z-2)/3 be the point (α, β, γ). The

Q: Why do we use the midpoint formula to find the image?

Great question! By definition, the image $Q$ of a point $P$ reflected across a line is such that the line is the perpendicular bisector of the segment $PQ$. This means the foot of the perpendicular $M$ is exactly halfway between $P$ and $Q$. Thus, $M = \frac{P+Q}{2}$, which rearranges to $Q = 2M - P$.

Q: How do I know if the angle is acute or obtuse when finding the direction cosine $l$?

The problem gives you the angles with the axes, which directly relate to the direction cosines $l, m, n$. The condition 'acute angle with the x-axis' means the cosine of that angle must be positive. Since $l = \cos(\theta_x)$, we must have $l > 0$. This is how we choose the positive root for $l$.

Q: Can I use the projection formula instead of the dot product?

Yes, you absolutely can! The projection of vector $\vec{AP}$ onto the line's direction vector $\vec{d}$ gives you the distance to the foot of the perpendicular. However, the dot product method is often faster and less prone to algebraic errors in 3D coordinate geometry. Both paths lead to the same beautiful destination.

Question

Let the image of the point $(1, 0, 7)$ in the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$ be the point $(\alpha, \beta, \gamma)$. Then which one of the following points lies on the line passing through $(\alpha, \beta, \gamma)$ and making angles $\frac{2\pi}{3}$ and $\frac{3\pi}{4}$ with $y$-axis and $z$-axis respectively and an acute angle with $x$-axis ?

Options:

(A) $(1, -2, 1 + \sqrt{2})$

(B) $(1, 2, 1 - \sqrt{2})$

(C) $(3, 4, 3 - 2\sqrt{2})$

(D) $(3, -4, 3 + 2\sqrt{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow explorer of the mathematical universe! Today, we are diving into the elegant world of 3D geometry. Imagine you are standing in a room, and there is a line suspended in the air.

You have a point, a specific coordinate in space, and you want to see its reflection—its image—as if that line were a mirror. This is exactly what we are tackling today. It is not just about crunching numbers; it is about visualizing the symmetry of space.

### Finding the Foot of the Perpendicular

We start with point $P(1, 0, 7)$ and the line $L_1$ defined by:

$$\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$$

To find the image $Q(\alpha, \beta, \gamma)$, we need a bridge. That bridge is the foot of the perpendicular, $M$. Think of $M$ as the point on the line closest to $P$.

Because $M$ lies on the line, we can describe it using a single parameter, $\lambda$. By setting the line equations equal to $\lambda$, we get the general coordinates of $M$ as $(\lambda, 1 + 2\lambda, 2 + 3\lambda)$.

Now, we invoke the power of vectors. The vector $\vec{PM}$ connects our original point to this mystery point $M$. Since $M$ is the foot of the perpendicular, $\vec{PM}$ must be perpendicular to the line itself.

The direction vector of our line is $\vec{d} = (1, 2, 3)$. The condition for perpendicularity is simple yet profound: the dot product $\vec{PM} \cdot \vec{d}$ must be zero.

Calculating $\vec{PM} = (\lambda - 1, 1 + 2\lambda, 3\lambda - 5)$, we set up the equation:

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

Solving this, we find $\lambda = 1$. Substituting this back, we find $M = (1, 3, 5)$.

### The Image Revealed

With $M$ found, the image $Q$ is just a step away. Since $M$ is the midpoint of $PQ$, we use the relation $Q = 2M - P$.

Plugging in our coordinates, we get:

$$Q = (2(1)-1, 2(3)-0, 2(5)-7)$$

This simplifies to **$Q(1, 6, 3)$**. We have successfully reflected our point!

### The New Path

Now, the problem shifts. We have a new line $L_2$ passing through $Q(1, 6, 3)$. We are given the angles it makes with the axes: $\frac{2\pi}{3}$ with the y-axis and $\frac{3\pi}{4}$ with the z-axis.

We need the direction cosines $l, m, n$. We know:

$$m = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$$

$$n = \cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}$$

The fundamental identity $l^2 + m^2 + n^2 = 1$ is our best friend here. Substituting our values, we get:

$$l^2 + \frac{1}{4} + \frac{1}{2} = 1 \Rightarrow l^2 = \frac{1}{4}$$

The problem tells us the line makes an acute angle with the x-axis, so $l$ must be positive. Thus, $l = \frac{1}{2}$.

### The Final Verification

The equation of our new line $L_2$ is:

$$\frac{x-1}{1/2} = \frac{y-6}{-1/2} = \frac{z-3}{-1/\sqrt{2}} = \mu$$

By testing the options, we find that for the point $(3, 4, 3 - 2\sqrt{2})$, the parameter $\mu$ is consistent across all coordinates. We have arrived!

The beauty of this problem lies in how it weaves together reflection, vector orthogonality, and the properties of direction cosines. Keep practicing, and soon, these 3D structures will feel as natural as walking in your own home.

Select Answer:

Visualized Solution

Visualizing the Setup

The Foot of the Perpendicular

General Point on $L_{1}$

Direction Vector $P M$

Applying Perpendicularity Condition

Exact Coordinates of $M$

Finding the Image Point $Q$

Calculating Coordinates of $Q$

The Second Line $L_{2}$

Direction Cosines $m$ and $n$

Finding Direction Cosine $l$

Applying the Acute Angle Condition

Equation of Line $L_{2}$

Verifying the Options

Final Conclusion

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

Finding the Foot of the Perpendicular

The Image Revealed

The New Path

The Final Verification

Similar Questions

Let the line $L$ pass through $(1, 1, 1)$ and intersect the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z}{1}$ . Then, which of the following points lies on the line $L$ ?

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 :

Let the line $L$ intersect the lines $x - 2 = - y = z - 1, 2 (x + 1) = 2 (y - 1) = z + 1$ and be parallel to the line $\frac{x - 2}{3} = \frac{y - 1}{1} = \frac{z - 2}{2}$ . Then which of the following points lies on L?

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 $(α, β, γ)$ be the image of the point $(8, 5, 7)$ in the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{5}$ . Then $α + β + γ$ is equal to :

Let $Q (a, b, c)$ be the image of the point $P (3, 2, 1)$ in the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ . Then the distance of $Q$ from the line $\frac{x - 9}{3} = \frac{y - 9}{2} = \frac{z - 5}{- 2}$ is

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

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 $P (1, 0, 3)$ 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

Select Answer:

Visualized Solution

Visualizing the Setup

The Foot of the Perpendicular

General Point on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

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

Applying Perpendicularity Condition

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

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

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

The Second Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Direction Cosines .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

Finding Direction Cosine .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l