The equation of the line through the point (0,1,2) and perpendicular to the line (x-1)/2 = (y+1)/3 =

Q: Wait, why do we assume point B exists on the line? How do we know it's the foot of the perpendicular?

That is a fantastic question! By definition, the 'perpendicular from a point to a line' must meet the line at a specific point. We call this point the 'foot of the perpendicular.' By setting the line equation equal to a parameter \lambda, we are essentially saying, 'Let B be any point on this line.' When we solve for \lambda using the dot product condition, we are forcing B to be the specific point that satisfies the 90-degree requirement. It is the most elegant way to lock down that intersection point.

Q: Can we use the cross product instead of the dot product to find the perpendicular line?

You are thinking in the right direction! The cross product of two vectors gives a third vector perpendicular to both. However, in this problem, we are looking for the equation of a line, which requires a point (which we have: A) and a direction vector. The dot product is the most direct tool to find the specific point B, which then gives us the direction vector of the line AB. Using the cross product would be more complex here, as it would require finding the plane containing the line and the point first.

Q: Why did we multiply the direction ratios by 17 and then divide by 7 at the end?

Remember, direction ratios are not unique; they are just proportional to the actual direction vector. Any set of numbers (a, b, c) that are proportional to the direction of the line will define the same line. We scaled the ratios to make them integers, which makes the final equation look much cleaner and easier to match with the given options. It is a standard practice in coordinate geometry to simplify direction ratios to their simplest integer form.

Question

The equation of the line through the point $(0,1,2)$ and perpendicular to the line $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{-2}$ is :

Options:

(A) $\frac{x}{-3} = \frac{y-1}{4} = \frac{z-2}{3}$

(B) $\frac{x}{3} = \frac{y-1}{4} = \frac{z-2}{3}$

(C) $\frac{x}{3} = \frac{y-1}{-4} = \frac{z-2}{3}$

(D) $\frac{x}{3} = \frac{y-1}{4} = \frac{z-2}{-3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing in a three-dimensional void. You are holding a fixed point $A(0, 1, 2)$, and floating below you is a line $L_1$ defined by the symmetric equation:

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

Your mission is to draw a line from $A$ that strikes $L_1$ at a perfect $90^{\circ}$ angle. This is the classic problem of finding the perpendicular from a point to a line, a fundamental geometric construction in 3D space analysis.

### The Power of the Parameter

To find this line, we need two points: $A$ and the 'foot of the perpendicular,' which we will call $B$. Since $B$ lies on $L_1$, it must satisfy the line's equation.

We introduce a scalar parameter $\lambda$ to unlock its coordinates. By setting:

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

We can express any point on the line as $B(2\lambda + 1, 3\lambda - 1, -2\lambda + 1)$. This parameter $\lambda$ acts as our bridge between the abstract line and the specific point $B$ we need.

### The Orthogonality Condition

Now, consider the vector $\vec{AB}$. Its direction ratios are the differences in coordinates: $(x_B - x_A, y_B - y_A, z_B - z_A)$.

Substituting our coordinates, we find the direction ratios of $\vec{AB}$ are $(2\lambda + 1, 3\lambda - 2, -2\lambda - 1)$. Because $\vec{AB}$ is perpendicular to $L_1$, the dot product of their direction vectors must vanish.

The direction vector of $L_1$ is $\vec{d}_{L_1} = (2, 3, -2)$. Thus, we set the dot product to zero:

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

### Solving for the Unknown

Expanding this equation requires precision. We obtain:

$$4\lambda + 2 + 9\lambda - 6 + 4\lambda + 2 = 0$$

Grouping the terms, we find $17\lambda - 2 = 0$, which yields:

$$\lambda = \frac{2}{17}$$

With $\lambda$ in hand, we can find the exact direction ratios of our line. Substituting $\lambda$ back into our expression for $\vec{AB}$, we get the ratios:

$$\left( \frac{21}{17}, -\frac{28}{17}, -\frac{21}{17} \right)$$

### Final Calculation

To make these manageable, we multiply by $17$ to get $(21, -28, -21)$, and then divide by $7$ to get $(3, -4, -3)$. Finally, to match the standard form, we multiply by $-1$ to obtain the direction ratios $(-3, 4, 3)$.

The equation of our line passing through $A(0, 1, 2)$ with direction ratios $(-3, 4, 3)$ is:

$$\frac{x}{-3} = \frac{y-1}{4} = \frac{z-2}{3}$$

You have successfully navigated the 3D space to construct the **perpendicular line**.

Animated Solution for Mathematics - Three Dimensional Geometry: The equation of the line through the point $(0, 1, 2)$ and perpendicular to the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{- 2}$ is :

Select Answer:

Visualized Solution

Visualizing the Geometry

Parametric Point $B$ on $L_{1}$

Coordinates of Point $B$

Direction Ratios of $A B$

Calculating DRs of $A B$

Perpendicularity Condition

Setting up the Dot Product

Expanding the Equation

Solving for $λ$

Substituting $λ$ back

Simplifying DRs

Finalizing DRs

Final Equation of Line $A B$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Power of the Parameter

The Orthogonality Condition

Solving for the Unknown

Final Calculation

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Select Answer:

Visualized Solution

Visualizing the Geometry

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

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

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