A line with positive direction cosines passes through the point P(2, -1, 2) and makes equal angles w

Q: Why is the distance $PQ$ simply equal to $|r|$? I thought $r$ was just a parameter.

That is a fantastic question! In the symmetric form of a line equation $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n} = r$, the denominator $(l, m, n)$ is a unit vector. When the direction vector is a unit vector, the parameter $r$ represents the exact distance from the fixed point $(x_1, y_1, z_1)$ to any point $(x, y, z)$ on the line. It is a beautiful property of unit vectors that simplifies our lives immensely!

Q: Why did we reject the negative root for the direction cosines?

The problem explicitly stated that the line has 'positive direction cosines.' In 3D geometry, a line can point in two opposite directions. By specifying positive cosines, the problem restricts our line to a specific orientation. Always read the constraints carefully—they are the guardrails of your solution, preventing you from wandering into the wrong octant.

Q: Could I have solved this using the vector equation of the line $\vec{r} = \vec{a} + t\vec{b}$?

Absolutely! You could define the line as $\vec{r} = (2, -1, 2) + t(1, 1, 1)$. However, remember that if you use the vector $(1, 1, 1)$, it is not a unit vector. The parameter $t$ would not be the distance $PQ$. You would have to calculate the distance using the distance formula $\sqrt{(x_2-x_1)^2 + ...}$ after finding $Q$. Using the unit vector approach (as we did) is faster because it gives you the distance directly.

Question

A line with positive direction cosines passes through the point $P(2, -1, 2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x + y + z = 9$ at point $Q$. The length of the line segment $PQ$ equals

Options:

(A) 1

(B) $\sqrt{2}$

(C) $\sqrt{3}$

(D) 2

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the length of a line segment $PQ$, where $P$ is the point $(2, -1, 2)$ and $Q$ is the intersection point of a line passing through $P$ with the plane $2x + y + z = 9$. The line is defined by the property that it makes equal angles with the coordinate axes.

### The Symmetry of the Line

Let the line make angles $\alpha, \beta,$ and $\gamma$ with the $x, y,$ and $z$ axes, respectively. Given the condition of equal angles, we have $\alpha = \beta = \gamma$.

The direction cosines of the line are $l = \cos \alpha$, $m = \cos \beta$, and $n = \cos \gamma$. Consequently, we have the equality $l = m = n$.

### The Fundamental Identity

We utilize the fundamental identity for direction cosines in 3D space:
$$l^2 + m^2 + n^2 = 1$$

Substituting $l = m = n$ into this identity, we obtain:
$$l^2 + l^2 + l^2 = 1 \Rightarrow 3l^2 = 1$$

Solving for $l$, we find $l^2 = \frac{1}{3}$. Given the constraint that the direction cosines are positive, we determine:
$$l = m = n = \frac{1}{\sqrt{3}}$$

### The Parametric Bridge

We define the line passing through $P(2, -1, 2)$ with direction ratios proportional to $(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$. We can represent any point $Q$ on this line using a parameter $r$, which represents the distance from $P$:
$$Q = \left( 2 + \frac{r}{\sqrt{3}}, -1 + \frac{r}{\sqrt{3}}, 2 + \frac{r}{\sqrt{3}} \right)$$

### The Collision

The point $Q$ must lie on the plane $2x + y + z = 9$. Substituting the parametric coordinates of $Q$ into the plane equation:

$$2\left(2 + \frac{r}{\sqrt{3}}\right) + \left(-1 + \frac{r}{\sqrt{3}}\right) + \left(2 + \frac{r}{\sqrt{3}}\right) = 9$$

Expanding the terms, we get:
$$4 + \frac{2r}{\sqrt{3}} - 1 + \frac{r}{\sqrt{3}} + 2 + \frac{r}{\sqrt{3}} = 9$$

Combining the constants and the $r$ terms:
$$5 + \frac{4r}{\sqrt{3}} = 9$$

### Final Calculation

Subtracting 5 from both sides yields:
$$\frac{4r}{\sqrt{3}} = 4$$

Solving for $r$:
$$r = \sqrt{3}$$

Since $r$ represents the distance between $P$ and $Q$ when the direction vector is normalized, the length of the segment $PQ$ is **$\sqrt{3}$**.

Animated Solution for Mathematics - Three Dimensional Geometry: A line with positive direction cosines passes through the point $P (2, - 1, 2)$ and makes equal angles with the coordinate axes. The line meets the plane $2 x + y + z = 9$ at point $Q$ . The length of the line segment $P Q$ equals

Select Answer:

Visualized Solution

Visualizing the Setup

Understanding Equal Angles

Direction Cosines Identity

Solving for $l, m, n$

Selecting Positive Direction Cosines

Equation of the Line

Parametric Coordinates of Point $Q$

Substituting $Q$ into the Plane Equation

Simplifying the Equation

Solving for $r$

Finding the Length $P Q$

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Symmetry of the Line

The Fundamental Identity

The Parametric Bridge

The Collision

Final Calculation

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Select Answer:

Visualized Solution

Visualizing the Setup

Understanding Equal Angles

Direction Cosines Identity

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

Selecting Positive Direction Cosines

Equation of the Line

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

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

Simplifying the Equation

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

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