Let P(α, β, γ) be the point on the line (x-1)/2 = (y+1)/-3 = z at a distance 4√14 from the point (1,

Q: Why do we get two values for $\lambda$ and how do we know which one to pick?

The equation $\lambda^2 = 16$ naturally yields $\pm 4$. The problem adds a constraint: 'nearer to the origin'. Always check the physical constraints given in the problem statement; they are there to guide you, not to confuse you. By calculating the distance from the origin for both potential points, you can easily identify the correct one.

Q: Is the order of subtraction in $\vec{a_2} - \vec{a_1}$ important?

Mathematically, no, because the formula uses the absolute value. Physically, it represents the vector connecting the two lines. As long as you are consistent, the magnitude will remain the same.

Q: Why can't we just use the projection of the vector connecting the lines onto the common perpendicular?

You absolutely can! That is the geometric derivation of the formula. The formula is just a shortcut. If you ever forget the formula, remember that the shortest distance is the projection of the vector connecting any two points on the lines onto the common normal vector $\vec{n} = \vec{b_1} \times \vec{b_2}$.

Question

Let $P(\alpha, \beta, \gamma)$ be the point on the line $\frac{x-1}{2} = \frac{y+1}{-3} = z$ at a distance $4\sqrt{14}$ from the point $(1, -1, 0)$ and nearer to the origin. Then the shortest distance, between the lines $\frac{x-\alpha}{1} = \frac{y-\beta}{2} = \frac{z-\gamma}{3}$ and $\frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1}$, is equal to

Options:

(A) $4\sqrt{\frac{7}{5}}$

(B) $2\sqrt{\frac{7}{4}}$

(C) $7\sqrt{\frac{5}{4}}$

(D) $4\sqrt{\frac{5}{7}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To navigate the 3D space, we first define the line given by the equation:

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

By setting this expression equal to a parameter $\lambda$, we can represent any arbitrary point $P$ on this line as:

$$P = (2\lambda + 1, -3\lambda - 1, \lambda)$$

### The Constraint of Distance

We are given that the distance between point $P$ and the fixed point $A(1, -1, 0)$ is $4\sqrt{14}$. The vector $\vec{AP}$ is calculated as:

$$\vec{AP} = (2\lambda, -3\lambda, \lambda)$$

The square of the distance is the sum of the squares of these components:

$$(2\lambda)^2 + (-3\lambda)^2 + \lambda^2 = 4\lambda^2 + 9\lambda^2 + \lambda^2 = 14\lambda^2$$

Setting this equal to the square of the given distance, $(4\sqrt{14})^2 = 16 	imes 14$, we obtain:

$$14\lambda^2 = 16 	imes 14 \implies \lambda^2 = 16 \implies \lambda = \pm 4$$

To find the point $P$ nearer to the origin, we test both values. For $\lambda = -4$, we get $P(-7, 11, -4)$, which is closer to the origin than the point resulting from $\lambda = 4$.

### The Challenge of Skew Lines

We now find the shortest distance between line $L_2$ (passing through $P(-7, 11, -4)$ with direction $\vec{b_1} = (1, 2, 3)$) and line $L_3$ (given by $\frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1}$ with direction $\vec{b_2} = (2, 1, 1)$).

The shortest distance $d$ between two skew lines is given by:

$$d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})|}{|\vec{b_1} 	imes \vec{b_2}|}$$

First, we calculate the cross product $\vec{b_1} 	imes \vec{b_2}$:

$$\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 2 & 3 \ 2 & 1 & 1 \end{vmatrix} = (-1, 5, -3)$$

The magnitude of this cross product is:

$$|\vec{b_1} 	imes \vec{b_2}| = \sqrt{(-1)^2 + 5^2 + (-3)^2} = \sqrt{1 + 25 + 9} = \sqrt{35}$$

### Final Calculation

We define $\vec{a_2} - \vec{a_1}$ as the vector connecting a point on $L_3$, which is $(-5, 10, 3)$, to our point $P(-7, 11, -4)$:

$$\vec{a_2} - \vec{a_1} = (-7 - (-5), 11 - 10, -4 - 3) = (-2, 1, -7)$$

The scalar triple product is the dot product of $(-2, 1, -7)$ and $(-1, 5, -3)$:

$$|(-2)(-1) + (1)(5) + (-7)(-3)| = |2 + 5 + 21| = 28$$

The shortest distance is therefore:

$$d = \frac{28}{\sqrt{35}} = \frac{4 	imes 7}{\sqrt{7} 	imes \sqrt{5}} = 4\sqrt{\frac{7}{5}}$$

The final shortest distance is **$4\sqrt{\frac{7}{5}}$**.

Select Answer:

Visualized Solution

Visualizing the Setup

Parametric Coordinates of $P$

Distance Constraint $A P = 414$

Solving for $λ$

Selecting $P$ Nearer to Origin

Identifying the New Lines $L_{2}$ and $L_{3}$

Formula for Shortest Distance

Direction Vector Cross Product

Magnitude of $b_{1} \times b_{2}$

Vector $a_{2} - a_{1}$

Scalar Triple Product Calculation

Final Result Simplification

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Constraint of Distance

The Challenge of Skew Lines

Final Calculation

Similar Questions

The shortest distance between the lines $\frac{x - 5}{1} = \frac{y - 2}{2} = \frac{z - 4}{- 3}$ and $\frac{x + 3}{1} = \frac{y + 5}{4} = \frac{z - 1}{- 5}$ is

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}$ is

Shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ is

The shortest distance between the lines $\frac{x - 1}{0} = \frac{y + 1}{- 1} = \frac{z}{1}$ and $x + y + z + 1 = 0, 2 x - y + z + 3 = 0$ is:

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 8}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is:

Let $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ $L_{2} : \frac{x - 2}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$ and be two lines. Then which of the following points lies on the line of the shortest distance between L₁ and $L_{2}$ ?

The shortest distance between the lines $\frac{x - 3}{4} = \frac{y + 7}{- 11} = \frac{z - 1}{5}$ and $\frac{x - 5}{3} = \frac{y - 9}{- 6} = \frac{z + 2}{1}$ is:

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{- 1}$ and $\frac{x + 3}{2} = \frac{y - 6}{1} = \frac{z - 5}{3}$ is:

The shortest distance between the lines $\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}$ is :

Select Answer:

Visualized Solution

Visualizing the Setup

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

Distance Constraint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AP=414​

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

Selecting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P Nearer to Origin

Identifying the New Lines .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L3​