Let a=i+2j+k and b=2i+7j+3k. Let L₁: r=(-i+2j+k)+λa λ ∈ R and L₂: r=(j+k)+μb μ ∈ R be two lines. If

Q: Wait, how do I know if the lines actually intersect? What if they are skew lines?

That is a brilliant question! In 3D geometry, two lines are not guaranteed to intersect. They could be parallel or skew. However, in this problem, we are told they intersect. To verify this, you must find a single pair of $(\lambda, \mu)$ that satisfies all three coordinate equations ($x$, $y$, and $z$) simultaneously. If you find a solution for two coordinates but it fails for the third, the lines are skew.

Q: Why can't I just use the cross product to find the intersection point?

The cross product of the direction vectors $\vec{a}$ and $\vec{b}$ gives you a vector perpendicular to both lines, which is useful for finding the shortest distance between skew lines or the normal to a plane containing them. It does not give you the intersection point. To find the intersection, you must use the parametric method to solve for the specific point where the lines meet.

Q: Why do we use a new parameter $t$ for line $L_3$?

Great observation! We use $t$ because $L_3$ is a completely new line. The parameters $\lambda$ and $\mu$ were specific to $L_1$ and $L_2$. Using a new parameter $t$ allows us to define any point on $L_3$ independently of the previous lines. It is a standard convention to avoid confusion.

Question

Let $\vec{a}=\hat{i}+2\hat{j}+\hat{k}$ and $\vec{b}=2\hat{i}+7\hat{j}+3\hat{k}$. Let $L_{1}:\vec{r}=(-\hat{i}+2\hat{j}+\hat{k})+\lambda\vec{a}$ $\lambda\in R$ and $L_{2}:\vec{r}=(\hat{j}+\hat{k})+\mu\vec{b}$ $\mu\in R$ be two lines. If the line $L_{3}$ passes through the point of intersection of $L_{1}$ and $L_{2}$, and is parallel to $\vec{a}+\vec{b}$ then $L_{3}$ passes through the point:

Options:

(A) (5, 17, 4)

(B) $(2,8,5)$

(C) $(8,26,12)$

(D) $(-1,-1,1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional void. You see two lines, $L_1$ and $L_2$, stretching out into infinity.

While they may appear to cross, in 3D space, appearances can be deceiving. Our mission is to find the exact point where these two lines meet and chart a new path, $L_3$, that emerges from that intersection.

### The Parametric Dance

To find where $L_1$ and $L_2$ meet, we translate their vector equations into parametric form. For $L_1$, given by $\vec{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} + 2\hat{j} + \hat{k})$, the coordinates are:

$$x = -1 + \lambda, \quad y = 2 + 2\lambda, \quad z = 1 + \lambda$$

Similarly, for $L_2$, defined by $\vec{r} = (\hat{j} + \hat{k}) + \mu(2\hat{i} + 7\hat{j} + 3\hat{k})$, the coordinates are:

$$x = 2\mu, \quad y = 1 + 7\mu, \quad z = 1 + 3\mu$$

At the point of intersection, the $x$, $y$, and $z$ coordinates must be identical for both lines.

### Solving the System

We begin with the $z$-coordinates, as they provide the simplest path to the solution. Equating them, we get $1 + \lambda = 1 + 3\mu$, which simplifies to:

$$\lambda = 3\mu$$

Next, we examine the $x$-coordinates: $-1 + \lambda = 2\mu$. Substituting our relation $\lambda = 3\mu$ into this equation, we get $-1 + 3\mu = 2\mu$, which leads directly to $\mu = 1$.

With $\mu = 1$, we find $\lambda = 3(1) = 3$. We must now perform a consistency check using the $y$-coordinates.

For $L_1$, $y = 2 + 2(3) = 8$. For $L_2$, $y = 1 + 7(1) = 8$. Since they match, the lines intersect at the point **$P(2, 8, 4)$**.

### Charting the New Path

Now that we have our intersection point $P(2, 8, 4)$, we define $L_3$. The problem states $L_3$ is parallel to the vector sum $\vec{a} + \vec{b}$.

We calculate the direction vector $\vec{v}$ as follows:

$$\vec{v} = (\hat{i} + 2\hat{j} + \hat{k}) + (2\hat{i} + 7\hat{j} + 3\hat{k}) = 3\hat{i} + 9\hat{j} + 4\hat{k}$$

Using point $P$ and direction $\vec{v}$, the vector equation for $L_3$ is:

$$\vec{r} = (2\hat{i} + 8\hat{j} + 4\hat{k}) + t(3\hat{i} + 9\hat{j} + 4\hat{k})$$

This yields the parametric equations:

$$x = 3t + 2, \quad y = 9t + 8, \quad z = 4t + 4$$

### The Final Verification

To verify a point such as $(8, 26, 12)$, we check if there exists a value of $t$ that satisfies all three equations. Setting $x = 8$:

$$3t + 2 = 8 \Rightarrow 3t = 6 \Rightarrow t = 2$$

Now, we test $t = 2$ for $y$ and $z$:

$$y = 9(2) + 8 = 26$$
$$z = 4(2) + 4 = 12$$

The point matches perfectly. We have successfully navigated the 3D space and confirmed the path.

Select Answer:

Visualized Solution

Visualizing the Lines $L_{1}$ and $L_{2}$

Parametric Coordinates of $L_{1}$

Parametric Coordinates of $L_{2}$

Equating the $z$ -Coordinates

Solving for $μ$ using $x$ -Coordinates

Finding $λ$ and Verifying

Coordinates of Intersection Point $P$

Direction Vector of $L_{3}$

Equation of Line $L_{3}$

Testing Option $(8, 26, 12)$

Verifying $y$ and $z$ Coordinates

Final Conclusion

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Parametric Dance

Solving the System

Charting the New Path

The Final Verification

Similar Questions

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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 $L$ ?

Let P be the foot of the perpendicular from the point (1, 2, 2) on the line $L : \frac{x - 1}{1} = \frac{y + 1}{- 1} = \frac{z - 2}{2}$ Let the line $r = (- \hat{i} + \hat{j} - 2 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$ , $λ \in R,$ intersect the line L at Q. Then $2 (P Q)^{2}$ is equal to:

Select Answer:

Visualized Solution

Visualizing the Lines .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

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

Parametric Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Equating the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z-Coordinates

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

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

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

Direction Vector of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L3​

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L3​

Testing Option .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (8,26,12)

Verifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z Coordinates