Let ℓ1 and ℓ2 be the lines r1 = λ(i + j + k) and r2 = (j - k) + μ(i + k), respectively. Let X be the

Q: Wait, why does the plane $H$ have to pass through the origin?

Great question! The problem states that the plane $H$ contains the line $\ell_1$. The equation of $\ell_1$ is $\vec{r}_1 = \lambda(\hat{i} + \hat{j} + \hat{k})$. If you set $\lambda = 0$, you get the point $(0, 0, 0)$. Since the line passes through the origin and the plane contains the line, the plane must also contain the origin. This is why the constant term $D$ in the plane equation $ax + by + cz + D = 0$ is zero.

Q: Why do we assume $\ell_2$ is parallel to the plane to maximize the distance?

Think about the geometry. If the line $\ell_2$ were not parallel to the plane, it would eventually intersect it. If it intersects, the distance between the line and the plane is zero. To get the *maximum* possible distance, we need the line to be as far away as possible, which happens when the line is parallel to the plane. In this configuration, the distance is constant for every point on the line.

Q: How do I know if the line is parallel or intersecting?

A line with direction vector $\vec{b}$ is parallel to a plane with normal vector $\vec{n}$ if and only if $\vec{n} \cdot \vec{b} = 0$. If the dot product is not zero, the line is not parallel and will intersect the plane. By setting the dot product to zero, we are forcing the parallel condition.

Q: Why is the distance of the origin from $H_0$ zero?

Look at the equation of the plane we derived: $x - z = 0$. If you plug in the coordinates of the origin $(0, 0, 0)$, you get $0 - 0 = 0$. Since the origin satisfies the equation of the plane, it lies on the plane. Therefore, the distance from the origin to the plane is zero.

Question

Let $\ell_1$ and $\ell_2$ be the lines $\vec{r}_1 = \lambda(\hat{i} + \hat{j} + \hat{k})$ and $\vec{r}_2 = (\hat{j} - \hat{k}) + \mu(\hat{i} + \hat{k})$, respectively. Let $X$ be the set of all the planes $H$ that contain the line $\ell_1$. For a plane $H$, let $d(H)$ denote the smallest possible distance between the points of $\ell_2$ and $H$. Let $H_0$ be plane in $X$ for which $d(H_0)$ is the maximum value of $d(H)$ as $H$ varies over all planes in $X$. Match each entry in List-I to the correct entries in List-II.

List-I:

P. (P) The value of $d(H_0)$ is

Q. (Q) The distance of the point $(0, 1, 2)$ from $H_0$ is

R. (R) The distance of origin from $H_0$ is

S. (S) The distance of origin from the point of intersection of planes $y = z, x = 1$ and $H_0$ is

List-II:

1. (1) $\sqrt{3}$

2. (2) $\frac{1}{\sqrt{3}}$

3. (3) 0

4. (4) $\sqrt{2}$

5. (5) $\frac{1}{\sqrt{2}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional coordinate system. You have two lines, $\ell_1$ and $\ell_2$, floating in space.

Your mission is to find a specific plane $H_0$ that contains $\ell_1$ and is positioned such that it is as far as possible from $\ell_2$. This is not just an algebraic exercise; it is a beautiful dance of vectors.

### The Origin Constraint

We begin with line $\ell_1$, defined by $\vec{r}_1 = \lambda(\hat{i} + \hat{j} + \hat{k})$. Notice that when $\lambda = 0$, the position vector is the zero vector.

This means $\ell_1$ passes directly through the origin $(0, 0, 0)$. If a plane $H$ contains this line, it must also contain the origin.

This is a massive simplification! Instead of the general plane equation $ax + by + cz + D = 0$, we can confidently write:

$$ax + by + cz = 0$$

Our normal vector is $\vec{n} = (a, b, c)$.

### The Parallelism Trap

Now, consider $\ell_2$, which passes through $(0, 1, -1)$ with direction $\vec{b}_2 = (1, 0, 1)$. We want to maximize the distance $d(H)$ between $\ell_2$ and $H$.

If the line intersects the plane, the distance is zero. To maximize this, we need the line to be parallel to the plane.

Geometrically, this means the normal vector $\vec{n}$ must be perpendicular to the direction vector of the line. We have two conditions for our normal vector $\vec{n}$:

1. It must be perpendicular to the direction of $\ell_1$, which is $\vec{b}_1 = (1, 1, 1)$, so $\vec{n} \cdot \vec{b}_1 = 0$. 
2. It must be perpendicular to the direction of $\ell_2$, so $\vec{n} \cdot \vec{b}_2 = 0$.

### The Algebraic Victory

Let us translate these conditions into equations. The first condition, $\vec{n} \cdot \vec{b}_1 = 0$, gives us:

$$a(1) + b(1) + c(1) = 0 \Rightarrow a + b + c = 0$$

The second condition, $\vec{n} \cdot \vec{b}_2 = 0$, gives us:

$$a(1) + b(0) + c(1) = 0 \Rightarrow a + c = 0$$

From the second equation, we see $c = -a$. Substituting this into the first equation, we get $a + b - a = 0$, which implies $b = 0$.

Our normal vector is $(a, 0, -a)$. We can choose $a = 1$ to get the simplest normal vector $(1, 0, -1)$. Thus, the equation of our optimal plane $H_0$ is:

$$x - z = 0$$

### Final Calculation

Now that we have $H_0: x - z = 0$, the rest is a series of elegant calculations. We use the distance formula:

$$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$$

**For (P):** The maximum distance $d(H_0)$ is the distance from any point on $\ell_2$ (like $(0, 1, -1)$) to the plane:

$$d = \frac{|(1)(0) + (0)(1) - (1)(-1)|}{\sqrt{1^2 + 0^2 + (-1)^2}} = \frac{1}{\sqrt{2}}$$

**For (Q):** The distance of $(0, 1, 2)$ from $x - z = 0$ is:

$$d = \frac{|0 - 2|}{\sqrt{1^2 + (-1)^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$$

**For (R):** The distance of the origin from $x - z = 0$ is clearly **0** because the origin lies on the plane.

**For (S):** We find the intersection of $y = z$, $x = 1$, and $x - z = 0$. Since $x = 1$ and $x = z$, we have $z = 1$. Since $y = z$, we have $y = 1$.

The intersection point is $(1, 1, 1)$. The distance of $(1, 1, 1)$ from the origin is:

$$\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$$

List-I

List-II

Select Matching Pairs:

Visualized Solution

Identify Lines $ℓ_{1}$ and $ℓ_{2}$

Plane $H$ Containing $ℓ_{1}$

Equation of Plane $H$

Condition for $ℓ_{1}$ lying on $H$

Maximizing Distance $d (H)$

Condition for $ℓ_{2}$ parallel to $H$

Second Relation for $a, b, c$

Equation of Plane $H_{0}$

Calculating (P): Maximum Distance $d (H_{0})$

Evaluating (P)

Calculating (Q): Distance of $(0, 1, 2)$ from $H_{0}$

Evaluating (Q)

Calculating (R): Distance of Origin from $H_{0}$

Calculating (S): Intersection of Planes

Evaluating (S) and Final Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Origin Constraint

The Parallelism Trap

The Algebraic Victory

Final Calculation

Similar Questions

If $L_{1}$ is the line of intersection of the planes $2 x - 2 y + 3 z - 2 = 0, x - y + z + 1 = 0$ and $L_{2}$ is the line of intersection of the planes $x + 2 y - z - 3 = 0, 3 x - y + 2 z - 1 = 0$ , then the distance of the origin from the plane, containing the lines $L_{1}$ and $L_{2}$ , is :

If the lines $r = (\hat{i} - \hat{j} + \hat{k}) + λ (3 \hat{j} - \hat{k})$ and $r = (α \hat{i} - \hat{j}) + μ (2 \hat{i} - 3 \hat{k})$ are co-planar, then distance of the plane containing these two lines from the point $(α, 0, 0)$ is :

Let the plane containing the line of intersection of the planes $P 1 : x + (λ + 4) y + z = 1$ and $P 2 : 2 x + y + z = 2$ pass through the points $(0, 1, 0)$ and $(1, 0, 1)$ . Then the distance of the point $(2 λ, λ, - λ)$ from the plane $P 2$ is

If the plane $2 x - y + 2 z + 3 = 0$ has the distances $1/3$ and $2/3$ units from the planes $4 x - 2 y + 4 z + λ = 0$ and $2 x - y + 2 z + μ = 0$ , respectively, then the maximum value of $λ + μ$ is equal to :

List-I

List-II

List-I

List-II

The largest value of $a$ , for which the perpendicular distance of the plane containing the lines $r = (\hat{i} + \hat{j}) + λ (\hat{i} + a \hat{j} - \hat{k})$ and $r = (\hat{i} + \hat{j}) + μ (- \hat{i} + \hat{j} - a \hat{k})$ from the point $(2, 1, 4)$ is $3$ , is ______.

If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

List-I

List-II

Select Matching Pairs:

Visualized Solution

Identify 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​

Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H Containing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ℓ1​

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

Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ℓ1​ lying on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H

Maximizing Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d(H)

Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ℓ2​ parallel to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H

Second Relation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a,b,c

Equation of Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H0​

Calculating (P): Maximum Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d(H0​)

Evaluating (P)

Calculating (Q): Distance of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (0,1,2) from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H0​

Evaluating (Q)

Calculating (R): Distance of Origin from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H0​

Calculating (S): Intersection of Planes

Evaluating (S) and Final Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Origin Constraint

The Parallelism Trap

The Algebraic Victory

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the plane 2x−y+2z+3=0 has the distances 1/3 and 2/3 units from the planes 4x−2y+4z+λ=0 and 2x−y+2z+μ=0, respectively, then the maximum value of λ+μ is equal to :

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List-II

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the distance between the plane Ax−2y+z=d and the plane containing the lines 2x−1​=3y−2​=4z−3​ and 3x−2​=4y−3​=5z−4​ is 6​, then find ∣d∣.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane which is perpendicular to two planes 2x−2y+z=0 and x−y+2z=4, passes through (1,−2,1). The distance of the plane from the point (1,2,2) is

Identify Lines $ℓ_{1}$ and $ℓ_{2}$

Plane $H$ Containing $ℓ_{1}$

Equation of Plane $H$

Condition for $ℓ_{1}$ lying on $H$

Maximizing Distance $d (H)$

Condition for $ℓ_{2}$ parallel to $H$

Second Relation for $a, b, c$

Equation of Plane $H_{0}$

Calculating (P): Maximum Distance $d (H_{0})$

Calculating (Q): Distance of $(0, 1, 2)$ from $H_{0}$

Calculating (R): Distance of Origin from $H_{0}$

If the plane $2 x - y + 2 z + 3 = 0$ has the distances $1/3$ and $2/3$ units from the planes $4 x - 2 y + 4 z + λ = 0$ and $2 x - y + 2 z + μ = 0$ , respectively, then the maximum value of $λ + μ$ is equal to :

If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is