The line of shortest distance between the lines (x-2)/0 = (y-1)/1 = z/1 and (x-3)/2 = (y-5)/2 = (z-1

Q: Wait, why are we using $\sin \theta$ for the angle between a line and a plane? Shouldn't it be $\cos \theta$?

This is the classic JEE trap! When you find the angle between two lines, you use the dot product (which involves $\cos \theta$). But for a line and a plane, the angle $\theta$ is between the line and its projection on the plane. The normal vector $\vec{n}$ is perpendicular to the plane, so the angle between the line and the normal is $90^\circ - \theta$. Thus, $\cos(90^\circ - \theta) = \sin \theta$. Always visualize the normal vector when dealing with planes!

Q: Why did we get $a^2 = -\frac{25}{11}$? Did I make a calculation error?

Take a deep breath—you didn't make a mistake! In JEE Advanced, sometimes the math leads you to a contradiction. A negative square for a real number $a$ simply means that no such plane exists. Trust your algebra; the 'no solution' result is a perfectly valid answer. It shows you have the courage to follow the logic to its conclusion, even when it defies intuition.

Q: Is there a shortcut to finding the direction vector of the shortest distance line?

The most reliable method is the cross product of the direction vectors of the two lines, $\vec{b_1} \times \vec{b_2}$. While there are formulas for the shortest distance itself, the direction vector is always the common perpendicular. Stick to the cross product; it is the most robust tool in your arsenal for 3D geometry.

Question

The line of shortest distance between the lines $\frac{x-2}{0} = \frac{y-1}{1} = \frac{z}{1}$ and $\frac{x-3}{2} = \frac{y-5}{2} = \frac{z-1}{1}$ makes an angle of $\cos^{-1} \sqrt{\frac{2}{27}}$ with the plane $P : ax - y - z = 0, (a > 0)$. If the image of the point $(1, 1, -5)$ in the plane $P$ is $(\alpha, \beta, \gamma)$, then $\alpha + \beta - \gamma$ is equal to ______. (Enter 0 if no such plane exists)

Sigma Percentile · Accepted Answer

### The Geometry of the Invisible

My dear student, welcome to the world of 3D geometry. Today, we are not just solving an equation; we are navigating the architecture of space.

We are dealing with two skew lines—lines that never meet, yet exist in the same universe. Our mission is to find the line of shortest distance between them and then relate it to a plane. This is a classic JEE Advanced problem that tests not just your calculation, but your geometric intuition.

### Phase 1: The Anatomy of the Lines

First, let us look at our lines, $L_1$ and $L_2$. They are given in symmetric form. The denominators are the keys to their orientation.

For $L_1$, the denominators are $(0, 1, 1)$, so our direction vector is $\vec{b_1} = 0\hat{i} + 1\hat{j} + 1\hat{k}$. For $L_2$, the denominators are $(2, 2, 1)$, giving us $\vec{b_2} = 2\hat{i} + 2\hat{j} + 1\hat{k}$.

Think of these vectors as arrows pointing in the direction of the lines. They define the 'flow' of the lines through space. Now, we need the line of shortest distance.

Geometrically, this line must be perpendicular to both $L_1$ and $L_2$. How do we find a vector that is perpendicular to two others? The cross product is our best friend here. We compute $\vec{d} = \vec{b_1} 	imes \vec{b_2}$.

$$
\vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 1 & 1 \ 2 & 2 & 1 \end{vmatrix}
$$

Expanding this determinant, we get $\vec{d} = \hat{i}(1-2) - \hat{j}(0-2) + \hat{k}(0-2)$, which simplifies to $\vec{d} = -\hat{i} + 2\hat{j} - 2\hat{k}$. This vector $\vec{d}$ is the backbone of our shortest distance line. Its magnitude is $|\vec{d}| = \sqrt{(-1)^2 + 2^2 + (-2)^2} = \sqrt{9} = 3$.

### Phase 2: The Plane and the Trap

Now, we introduce the plane $P: ax - y - z = 0$. The normal vector to this plane is $\vec{n} = (a, -1, -1)$. The magnitude of this normal is $|\vec{n}| = \sqrt{a^2 + (-1)^2 + (-1)^2} = \sqrt{a^2 + 2}$.

Here is where many students stumble. We are given the angle $	heta$ between the line and the plane, where $\cos 	heta = \sqrt{\frac{2}{27}}$. Remember, the angle between a line and a plane is defined by the sine function, not the cosine.

Why? Because the angle between the line and the plane's normal is the complement of the angle between the line and the plane itself. So, we must convert our given cosine value to sine:

$$
\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{2}{27} = \frac{25}{27} \implies \sin 	heta = \frac{5}{3\sqrt{3}}
$$

### Phase 3: The Algebraic Reality Check

We now use the formula for the angle between a line and a plane:

$$
\sin 	heta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}| |\vec{n}|}
$$

Substituting our values, we get:

$$
\frac{5}{3\sqrt{3}} = \frac{|(-1)(a) + (2)(-1) + (-2)(-1)|}{3 \cdot \sqrt{a^2 + 2}}
$$

Simplifying the numerator, we get $|-a - 2 + 2| = |-a| = a$ (since $a > 0$). The equation becomes:

$$
\frac{5}{3\sqrt{3}} = \frac{a}{3\sqrt{a^2 + 2}}
$$

The $3$ in the denominator cancels out beautifully. We are left with $\frac{5}{\sqrt{3}} = \frac{a}{\sqrt{a^2 + 2}}$. Squaring both sides to eliminate the radicals gives us:

$$
\frac{25}{3} = \frac{a^2}{a^2 + 2}
$$

Cross-multiplying, we get $25(a^2 + 2) = 3a^2$, which simplifies to $25a^2 + 50 = 3a^2$, or $22a^2 = -50$.

### Conclusion

Look at that result: $a^2 = -\frac{50}{22}$. A square of a real number cannot be negative. This is not a failure of your math; it is a discovery of the problem's nature.

It tells us that no such plane $P$ exists that satisfies the given condition. In the high-stakes environment of JEE Advanced, trusting your derivation is just as important as the calculation itself. You have proven that the configuration is impossible. Therefore, the answer is **0**. Well done on navigating this complex path!

Enter Numerical Value:

Visualized Solution

Analyze Line $L_{1}$

Analyze Line $L_{2}$

Direction of Shortest Distance Line

Calculate Cross Product

Magnitude of Vector $d$

Analyze the Plane $P$

Angle Between Line and Plane

Convert $cos θ$ to $sin θ$

Dot Product Calculation

Set up the Equation for $a$

Solve for $a$

Conclusion

The Sigma Insight: Shortest Distance Between Two Skew Lines

The Geometry of the Invisible

Phase 1

Phase 2

Phase 3

Conclusion

Similar Questions

Comprehension Passage

The unit vector perpendicular to both $L_{1}$ and $L_{2}$ is

The shortest distance between $L_{1}$ and $L_{2}$ is

The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_{1}$ and $L_{2}$ is

Let the point $(- 1, α, β)$ lie on the line of the shortest distance between the lines $\frac{x + 2}{- 3} = \frac{y - 2}{4} = \frac{z - 5}{2}$ and $\frac{x + 2}{- 1} = \frac{y + 6}{2} = \frac{z - 1}{0}$ . Then $(α - β)^{2}$ is equal to___________

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

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ is $\frac{5}{6}$ , then the sum of all possible values of $α$ is

If the shortest distance between the lines $\frac{x - λ}{- 2} = \frac{y - 2}{1} = \frac{z - 1}{1}$ and $\frac{x - 3}{1} = \frac{y - 1}{- 2} = \frac{z - 2}{1}$ is 1, then the sum of all possible values of $λ$ 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 $L_{2}$ ?

If the shortest distance between the lines $\frac{x - λ}{2} = \frac{y - 4}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{4} = \frac{y - 4}{6} = \frac{z - 7}{8}$ is $\frac{13}{29}$ , then a value of $λ$ is :

The shortest distance between the lines $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$ and $\frac{x - 6}{3} = \frac{1 - y}{2} = \frac{z + 8}{0}$ is equal to ______.

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:

Enter Numerical Value:

Visualized Solution

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

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

Direction of Shortest Distance Line

Calculate Cross Product

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

Analyze the Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Angle Between Line and Plane

Convert .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosθ to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sinθ

Dot Product Calculation

Set up the Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

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

Conclusion

The Sigma Insight: Shortest Distance Between Two Skew Lines

The Geometry of the Invisible

Phase 1

Phase 2

Phase 3

Conclusion

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The unit vector perpendicular to both L1​ and L2​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance between L1​ and L2​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines L1​ and L2​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance between the lines 1x−5​=2y−2​=−3z−4​ and 1x+3​=4y+5​=−5z−1​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the shortest distance between the lines 2x−1​=3y−2​=4z−3​ and 1x​=αy​=1z−5​ is 6​5​, then the sum of all possible values of α is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the shortest distance between the lines −2x−λ​=1y−2​=1z−1​ and 1x−3​​=−2y−1​=1z−2​ is 1, then the sum of all possible values of λ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the shortest distance between the lines 2x−λ​=3y−4​=4z−3​ and 4x−2​=6y−4​=8z−7​ is 29​13​, then a value of λ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance between the lines 3x−2​=2y+1​=2z−6​ and 3x−6​=21−y​=0z+8​ is equal to ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance between the lines 2x−3​=3y−2​=−1z−1​ and 2x+3​=1y−6​=3z−5​ is:

Analyze Line $L_{1}$

Analyze Line $L_{2}$

Magnitude of Vector $d$

Analyze the Plane $P$

Convert $cos θ$ to $sin θ$

Set up the Equation for $a$

Solve for $a$

The unit vector perpendicular to both $L_{1}$ and $L_{2}$ is

The shortest distance between $L_{1}$ and $L_{2}$ is

The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_{1}$ and $L_{2}$ is

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

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ is $\frac{5}{6}$ , then the sum of all possible values of $α$ is

If the shortest distance between the lines $\frac{x - λ}{- 2} = \frac{y - 2}{1} = \frac{z - 1}{1}$ and $\frac{x - 3}{1} = \frac{y - 1}{- 2} = \frac{z - 2}{1}$ is 1, then the sum of all possible values of $λ$ is :

If the shortest distance between the lines $\frac{x - λ}{2} = \frac{y - 4}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{4} = \frac{y - 4}{6} = \frac{z - 7}{8}$ is $\frac{13}{29}$ , then a value of $λ$ is :

The shortest distance between the lines $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 6}{2}$ and $\frac{x - 6}{3} = \frac{1 - y}{2} = \frac{z + 8}{0}$ is equal to ______.

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: