How do you find a point on a line at a specific distance?

Use the parametric form of the line (x = x₁+lr, y = y₁+mr, z = z₁+nr) where (l,m,n) are unit direction cosines and r is the distance.

What is the formula for the shortest distance between two skew lines?

S.D. = |(a₂ - a₁)·(b₁ × b₂)| / |b₁ × b₂|, where a are position vectors and b are direction vectors.

How do you determine which point is 'nearer to the origin'?

Calculate the distance from the origin √(x²+y²+z²) for both possible points found using the distance formula.

What are direction ratios (DRs)?

DRs are numbers proportional to the direction cosines of a line, representing its orientation in 3D space.

What does 'skew lines' mean?

Skew lines are lines in 3D space that are neither parallel nor intersecting.

How do you calculate the cross product b₁ × b₂?

Use a determinant with unit vectors i, j, k in the first row and the DRs of the two lines in the subsequent rows.

Why use unit direction cosines for distance?

Unit direction cosines ensure that the parameter 'r' represents the actual linear distance along the line.

What is the significance of the origin (0,0,0) in this problem?

It acts as a reference point to choose between two mathematically valid points P on the first line.

Can the shortest distance be zero?

Yes, if the shortest distance is zero, it means the two lines intersect.

What is the final answer for this problem?

The shortest distance is 4√(7/5).

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, -1, 0) and nearer to the origin. Then the shortest distance, between the lines (x-α)/1 = (y-β)/2 = (z-γ)/3 and (x+5)/2 = (y-10)/1 = (z-3)/1, is equal to

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, -1, 0) and nearer to the origin. Then the shortest distance, between the lines (x-α)/1 = (y-β)/2 = (z-γ)/3 and (x+5)/2 = (y-10)/1 = (z-3)/1, is equal to | JEE Main Mathematics

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Math 3D Geometry

Q MCQ Shortest Distance

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

✅ Correct Answer

4√(7/5)

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Solution Steps

Find Parametric Point P

The given line is $\frac{x-1}{2} = \frac{y+1}{-3} = \frac{z-0}{1} = \lambda$.

Any point on this line is $P(2\lambda+1, -3\lambda-1, \lambda)$.

Distance of $P$ from $A(1, -1, 0)$ is $4\sqrt{14}$.

$\sqrt{(2\lambda)^2 + (-3\lambda)^2 + \lambda^2} = 4\sqrt{14}$

$\sqrt{14\lambda^2} = 4\sqrt{14} \implies |\lambda|\sqrt{14} = 4\sqrt{14} \implies \lambda = \pm 4$.

Determine Point Nearer to Origin

If $\lambda = 4$, $P_1 = (9, -13, 4)$. Distance from origin $OP_1 = \sqrt{81 + 169 + 16} = \sqrt{266}$.

If $\lambda = -4$, $P_2 = (-7, 11, -4)$. Distance from origin $OP_2 = \sqrt{49 + 121 + 16} = \sqrt{186}$.

Since $\sqrt{186} < \sqrt{266}$, we choose $P(-7, 11, -4)$, so $\alpha = -7, \beta = 11, \gamma = -4$.

Define the Two Skew Lines

Line 1 ($L_1$): $\frac{x+7}{1} = \frac{y-11}{2} = \frac{z+4}{3}$. Point $\vec{a}_1 = -7\hat{i} + 11\hat{j} – 4\hat{k}$, DRs $\vec{b}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$.

Line 2 ($L_2$): $\frac{x+5}{2} = \frac{y-10}{1} = \frac{z-3}{1}$. Point $\vec{a}_2 = -5\hat{i} + 10\hat{j} + 3\hat{k}$, DRs $\vec{b}_2 = 2\hat{i} + \hat{j} + \hat{k}$.

Calculate $\vec{b}_1 \times \vec{b}_2$

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix} = \hat{i}(2-3) – \hat{j}(1-6) + \hat{k}(1-4) = -\hat{i} + 5\hat{j} – 3\hat{k}$.

Magnitude $|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-1)^2 + 5^2 + (-3)^2} = \sqrt{1 + 25 + 9} = \sqrt{35}$.

Calculate $(\vec{a}_2 – \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)$

$\vec{a}_2 – \vec{a}_1 = (-5 – (-7))\hat{i} + (10 – 11)\hat{j} + (3 – (-4))\hat{k} = 2\hat{i} – \hat{j} + 7\hat{k}$.

Dot product: $(2)(-1) + (-1)(5) + (7)(-3) = -2 – 5 – 21 = -28$.

Find Shortest Distance (S.D.)

$S.D. = \frac{|(\vec{a}_2 – \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} = \frac{|-28|}{\sqrt{35}} = \frac{28}{\sqrt{35}}$.

Simplifying: $\frac{28}{\sqrt{35}} = \frac{28 \times \sqrt{35}}{35} = \frac{4\sqrt{35}}{5} = 4\sqrt{\frac{35}{25}} = 4\sqrt{\frac{7}{5}}$.

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Key Insight

Shortest distance calculation depends heavily on correctly identifying the position vectors and direction vectors of both lines after locating point P.

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3D Geometry Theory

1. Point at a Distance r on a Line

Given a line passing through $(x_1, y_1, z_1)$ with direction cosines $(l, m, n)$, any point at a distance $r$ from the given point is $(x_1 \pm lr, y_1 \pm mr, z_1 \pm nr)$. If direction ratios $(a, b, c)$ are provided instead of unit direction cosines, one must first normalize them by dividing each by $\sqrt{a^2 + b^2 + c^2}$. In this problem, the DRs were $(2, -3, 1)$ and the distance was $4\sqrt{14}$, which simplified the calculation as $\sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{14}$.

2. Skew Lines and Shortest Distance

Skew lines are lines that are not parallel and do not intersect. They lie in different planes. The shortest distance between two such lines is the length of the common perpendicular segment joining them. Vectorially, if lines are $\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$, the S.D. is the projection of the vector $(\vec{a}_2 – \vec{a}_1)$ onto the direction of the common perpendicular $(\vec{b}_1 \times \vec{b}_2)$. This is a standard and frequently asked concept in JEE Main.

3. Vector Cross Product in 3D

The cross product $\vec{b}_1 \times \vec{b}_2$ yields a vector that is perpendicular to both $\vec{b}_1$ and $\vec{b}_2$. In 3D geometry, this is used to find the direction of the common perpendicular to two lines. The magnitude of this cross product represents the area of a parallelogram formed by the two vectors. For shortest distance, we divide the scalar triple product by this magnitude to “normalize” the projection length. Calculation involves a $3 \times 3$ determinant involving $\hat{i}, \hat{j}, \hat{k}$ and the components of the two vectors.

4. Parametric Equations and Origin Distance

Parametric representation of a line expresses the coordinates of any point on the line as a function of a single parameter ($\lambda$). This allows for solving constraints like “distance from a point” or “nearest to origin.” The distance from the origin to a point $(x, y, z)$ is given by $\sqrt{x^2 + y^2 + z^2}$. In geometry problems, “nearer to origin” acts as a filter to select one of multiple possible solutions. It requires a final comparison of magnitudes after finding potential candidates.

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FAQs

Why did we get two values for λ?

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Because the distance formula involves a square root (or squaring), there are two directions along the line you can travel to reach the specified distance.

What if the lines were parallel?

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The cross product $\vec{b}_1 \times \vec{b}_2$ would be zero. You would then use the parallel lines formula: $d = \frac{|\vec{b} \times (\vec{a}_2 – \vec{a}_1)|}{|\vec{b}|}$.

How to simplify 28/√35?

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Multiply and divide by √35: (28√35)/35. Simplify 28/35 to 4/5. So 4√35/5 = 4√(35/25) = 4√(7/5).

Is DRs the same as direction cosines?

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No, DRs are proportional to cosines. Cosines must satisfy $l^2 + m^2 + n^2 = 1$.

What if P was further from origin?

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You would have used λ = 4, which gave P(9, -13, 4). The S.D. calculation steps would then change significantly.

What is a unit vector?

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A vector with magnitude 1, obtained by dividing a vector by its own magnitude.

Can I use the determinant formula for S.D.?

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Yes, $S.D. = \frac{\text{Determinant}(x_2-x_1, y_2-y_1, z_2-z_1, \text{DRs of } L_1, \text{DRs of } L_2)}{\text{Magnitude of Cross Product}}$. It is the same result.

Why is shortest distance important?

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It determines if lines intersect (S.D.=0) or are skew, and is a key concept in vector calculus and physics.

Is the point P unique?

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Given the “nearer to origin” condition, the point P becomes unique.

What coordinate system is used?

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Standard 3D Cartesian Coordinate system with axes X, Y, and Z.

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