JEE Advanced 2013

LEVELJEE Advanced

Animated Solution for Mathematics - Three Dimensional Geometry: A line $l$ passing through the origin is perpendicular to the lines $l_{1} : (3 + t) \hat{i} + (- 1 + 2 t) \hat{j} + (4 + 2 t) \hat{k}, - \infty < t < \infty$ and $l_{2} : (3 + 2 s) \hat{i} + (3 + 2 s) \hat{j} + (2 + s) \hat{k}, - \infty < s < \infty$ . Then, the coordinate(s) of the point(s) on $l_{2}$ at a distance of $17$ from the point of intersection of $l$ and $l_{1}$ is (are)

Q: Wait, why do we use the cross product to find the direction of line $l$?

That is a fantastic question! Think of the cross product as the 'orthogonal anchor' in 3D space. When you take the cross product of two vectors, $\vec{d_1}$ and $\vec{d_2}$, the resulting vector is, by definition, perpendicular to the plane containing both $\vec{d_1}$ and $\vec{d_2}$. Since line $l$ must be perpendicular to both lines, its direction must align with this normal vector. It is the most elegant way to find a common perpendicular direction!

Q: How do I know if I should equate the parameters or just the coordinates when finding the intersection?

Always equate the coordinates! Remember, a line is a set of points. When we say two lines intersect at a point $P$, it means there exists a specific point $(x, y, z)$ that lies on both lines simultaneously. By setting the general coordinate expressions equal to each other, you are forcing the lines to share that specific point. The parameters (like $t$ and $\lambda$) are just labels for where you are on each line; they don't have to be the same value at the intersection point.

Q: Is it possible for the distance equation to have no real solutions?

Absolutely! If the quadratic equation we derived, $9s^2 + 28s + 20 = 0$, had a negative discriminant ($b^2 - 4ac < 0$), it would mean that no point on line $l_2$ is at a distance of $\sqrt{17}$ from $P$. In that case, the lines would be 'too far apart' or 'too close' to satisfy the condition. Always check your discriminant if you feel stuck, but here, the math works out beautifully to give us two distinct points.

Select Answer:

* Multiple Correct

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the geometry. A line

l

is perpendicular to two other lines,

l_{1}

and

l_{2}

. In 3D space, if a line is perpendicular to two others, its direction vector must be parallel to the cross product of the direction vectors of those two lines. Start by finding

d_{1}

and

d_{2}

from the given equations.

Hint 2

Hint 2 (The Arsenal): Once you have the direction vector

d = d_{1} \times d_{2}

, you can write the equation of line

l

since it passes through the origin. To find the intersection point

P

l

and

l_{1}

, equate their general coordinates. This will give you a system of equations to solve for the parameters

λ

and

t

Hint 3

Hint 3 (The Execution): After finding the coordinates of point

P

, let

Q

be a general point on

l_{2}

with parameter

s

. Use the 3D distance formula to express the distance

P Q

(x_{Q} - x_{P})^{2} + (y_{Q} - y_{P})^{2} + (z_{Q} - z_{P})^{2} = 17

. Square both sides to eliminate the radical and solve the resulting quadratic equation for

s

Visualized Solution

Direction Vectors of $l_{1}$ and $l_{2}$

Given line $l_{1}$ : $r = (3 \hat{i} - \hat{j} + 4 \hat{k}) + t (\hat{i} + 2 \hat{j} + 2 \hat{k})$
Given line $l_{2}$ : $r = (3 \hat{i} + 3 \hat{j} + 2 \hat{k}) + s (2 \hat{i} + 2 \hat{j} + \hat{k})$
Direction of $l_{1}$ : $d_{1} = \hat{i} + 2 \hat{j} + 2 \hat{k}$
Direction of $l_{2}$ : $d_{2} = 2 \hat{i} + 2 \hat{j} + \hat{k}$

Direction of Perpendicular Line $l$

Line $l$ is perpendicular to both $l_{1}$ and $l_{2}$ .
Therefore, its direction $d$ is given by the cross product: $d = d_{1} \times d_{2}$

Computing the Cross Product

$d = \hat{i} 12 \hat{j} 22 \hat{k} 21$
$d = \hat{i} (2 - 4) - \hat{j} (1 - 4) + \hat{k} (2 - 4)$
$d = - 2 \hat{i} + 3 \hat{j} - 2 \hat{k}$

Equation of Line $l$

Line $l$ passes through the origin $(0, 0, 0)$ .
Equation of $l$ : $r = λ (- 2 \hat{i} + 3 \hat{j} - 2 \hat{k})$
General point on $l$ : $(- 2 λ, 3 λ, - 2 λ)$

Intersection of $l$ and $l_{1}$

Let $P$ be the point of intersection of $l$ and $l_{1}$ .
General point on $l_{1}$ : $(3 + t, - 1 + 2 t, 4 + 2 t)$
Equating coordinates of $l$ and $l_{1}$ :
$- 2 λ = 3 + t$ ...(i)
$3 λ = - 1 + 2 t$ ...(ii)
$- 2 λ = 4 + 2 t$ ...(iii)

Solving for Parameters

From (i) and (iii), the LHS is the same ( $- 2 λ$ ).
Equating RHS: $3 + t = 4 + 2 t$
$t = - 1$
Substitute $t = - 1$ into (i):
$- 2 λ = 3 + (- 1) = 2 ⟹ λ = - 1$

Coordinates of Point $P$

Substitute $λ = - 1$ into the general point of $l$ :
$P = (- 2 (- 1), 3 (- 1), - 2 (- 1))$
$P = (2, - 3, 2)$

Point $Q$ on $l_{2}$ and Distance

Let $Q$ be a point on $l_{2}$ .
General coordinates of $Q$ : $(3 + 2 s, 3 + 2 s, 2 + s)$
Given distance $P Q = 17$
Squaring both sides: $P Q^{2} = 17$

Applying the Distance Formula

Using distance formula for $P (2, - 3, 2)$ and $Q (3 + 2 s, 3 + 2 s, 2 + s)$ :
$P Q^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}$
$17 = (3 + 2 s - 2)^{2} + (3 + 2 s - (- 3))^{2} + (2 + s - 2)^{2}$
$17 = (1 + 2 s)^{2} + (6 + 2 s)^{2} + (s)^{2}$

Expanding the Squared Terms

Expand $(1 + 2 s)^{2} = 1 + 4 s + 4 s^{2}$
Expand $(6 + 2 s)^{2} = 36 + 24 s + 4 s^{2}$
Substitute back:
$(1 + 4 s + 4 s^{2}) + (36 + 24 s + 4 s^{2}) + s^{2} = 17$

Forming the Quadratic Equation

Combine like terms:
$s^{2}$ terms: $4 s^{2} + 4 s^{2} + s^{2} = 9 s^{2}$
$s$ terms: $4 s + 24 s = 28 s$
Constant terms: $1 + 36 = 37$
Equation becomes: $9 s^{2} + 28 s + 37 = 17$
Bring 17 to LHS: $9 s^{2} + 28 s + 20 = 0$

Solving the Quadratic Equation

Factorize $9 s^{2} + 28 s + 20 = 0$
Split the middle term: $28 s = 18 s + 10 s$
$9 s^{2} + 18 s + 10 s + 20 = 0$
$9 s (s + 2) + 10 (s + 2) = 0$
$(9 s + 10) (s + 2) = 0$
Roots: $s = - 2$ or $s = - \frac{10}{9}$

Coordinates for $s = - 2$

Substitute $s = - 2$ into $Q (3 + 2 s, 3 + 2 s, 2 + s)$ :
$x = 3 + 2 (- 2) = 3 - 4 = - 1$
$y = 3 + 2 (- 2) = 3 - 4 = - 1$
$z = 2 + (- 2) = 0$
First point: $Q_{1} = (- 1, - 1, 0)$

Coordinates for $s = - \frac{10}{9}$

Substitute $s = - \frac{10}{9}$ into $Q (3 + 2 s, 3 + 2 s, 2 + s)$ :
$x = 3 + 2 (- \frac{10}{9}) = 3 - \frac{20}{9} = \frac{7}{9}$
$y = 3 + 2 (- \frac{10}{9}) = 3 - \frac{20}{9} = \frac{7}{9}$
$z = 2 + (- \frac{10}{9}) = \frac{18 - 10}{9} = \frac{8}{9}$
Second point: $Q_{2} = (\frac{7}{9}, \frac{7}{9}, \frac{8}{9})$

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The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Setup

To define line

l

, we first identify its direction. We are given the lines:

l_{1} : r = (3 \hat{i} - \hat{j} + 4 \hat{k}) + t (\hat{i} + 2 \hat{j} + 2 \hat{k})

l_{2} : r = (3 \hat{i} + 3 \hat{j} + 2 \hat{k}) + s (2 \hat{i} + 2 \hat{j} + \hat{k})

The direction vectors are

d_{1} = \hat{i} + 2 \hat{j} + 2 \hat{k}

and

d_{2} = 2 \hat{i} + 2 \hat{j} + \hat{k}

. Since line

l

is perpendicular to both, its direction

d

must be the cross product

d_{1} \times d_{2}

Calculating this determinant:

d = \hat{i} 12 \hat{j} 22 \hat{k} 21 = \hat{i} (2 - 4) - \hat{j} (1 - 4) + \hat{k} (2 - 4) = - 2 \hat{i} + 3 \hat{j} - 2 \hat{k}

Since

l

passes through the origin, its equation is simply

r = λ (- 2 \hat{i} + 3 \hat{j} - 2 \hat{k})

The Intersection Point

We seek the intersection point

P

l

and

l_{1}

. We equate the general points:

(- 2 λ, 3 λ, - 2 λ) = (3 + t, - 1 + 2 t, 4 + 2 t)

This yields the following system of equations: 1)

- 2 λ = 3 + t

3 λ = - 1 + 2 t

- 2 λ = 4 + 2 t

By comparing the first and third equations, we see

3 + t = 4 + 2 t

, which implies

t = - 1

. Substituting

t = - 1

into the first equation, we get

- 2 λ = 3 - 1 = 2

, so

λ = - 1

Plugging

λ = - 1

back into the coordinates of

l

, we find the point

P = (2, - 3, 2)

The Final Calculation

We look for point

Q

l_{2}

such that the distance

P Q = 17

. The general point

Q

l_{2}

(3 + 2 s, 3 + 2 s, 2 + s)

Using the distance formula

P Q^{2} = 17

(3 + 2 s - 2)^{2} + (3 + 2 s - (- 3))^{2} + (2 + s - 2)^{2} = 17

This simplifies to

(1 + 2 s)^{2} + (6 + 2 s)^{2} + s^{2} = 17

. Expanding these terms:

(1 + 4 s + 4 s^{2}) + (36 + 24 s + 4 s^{2}) + s^{2} = 17

9 s^{2} + 28 s + 37 = 17 \Rightarrow 9 s^{2} + 28 s + 20 = 0

Factoring the quadratic equation

(9 s + 10) (s + 2) = 0

, we find

s = - 2

s = - \frac{10}{9}

Substituting these values back into the coordinates of

Q

, we find our two points: $Q_{1} = (- 1, - 1, 0)$ and $Q_{2} = (\frac{7}{9}, \frac{7}{9}, \frac{8}{9})$ .

Common Traps & FAQs

Wait, why do we use the cross product to find the direction of line

l

That is a fantastic question! Think of the cross product as the 'orthogonal anchor' in 3D space. When you take the cross product of two vectors,

d_{1}

and

d_{2}

, the resulting vector is, by definition, perpendicular to the plane containing both

d_{1}

and

d_{2}

. Since line

l

must be perpendicular to both lines, its direction must align with this normal vector. It is the most elegant way to find a common perpendicular direction!

How do I know if I should equate the parameters or just the coordinates when finding the intersection?

Always equate the coordinates! Remember, a line is a set of points. When we say two lines intersect at a point

P

, it means there exists a specific point

(x, y, z)

that lies on both lines simultaneously. By setting the general coordinate expressions equal to each other, you are forcing the lines to share that specific point. The parameters (like

t

and

λ

) are just labels for where you are on each line; they don't have to be the same value at the intersection point.

Is it possible for the distance equation to have no real solutions?

Absolutely! If the quadratic equation we derived,

9 s^{2} + 28 s + 20 = 0

, had a negative discriminant (

b^{2} - 4 a c < 0

), it would mean that no point on line

l_{2}

is at a distance of

17

from

P

. In that case, the lines would be 'too far apart' or 'too close' to satisfy the condition. Always check your discriminant if you feel stuck, but here, the math works out beautifully to give us two distinct points.

Select Answer:

Visualized Solution

Direction Vectors of $l_{1}$ and $l_{2}$

Direction of Perpendicular Line $l$

Computing the Cross Product

Equation of Line $l$

Intersection of $l$ and $l_{1}$

Solving for Parameters

Coordinates of Point $P$

Point $Q$ on $l_{2}$ and Distance

Applying the Distance Formula

Expanding the Squared Terms

Forming the Quadratic Equation

Solving the Quadratic Equation

Coordinates for $s = - 2$

Coordinates for $s = - \frac{10}{9}$

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Intersection Point

The Final Calculation

Similar Questions

Let $A$ and $B$ be two distinct points on the line $L : \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ . Both $A$ and $B$ are at a distance $217$ from the foot of perpendicular drawn from the point $(1, 2, 3)$ on the line $L$ . If $O$ is the origin, then $O A \cdot O B$ is equal to:

Let $P$ be the point of intersection of the lines $\frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1}$ and $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2}$ . Then, the shortest distance of $P$ from the line $4 x = 2 y = z$ is

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Three lines $L_{1} : r = λ \hat{i}, λ \in R$ , $L_{2} : r = \hat{k} + μ \hat{j}, μ \in R$ and $L_{3} : r = \hat{i} + \hat{j} + ν \hat{k}, ν \in R$ are given. For which point(s) $Q$ on $L_{2}$ can we find a point $P$ on $L_{1}$ and a point $R$ on $L_{3}$ so that $P, Q$ and $R$ are collinear ?

Select Answer:

Visualized Solution

Direction Vectors of .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​

Direction of Perpendicular Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l

Computing the Cross Product

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

Intersection of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l1​

Solving for Parameters

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

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l2​ and Distance

Applying the Distance Formula

Expanding the Squared Terms

Forming the Quadratic Equation

Solving the Quadratic Equation

Coordinates for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } s=−2

Coordinates for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } s=−910​

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Intersection Point

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P be the point of intersection of the lines 1x−2​=5y−4​=1z−2​ and 2x−3​=3y−2​=2z−3​. Then, the shortest distance of P from the line 4x=2y=z is

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Direction Vectors of $l_{1}$ and $l_{2}$

Direction of Perpendicular Line $l$

Equation of Line $l$

Intersection of $l$ and $l_{1}$

Coordinates of Point $P$

Point $Q$ on $l_{2}$ and Distance

Coordinates for $s = - 2$

Coordinates for $s = - \frac{10}{9}$

Let $P$ be the point of intersection of the lines $\frac{x - 2}{1} = \frac{y - 4}{5} = \frac{z - 2}{1}$ and $\frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 3}{2}$ . Then, the shortest distance of $P$ from the line $4 x = 2 y = z$ is

The distance of the line $\frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}$ from the point (1,4,0) along the line $\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$ is: