JEE Main 2021 (27 Aug Shift 2)

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Animated Solution for Mathematics - Three Dimensional Geometry: The angle between the straight lines, whose direction cosines are given by the equations $2 l + 2 m - n = 0$ and $mn + n l + l m = 0$ , is :

Q: Why did we divide by $m^2$ in the quadratic equation? Is that always allowed?

Great observation. We divide by $m^2$ because the equation is homogeneous. As long as $m \neq 0$, this is perfectly valid. If $m$ were zero, we would check the equation $2l^2 = 0$, which implies $l=0$, and consequently $n=0$. But $(0,0,0)$ cannot be a direction vector. Thus, $m$ cannot be zero, and our division is mathematically sound.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Look closely at the system of equations. You have one linear constraint and one quadratic constraint involving the direction cosines

l, m, n

. This is not just algebra; it is a geometric intersection. You are essentially finding the intersection of a plane and a cone in the space of direction ratios.

Hint 2

Hint 2: To solve for the direction ratios, you don't need the absolute values of

l, m, n

. You only need their ratios. Use the linear equation to express one variable (like

n

) in terms of the others (

l

and

m

), and substitute this into the quadratic equation to create a homogeneous equation in

l

and

m

Hint 3

Hint 3: Once you have the homogeneous equation

2 l^{2} + 5 l m + 2 m^{2} = 0

, divide by

m^{2}

to get a quadratic in the variable

t = \frac{l}{m}

. Solving this will give you the two distinct slopes (ratios) that define the direction vectors of your two lines.

Visualized Solution

$2 l + 2 m - n = 0$

Given equations for direction cosines $(l, m, n)$ :
1) $2 l + 2 m - n = 0$
2) $mn + n l + l m = 0$
Goal: Find the angle $θ$ between the two lines.

$Elimination Strategy$

We have a linear equation and a quadratic equation.
Strategy: Isolate one variable from the linear equation and substitute it into the quadratic one.

$Isolating n$

From equation (1):
$n = 2 l + 2 m$
$n = 2 (l + m)$

$Substituting n$

Substitute $n = 2 (l + m)$ into $mn + n l + l m = 0$ :
$m (2 l + 2 m) + (2 l + 2 m) l + l m = 0$

$Expanding Terms$

Expanding the brackets:
$2 l m + 2 m^{2} + 2 l^{2} + 2 l m + l m = 0$

$Grouping Terms$

Grouping like terms:
$2 l^{2} + (2 l m + 2 l m + l m) + 2 m^{2} = 0$
$2 l^{2} + 5 l m + 2 m^{2} = 0$

$Forming a Quadratic$

Divide the entire equation by $m^{2}$ :
$2 (\frac{l}{m})^{2} + 5 (\frac{l}{m}) + 2 = 0$

$Factoring the Quadratic$

Let $t = \frac{l}{m}$ . The equation becomes:
$2 t^{2} + 5 t + 2 = 0$
Splitting the middle term:
$2 t^{2} + 4 t + t + 2 = 0$
$(2 t + 1) (t + 2) = 0$

$Solving for Ratios$

Setting factors to zero:
$t_{1} = - 2 ⟹ \frac{l}{m} = - 2$
$t_{2} = - \frac{1}{2} ⟹ \frac{l}{m} = - \frac{1}{2}$

$Direction Ratios: Line 1$

Case 1: $\frac{l}{m} = - 2 ⟹ l = - 2 m$
Substitute into $n = 2 (l + m)$ :
$n = 2 (- 2 m + m) = - 2 m$
Direction Ratios $(a_{1}, b_{1}, c_{1}) = (- 2 m, m, - 2 m) \propto (- 2, 1, - 2)$

$Direction Ratios: Line 2$

Case 2: $\frac{l}{m} = - \frac{1}{2} ⟹ m = - 2 l$
Substitute into $n = 2 (l + m)$ :
$n = 2 (l - 2 l) = - 2 l$
Direction Ratios $(a_{2}, b_{2}, c_{2}) = (l, - 2 l, - 2 l) \propto (1, - 2, - 2)$

$Angle Formula$

The angle $θ$ between two lines with direction ratios $(a_{1}, b_{1}, c_{1})$ and $(a_{2}, b_{2}, c_{2})$ is:
$cos θ = \frac{∣ a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2} ∣}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}$

$Calculating Dot Product$

Vectors: $v_{1} = (- 2, 1, - 2)$ and $v_{2} = (1, - 2, - 2)$
Dot Product $= (- 2) (1) + (1) (- 2) + (- 2) (- 2)$
Dot Product $= - 2 - 2 + 4 = 0$

$Final Angle$

Since the dot product is $0$ , $cos θ = 0$ .
Therefore, the angle is:
$θ = \frac{π}{2}$
The lines are perpendicular.

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The Sigma Insight: Direction Cosines and Direction Ratios

Solution Diagram

Analyzing the Setup

We are given two constraints on the direction cosines

(l, m, n)

of two lines: 1. The linear equation:

2 l + 2 m - n = 0

2. The quadratic equation:

mn + n l + l m = 0

Our objective is to determine the angle

θ

between these two lines.

The Algebraic Dance

We begin by isolating

n

from the linear equation:

n = 2 (l + m)

Next, we substitute this expression into the quadratic equation

mn + n l + l m = 0

to reduce the system to variables

l

and

m

m (2 l + 2 m) + (2 l + 2 m) l + l m = 0

Homogenization and the Quadratic Reveal

Expanding the terms, we obtain:

2 l m + 2 m^{2} + 2 l^{2} + 2 l m + l m = 0

Grouping the like terms results in the homogeneous equation:

2 l^{2} + 5 l m + 2 m^{2} = 0

Dividing the entire expression by

m^{2}

, we define the ratio

t = \frac{l}{m}

, which yields the quadratic equation:

2 t^{2} + 5 t + 2 = 0

Factoring the quadratic, we find:

(2 t + 1) (t + 2) = 0

This provides two distinct ratios:

t_{1} = - 2

and

t_{2} = - \frac{1}{2}

The Final Climax

These ratios correspond to the direction vectors of the two lines.

For

t_{1} = - 2

, we have

l = - 2 m

. Substituting this into

n = 2 (l + m)

, we get

n = 2 (- 2 m + m) = - 2 m

. Thus, the first direction vector is

v_{1} = (- 2, 1, - 2)

For

t_{2} = - \frac{1}{2}

, we have

m = - 2 l

. Substituting this into

n = 2 (l + m)

, we get

n = 2 (l - 2 l) = - 2 l

. Thus, the second direction vector is

v_{2} = (1, - 2, - 2)

To find the angle

θ

, we calculate the dot product of

v_{1}

and

v_{2}

v_{1} \cdot v_{2} = (- 2) (1) + (1) (- 2) + (- 2) (- 2) = - 2 - 2 + 4 = 0

Since the dot product is zero, the lines are perpendicular. Therefore, the angle between the lines is:

θ = \frac{π}{2}

Common Traps & FAQs

Wait, why are we solving for the ratio

l / m

instead of finding the exact values of

l, m, n

That is a fantastic question! Remember, direction cosines are not unique; they are proportional to the direction ratios. Any line in 3D space is defined by its direction, not its specific coordinates. By finding the ratio

l / m

, we are identifying the 'slope' of the line in the

l m

-plane, which is sufficient to determine the direction vector

(a, b, c)

Why did we divide by

m^{2}

in the quadratic equation? Is that always allowed?

Great observation. We divide by

m^{2}

because the equation is homogeneous. As long as

m \neq = 0

, this is perfectly valid. If

m

were zero, we would check the equation

2 l^{2} = 0

, which implies

l = 0

, and consequently

n = 0

. But

(0, 0, 0)

cannot be a direction vector. Thus,

m

cannot be zero, and our division is mathematically sound.

What if the dot product hadn't turned out to be zero? How would I find the angle then?

If the dot product were non-zero, you would simply use the standard formula:

cos θ = \frac{∣ a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2} ∣}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}

. The result would be an inverse cosine value, like

cos^{- 1} (something)

. Always remember to take the absolute value in the numerator to ensure you get the acute angle between the lines!

Animated Solution for Mathematics - Three Dimensional Geometry: The angle between the straight lines, whose direction cosines are given by the equations $2 l + 2 m - n = 0$ and $mn + n l + l m = 0$ , is :

Select Answer:

Visualized Solution

$2 l + 2 m - n = 0$

$Elimination Strategy$

$Isolating n$

$Substituting n$

$Expanding Terms$

$Grouping Terms$

$Forming a Quadratic$

$Factoring the Quadratic$

$Solving for Ratios$

$Direction Ratios: Line 1$

$Direction Ratios: Line 2$

$Angle Formula$

$Calculating Dot Product$

$Final Angle$

The Sigma Insight: Direction Cosines and Direction Ratios

Analyzing the Setup

The Algebraic Dance

Homogenization and the Quadratic Reveal

The Final Climax

Similar Questions

The angle between the lines whose direction cosines satisfy the equations $l + m + n = 0$ and $l^{2} = m^{2} + n^{2}$ is

Let the direction cosines of two lines satisfy the equations : $4 l + m - n = 0$ and $2 mn + 10 n l + 3 l m = 0$ . Then the cosine of the acute angle between these lines is :

Let $α$ be the angle between the lines whose direction cosines satisfy the equations $l + m - n = 0$ and $l^{2} + m^{2} - n^{2} = 0$ . Then the value of $sin^{4} α + cos^{4} α$ is :

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A line $A B$ in three-dimensional space makes angles $4 5^{\circ}$ and $12 0^{\circ}$ with the positive $x$ -axis and the positive $y$ -axis respectively. If $A B$ makes an acute angle $θ$ with the positive $z$ -axis, then $θ$ equals

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2l+2m−n=0

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

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

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

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

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Direction Ratios: Line 2

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Dot Product

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

The Sigma Insight: Direction Cosines and Direction Ratios

Analyzing the Setup

The Algebraic Dance

Homogenization and the Quadratic Reveal

The Final Climax

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The angle between the lines whose direction cosines satisfy the equations l+m+n=0 and l2=m2+n2 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the direction cosines of two lines satisfy the equations : 4l+m−n=0 and 2mn+10nl+3lm=0. Then the cosine of the acute angle between these lines is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α be the angle between the lines whose direction cosines satisfy the equations l+m−n=0 and l2+m2−n2=0. Then the value of sin4α+cos4α is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If two straight lines whose direction cosines are given by the relations l+m−n=0,3l2+m2+cnl=0 are parallel, then the positive value of c is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a line makes an angle of π/4 with the positive directions of each of x-axis and y-axis, then the angle that the line makes with the positive direction of the z-axis is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A line makes the same angle θ, with each of the x and z axis. If the angle β, which it makes with y-axis, is such that sin2β=3sin2θ, then cos2θ equals

$2 l + 2 m - n = 0$

$Elimination Strategy$

$Isolating n$

$Substituting n$

$Expanding Terms$

$Grouping Terms$

$Forming a Quadratic$

$Factoring the Quadratic$

$Solving for Ratios$

$Direction Ratios: Line 1$

$Direction Ratios: Line 2$

$Angle Formula$

$Calculating Dot Product$

$Final Angle$

The angle between the lines whose direction cosines satisfy the equations $l + m + n = 0$ and $l^{2} = m^{2} + n^{2}$ is

Let the direction cosines of two lines satisfy the equations : $4 l + m - n = 0$ and $2 mn + 10 n l + 3 l m = 0$ . Then the cosine of the acute angle between these lines is :

Let $α$ be the angle between the lines whose direction cosines satisfy the equations $l + m - n = 0$ and $l^{2} + m^{2} - n^{2} = 0$ . Then the value of $sin^{4} α + cos^{4} α$ is :

If two straight lines whose direction cosines are given by the relations $l + m - n = 0, 3 l^{2} + m^{2} + c n l = 0$ are parallel, then the positive value of $c$ is :

If a line makes an angle of $π /4$ with the positive directions of each of $x$ -axis and $y$ -axis, then the angle that the line makes with the positive direction of the $z$ -axis is

A line makes the same angle $θ$ , with each of the $x$ and $z$ axis. If the angle $β$ , which it makes with $y$ -axis, is such that $sin^{2} β = 3 sin^{2} θ$ , then $cos^{2} θ$ equals