Angle between two lines MCQ Quiz - Objective Question with Answer for Angle between two lines - Download Free PDF

Concept used 

Two lines with direction ratios a₁, b₁, c₁ and a₂, b₂, c₂ are perpendicular

if  a₁a₂ + b₁b₂ + c₁c₂ = 0

Calculation

Lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$  and $\frac{x-1}{3k}=\frac{y-5}{1}=\frac{z-6}{-5}$ are perpendicular then

⇒ -3(2k) + 2k + 2(-5) = 0

 ∴  k = -$\frac{10}{7}$

Explanation:

Let the direction ratios of the two lines be:

Vector A = (1, 1, -2)

Vector B = (√3 - 1, -√3 - 1, -4)

The angle θ between two vectors A and B is given by:

cos θ = (a₁a₂ + b₁b₂ + c₁c₂) / (√(a₁² + b₁² + c₁²) × √(a₂² + b₂² + c₂²))

Numerator (dot product):

(1)(√3 - 1) + (1)(-√3 - 1) + (-2)(-4) = (√3 - 1) + (-√3 - 1) + 8 = -2 + 8 = 6

Denominator (product of magnitudes):

|A| = √(1² + 1² + (-2)²) = √(1 + 1 + 4) = √6

|B| = √((√3 - 1)² + (-√3 - 1)² + (-4)²)

 = √((4 - 2√3) + (4 + 2√3) + 16) = √24 = 2√6

cos θ = 6 / (√6 × 2√6) = 6 / (2 × 6) = 1 / 2

θ = cos^-1(1 / 2) = 60° = $\frac{\pi }{3}$ 

Hence Option 1 is the correct answer.

Calculation

$\tan \theta =\frac{\mathrm{m}-\frac{1}{2}}{1+\frac{1}{2}\cdot \mathrm{m}}=\frac{-1-\mathrm{m}}{1-\mathrm{m}}=\frac{\mathrm{m}+1}{\mathrm{m}-1}$

$\frac{2m-1}{2+m}=\frac{m+1}{m-1}$

2m² – 3m + 1 = m² + 3m + 2

m² – 6m –1 = 0

sum of root = 6

sum is 6  

Hence option 4 is correct

Concept:

The angle between the pair of lines

$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$

and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ is given by

$\mathrm{c}\mathrm{o}\mathrm{s}\theta =\left|\frac{{\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2+{\mathrm{c}}_1{\mathrm{c}}_2}{\sqrt{{\mathrm{a}}_1^2+{\mathrm{b}}_1^2+{\mathrm{c}}_1^2}\sqrt{{\mathrm{a}}_2^2+{\mathrm{b}}_2^2+{\mathrm{c}}_2^2}}\right|$

​Calculation:

Given direction ratios satisfy the equations  l + m + n = 0 and l2 = m2 + n2.

Substituting l = - (m + n) in the equation l2 = m2 + n2.

⇒ (- (m + n))² = m2 + n2.

⇒ m² + n² + 2mn = m² + n².

⇒ mn = 0

m = 0 or n = 0.

If m  = 0

⇒ l + n = 0

or l = - n

⇒ $\frac{l}{1}=\frac{m}{0}=\frac{n}{-1}$

If n = 0,

⇒ m + l = 0

or m = - l.

⇒ $\frac{l}{1}=\frac{m}{-1}=\frac{n}{0}$

∴ Direction cosines of the lines are <1, 0, - 1> and <1, - 1, 0>

⇒ cos θ = $|\frac{1\times 1+0\times (-1)+(-1)\times 0}{\sqrt{1^2+1^2}\sqrt{1^2+1^2}}|$

⇒ cos θ = $\frac{1}{\sqrt{2}\sqrt{2}}$

⇒ cos θ = $\frac{1}{2}$

∴  θ = $\frac{\pi }{3}$

The angle between the lines is $\frac{\pi }{3}$.

Concept:

The angle between the lines with direction ratios (a1,b1,c1) and (a2,b2,c2) is given by: 

$\cos \mathbf{\theta }=\frac{{\mathbf{a}}_1{\mathbf{a}}_2+{\mathbf{b}}_1{\mathbf{b}}_2+{\mathbf{c}}_1{\mathbf{c}}_2}{\sqrt{{\mathbf{a}}_1^2+{\mathbf{b}}_1^2+{\mathbf{c}}_1^2}\sqrt{{\mathbf{a}}_2^2+{\mathbf{b}}_2^2+{\mathbf{c}}_2^2}}$

Calculation:

Given:  $\frac{x-3}{1}=\frac{y-2}{2}=\frac{z+1}{2}$ and $\frac{x-0}{3}=\frac{y-5}{2}=\frac{z-2}{6}$

As we know that, if $\frac{\mathrm{x}-{\mathrm{x}}_1}{{\mathrm{a}}_1}=\frac{\mathrm{y}-{\mathrm{y}}_1}{{\mathrm{b}}_1}=\frac{\mathrm{z}-{\mathrm{z}}_1}{{\mathrm{c}}_1}\mathrm{a}\mathrm{n}\mathrm{d}\frac{\mathrm{x}-{\mathrm{x}}_2}{{\mathrm{a}}_2}=\frac{\mathrm{y}-{\mathrm{y}}_2}{{\mathrm{b}}_2}=\frac{\mathrm{z}-{\mathrm{z}}_2}{{\mathrm{c}}_2}$ are two lines

then the angle between them is given by: $\cos \mathbf{\theta }=\frac{{\mathbf{a}}_1{\mathbf{a}}_2+{\mathbf{b}}_1{\mathbf{b}}_2+{\mathbf{c}}_1{\mathbf{c}}_2}{\sqrt{{\mathbf{a}}_1^2+{\mathbf{b}}_1^2+{\mathbf{c}}_1^2}\sqrt{{\mathbf{a}}_2^2+{\mathbf{b}}_2^2+{\mathbf{c}}_2^2}}$

Here, a1 = 1, b1 = 2, c1 = 2, a2 = 3, b2 = 2 and c2 = 6

⇒ $\cos \mathbf{\theta }=\frac{(1)(3)+(2)(2)+(2)(6)}{\sqrt{1^2+2^2+2^2}\sqrt{3^2+2^2+6^2}}$

$\cos \mathbf{\theta }=\frac{19}{\sqrt{9}\sqrt{49}}=\frac{19}{21}$

∴ θ = cos^-1 $\frac{19}{21}$

Concept:

Let θ be the angle between two lines of slope m₁ and m₂, then the acute angle between the lines is given by:

$\tan \theta =\left|\frac{m_2-m_1}{1+m_1\cdot m_2}\right|$

Calculations:

Consider, the slope of the line (x - 2 = 0) is m₁

So, m₁ = ∞ .

And, the slope of the line (√3x - y - 2 = 0) is m₂

So, m₂ = √3.

Now, the angle between the given lines is θ.

⇒ $\tan \theta =\left|\frac{m_2-m_1}{1+m_1\cdot m_2}\right|$

⇒ $\tan \theta =\left|\frac{\frac{m_2}{m_1}-1}{\frac{1}{m_1}+m_2}\right|$

⇒ $tan\theta =\frac{1}{\sqrt{3}}$

⇒ θ = 30°

Hence, the correct option is 2.

Alternate Method

For line 1:x – 2 = 0Hence, x = 2

Since it is a vertical line, the slope is undefined.

Hence, θ₁ = 90°

For line 2:

√3x - y - 2 = 0

Compare this equation with, y = mx + c

It becomes, y = √3x + 2

Hence, m = √3 = tan θ₂

Hence, θ₂ = $ta{n}^{-1}\sqrt{3}$ = 60°

Now, the graph of intersection between two lines can be drawn as

Hence, acute angle between the lines = θ₁ – θ₂ = 90° – 60° = 30°

Concept:

Let two lines having direction ratios a_1, b₁, c_1, and a_2, b₂, c₂ respectively.

Condition for perpendicular lines: a₁a₂ + b₁b₂ + c₁c₂ = 0

Calculation:

Given lines are  $\frac{2\mathrm{x}-2}{2\mathrm{k}}=\frac{4-\mathrm{y}}{3}=\frac{\mathrm{z}+2}{-1}$ and $\frac{\mathrm{x}-5}{1}=\frac{\mathrm{y}}{\mathrm{k}}=\frac{\mathrm{z}+6}{4}$ 

Write the above equation of a line in the standard form of lines

$\Rightarrow \frac{2\left(\mathrm{x}-1\right)}{2\mathrm{k}}=\frac{-\left(\mathrm{y}-4\right)}{3}=\frac{\mathrm{z}+2}{-1}\iff \frac{\left(\mathrm{x}-1\right)}{\mathrm{k}}=\frac{\mathrm{y}-4}{-3}=\frac{\mathrm{z}+2}{-1}$

So, the direction ratio of the first line is (k, -3, -1)

$\frac{\mathrm{x}-5}{1}=\frac{\mathrm{y}}{\mathrm{k}}=\frac{\mathrm{z}+6}{4}$

So, direction ratio of second line is (1, k, 4)

Lines are perpendicular,

∴ (k × 1) + (-3 × k) + (-1 × 4) = 0

⇒ k – 3k – 4 = 0

⇒ -2k – 4 = 0

∴ k = -2

Concept:

The angle between the lines:

The angle between the lines $\frac{\mathrm{x}-{\mathrm{x}}_1}{{\mathrm{a}}_1}=\frac{\mathrm{y}-{\mathrm{y}}_1}{{\mathrm{b}}_1}=\frac{\mathrm{z}-{\mathrm{z}}_1}{{\mathrm{c}}_1}\mathrm{a}\mathrm{n}\mathrm{d}\frac{\mathrm{x}-{\mathrm{x}}_2}{{\mathrm{a}}_2}=\frac{\mathrm{y}-{\mathrm{y}}_2}{{\mathrm{b}}_2}=\frac{\mathrm{z}-{\mathrm{z}}_2}{{\mathrm{c}}_2}$ is given by:

$\cos \theta =\frac{{\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2+{\mathrm{c}}_1{\mathrm{c}}_2}{\left(\sqrt{{\mathrm{a}}_1^2+{\mathrm{b}}_1^2+{\mathrm{c}}_1^2}\right)\cdot \left(\sqrt{{\mathrm{a}}_2^2+{\mathrm{b}}_2^2+{\mathrm{c}}_2^2}\right)}$, where a₁, b₁, c₁, a₂, b₂ and c₂ are the direction ratios

Calculation:

Given:  $\frac{x+1}{2}=\frac{y-2}{5}=\frac{z+3}{4}$ and $\frac{x-1}{1}=\frac{y+2}{2}=\frac{z-3}{-3}$

Direction ratios of lines are a₁ = 2, b₁ = 5, c₁ = 4 and a₂ = 1, b₂ = 2 , c₂ = -3

As we know, The angle between the lines is given by $\cos \theta =\frac{{\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2+{\mathrm{c}}_1{\mathrm{c}}_2}{\left(\sqrt{{\mathrm{a}}_1^2+{\mathrm{b}}_1^2+{\mathrm{c}}_1^2}\right)\cdot \left(\sqrt{{\mathrm{a}}_2^2+{\mathrm{b}}_2^2+{\mathrm{c}}_2^2}\right)}$

$\Rightarrow \cos \theta =\frac{2\times 1+5\times 2+4\times -3}{\left(\sqrt{2^2+5^2+4^2}\right)\cdot \left(\sqrt{1^2+2^2+(-3{)}^2}\right)}=0$

∴ θ = 90° 

CONCEPT:

The angle between the lines with direction ratios $⟨a_1,b_1,c_1⟩$ and $⟨a_2,b_2,c_2⟩$ is given by: $\cos \theta =\frac{{\mathbf{a}}_1{\mathbf{a}}_2+{\mathbf{b}}_1{\mathbf{b}}_2+{\mathbf{c}}_1{\mathbf{c}}_2}{\sqrt{{\mathbf{a}}_1^2+{\mathbf{b}}_1^2+{\mathbf{c}}_1^2}\sqrt{{\mathbf{a}}_2^2+{\mathbf{b}}_2^2+{\mathbf{c}}_2^2}}$

Concept:

Let the two lines have direction ratio’s a1, b1, c1 and a2, b2, c2 respectively.

Condition for perpendicular lines: a1a2 + b1b2 + c1c2 = 0

Calculation:

Given lines are  $\frac{\mathrm{x}+4}{2}=\frac{4-\mathrm{y}}{-2}=\frac{2\mathrm{z}-4}{2\mathrm{k}}$ and $\frac{\mathrm{x}+3}{-\mathrm{k}}=\frac{\mathrm{y}-3}{2}=\frac{\mathrm{z}+1}{5}$ 

Write the above equation of a line in the standard form of lines

$\Rightarrow \frac{\mathrm{x}+4}{2}=\frac{-(\mathrm{y}-4)}{-2}=\frac{2(\mathrm{z}-2)}{2\mathrm{k}}\iff \frac{\left(\mathrm{x}+4\right)}{2}=\frac{\mathrm{y}-4}{2}=\frac{\mathrm{z}-2}{\mathrm{k}}$

So, the direction ratio of the first line is (2, 2, k)

$\frac{\mathrm{x}+3}{-\mathrm{k}}=\frac{\mathrm{y}-3}{2}=\frac{\mathrm{z}+1}{5}$

So, direction ratio of second line is (-k, 2, 5)

Lines are perpendicular,

∴ (2 × -k) + (2 × 2) + (k × 5) = 0

⇒ -2k + 4 + 5k = 0

⇒ 3k + 4 = 0

∴ k = -4/3

Concept:

The angle between the pair of lines

$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$

and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ is given by

$cos\theta =\left|\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\right|$

Solution:

Given the pair of lines $\frac{x-4}{2}=\frac{y}{1}=\frac{z+1}{-2}$ and 

$\frac{x-1}{4}=\frac{y+1}{-4}=\frac{z-2}{2}$

a₁ = 2, b₁ = 1, c₁ = -2

a₂ = 4, b₂ = -4, c₂ = 2

$cos\theta =\left|\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\right|$

$cos\theta =\left|\frac{2\times 4-1\times 4-2\times 2}{\sqrt{2^2+1^2+(-2{)}^2}\sqrt{4^2+(-4{)}^2+2^2}}\right|$

$cos\theta =\left|\frac{8-8}{\sqrt{9}\sqrt{36}}\right|$

$cos\theta =\left|\frac{0}{3\times 6}\right|$

cosθ = 0

θ = $\frac{\pi }{2}$

CONCEPT:

The angle θ between two lines whose direction ratios are proportional to a₁, b₁, c₁ and a₂, b₂, c₂ respectively is given by: $\cos \theta =\left|\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\right|$

CALCULATION:

Here, we have to find the angle between the two lines having direction ratios (6, 3, 6) and (3, 3, 0).

Here, a₁ = 6, b₁ = 3, c₁ = 6, a₂ = 3, b₂ = 3 and c₂ = 0

As we know that, if θ is the angle between two lines having direction ratios proportional to a1, b1, c1 and a2, b2, c₂ is given by: $\cos \theta =\left|\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\right|$

⇒ $\cos \theta =\left|\frac{6\cdot 3+3\cdot 3+6\cdot 0}{\sqrt{6^2+3^2+6^2}\sqrt{3^2+3^2+0^2}}\right|$

⇒ $cos\theta =\frac{1}{\sqrt{2}}$

⇒ θ = π/4

Hence, correct option is 2.

CONCEPT:

The angle between the lines with direction ratios $⟨a_1,b_1,c_1⟩$ and $⟨a_2,b_2,c_2⟩$ is given by: $\cos \theta =\frac{{\mathbf{a}}_1{\mathbf{a}}_2+{\mathbf{b}}_1{\mathbf{b}}_2+{\mathbf{c}}_1{\mathbf{c}}_2}{\sqrt{{\mathbf{a}}_1^2+{\mathbf{b}}_1^2+{\mathbf{c}}_1^2}\sqrt{{\mathbf{a}}_2^2+{\mathbf{b}}_2^2+{\mathbf{c}}_2^2}}$

Concept:

The angle θ between the two vectors A and B is given by:

cos θ = $\frac{\overrightarrow{\mathrm{A}}\cdot \overrightarrow{\mathrm{B}}}{|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{B}}|}$

Calculation:

Given A = 3i + 5j - 4k and B = -5i + 11j + 10k

Let the angle between them be θ 

cos θ = $\frac{(3\mathrm{i}+5\mathrm{j}-4\mathrm{k})\cdot (-5\mathrm{i}+11\mathrm{j}+10\mathrm{k})}{\sqrt{(3{)}^2+(5{)}^2+(-4{)}^2}\times \sqrt{(5{)}^2+(11{)}^2+(10{)}^2}}$

cos θ = $\frac{(-15+55-40)}{\sqrt{50}\times \sqrt{246}}$

cos θ = 0

∴ θ = $\text{boldsymbol}\frac{\pi }{2}$

Concept:

$\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}|\cos \theta$

$$\begin{gathered} \text{If A = (x, y, z) and B = (x', y', z') then } \\ \overrightarrow{\mathrm{A}\mathrm{B}}=({\mathrm{x}}^{\prime }-\mathrm{x})\hat{\mathrm{i}}+({\mathrm{y}}^{\prime }-\mathrm{y})\hat{\mathrm{j}}+({\mathrm{z}}^{\prime }-\mathrm{z})\hat{\mathrm{k}} \\ |\overrightarrow{\mathrm{A}\mathrm{B}}|=\sqrt{({\mathrm{x}}^{\prime }-\mathrm{x}{)}^2+({\mathrm{y}}^{\prime }-\mathrm{y}{)}^2+({\mathrm{z}}^{\prime }-\mathrm{z}{)}^2} \end{gathered}$$

Calculation:

Here, OA and BC are diagonals of cube 

O (0, 0, 0), A = (a, a, a), B = (0, 0, a), C = (a, a, 0)

$$\begin{gathered} \overrightarrow{\mathrm{O}\mathrm{A}}=(\mathrm{a}-0)\hat{\mathrm{i}}+(\mathrm{a}-0)\hat{\mathrm{j}}+(\mathrm{a}-0)\hat{\mathrm{k}} \\ =\mathrm{a}\hat{\mathrm{i}}+\mathrm{a}\hat{\mathrm{j}}+\mathrm{a}\hat{\mathrm{k}} \\ |\overrightarrow{\mathrm{O}\mathrm{A}}|=\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+{\mathrm{a}}^2}=\sqrt{3}\mathrm{a} \\ \overrightarrow{\mathrm{B}\mathrm{C}}=(\mathrm{a}-0)\hat{\mathrm{i}}+(\mathrm{a}-0)\hat{\mathrm{j}}+(0-\mathrm{a})\hat{\mathrm{k}} \\ =\mathrm{a}\hat{\mathrm{i}}+\mathrm{a}\hat{\mathrm{j}}-\mathrm{a}\hat{\mathrm{k}} \\ |\overrightarrow{\mathrm{B}\mathrm{C}}|=\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2+(-\mathrm{a}{)}^2}=\sqrt{3}\mathrm{a} \end{gathered}$$

Now,

 $$\begin{gathered} \overrightarrow{\mathrm{O}\mathrm{A}}.\overrightarrow{\mathrm{B}\mathrm{C}}=|\overrightarrow{\mathrm{O}\mathrm{A}}||\overrightarrow{\mathrm{B}\mathrm{C}}|\cos \theta \\ \Rightarrow {\mathrm{a}}^2+{\mathrm{a}}^2-{\mathrm{a}}^2=\sqrt{3}\mathrm{a}\times \sqrt{3}\mathrm{a}\cos \theta \\ \Rightarrow {\mathrm{a}}^2=3{\mathrm{a}}^2\cos \theta \\ \Rightarrow 3\cos \theta =1 \end{gathered}$$

Hence, option (4) is correct.