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 \(\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}\)  and \(\frac{x-1}{3 k}=\frac{y-5}{1}=\frac{z-6}{-5}\) 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° = \(\frac{π}{3} \) 

Hence Option 1 is the correct answer.

Calculation

\(\tan \theta=\frac{\mathrm{m}-\frac{1}{2}}{1+\frac{1}{2} \cdot \mathrm{~m}}=\frac{-1-\mathrm{m}}{1-\mathrm{m}}=\frac{\mathrm{m}+1}{\mathrm{~m}-1}\)

\(\frac{2 m-1}{2+m}=\frac{m+1}{m-1}\)

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

\(\rm cosθ =\left|\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^{2}+b_1^{2}+c_1^{2} }\sqrt{a_2^{2}+b_2^{2}+c_2^{2} }} \right| \)

​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 = \(\Big|\frac{1\times 1+0\times (-1)+(-1)\times 0}{\sqrt{1^2+1^2}\sqrt{1^2+1^2}}\Big|\)

⇒ 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: 

\(\rm \cos {\bf{θ }} = \;\frac{{{{\bf{a}}_1}{{\bf{a}}_2} + \;{{\bf{b}}_1}{{\bf{b}}_2} + {{\bf{c}}_1}{{\bf{c}}_2}\;}}{{\sqrt {{\bf{a}}_{1\;}^2 + \;{\bf{b}}_1^2 + \;{\bf{c}}_1^2} \sqrt {{\bf{a}}_{2\;}^2 + \;{\bf{b}}_2^2 + \;{\bf{c}}_2^2} }}\)

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 \(\rm \frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\;and\frac{{x - {x_2}}}{{{a_2}}} = \frac{{y - {y_2}}}{{{b_2}}} = \frac{{z - {z_2}}}{{{c_2}}}\) are two lines

then the angle between them is given by: \(\rm \cos {\bf{θ }} = \;\frac{{{{\bf{a}}_1}{{\bf{a}}_2} + \;{{\bf{b}}_1}{{\bf{b}}_2} + {{\bf{c}}_1}{{\bf{c}}_2}\;}}{{\sqrt {{\bf{a}}_{1\;}^2 + \;{\bf{b}}_1^2 + \;{\bf{c}}_1^2} \sqrt {{\bf{a}}_{2\;}^2 + \;{\bf{b}}_2^2 + \;{\bf{c}}_2^2} }}\)

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

⇒ \(\rm \cos {\bf{θ }} = \;\frac{ (1)(3)+(2)(2)+(2)(6) }{{\sqrt {{1}^2 + 2^2 + 2^2} \sqrt {3^2 + 2^2 + 6^2} }}\)

\(\rm \cos {\bf{θ }} = \;\frac{19}{{\sqrt {9} \sqrt {49}}} = \frac{19}{21}\)

∴ 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:

\(\tan θ = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1} \cdot \ {m_2}}}} \right|\)

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.

⇒ \(\tan θ = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1} \cdot \ {m_2}}}} \right|\)

⇒ \(\tan θ = \left| {\frac{{\frac{{{m_2}}}{{{m_1}}} - 1}}{{\frac{1}{{{m_1}}} + {m_2}}}} \right|\)

⇒ \(tan \ θ = \frac{1}{\sqrt 3}\)

⇒ 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 = \(tan^{-1}{\sqrt 3}\) = 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:

The angle between the lines:

The angle between the lines \(\rm \frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\;and\frac{{x - {x_2}}}{{{a_2}}} = \frac{{y - {y_2}}}{{{b_2}}} = \frac{{z - {z_2}}}{{{c_2}}}\) is given by:

\(\rm \cos θ = \frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\left( {\sqrt {a_1^2 + b_1^2 + c_1^2} } \right) ⋅ \left( {\sqrt {a_2^2 + b_2^2 + c_2^2} } \right)}}\), 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 \(\rm \cos θ = \frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\left( {\sqrt {a_1^2 + b_1^2 + c_1^2} } \right) ⋅ \left( {\sqrt {a_2^2 + b_2^2 + c_2^2} } \right)}}\)

\(\Rightarrow \rm \cos θ = \frac{{{2} \times {1} + {5}\times{2} + {4}\times{-3}}}{{\left( {\sqrt {2^2 + 5^2 + 4^2} } \right) ⋅ \left( {\sqrt {1^2 + 2^2 + (-3)^2} } \right)}} = 0\)

∴ θ = 90° 

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  \(\frac{{2{\rm{x}} - 2}}{{2{\rm{k}}}} = \frac{{4 - {\rm{y}}}}{3} = \frac{{{\rm{z}} + 2}}{{ - 1}}\) and \(\frac{{{\rm{x}} - 5}}{1} = \frac{{\rm{y}}}{{\rm{k}}} = \frac{{{\rm{z}} + 6}}{4}\) 

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

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

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

\(\frac{{{\rm{x}} - 5}}{1} = \frac{{\rm{y}}}{{\rm{k}}} = \frac{{{\rm{z}} + 6}}{4}\)

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 with direction ratios \(\left\langle {{a_1},\;{b_1},\;{c_1}} \right\rangle \) and \(\left\langle {{a_2},\;{b_2},\;{c_2}} \right\rangle \) is given by: \(\cos {\bf{\theta }} = \;\frac{{{{\bf{a}}_1}{{\bf{a}}_2} + \;{{\bf{b}}_1}{{\bf{b}}_2} + {{\bf{c}}_1}{{\bf{c}}_2}\;}}{{\sqrt {{\bf{a}}_{1\;}^2 + \;{\bf{b}}_1^2 + \;{\bf{c}}_1^2} \sqrt {{\bf{a}}_{2\;}^2 + \;{\bf{b}}_2^2 + \;{\bf{c}}_2^2} }}\)

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  \(\frac{{{\rm{x}} + 4}}{{2{\rm{}}}} = \frac{{4 - {\rm{y}}}}{-2} = \frac{{{\rm{2z}} - 4}}{{\rm 2k}}\) and \(\frac{{{\rm{x}} +3}}{\rm -k} = \frac{{\rm{y-3}}}{{\rm{2}}} = \frac{{{\rm{z}} + 1}}{5}\) 

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

\( \Rightarrow \frac{{{\rm{x}} + 4}}{{2{\rm{}}}} = \frac{-{(\rm y - {\rm{4})}}}{-2} = \frac{2{{(\rm{z}} - 2)}}{{\rm 2k}} \Leftrightarrow \frac{{\left( {{\rm{x}} +4 } \right)}}{{\rm{2}}} = \frac{{{\rm{y}} - 4}}{{ 2}} = \frac{{{\rm{z}} - 2}}{{ \rm k}}\)

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

\(\frac{{{\rm{x}} +3}}{\rm -k} = \frac{{\rm{y-3}}}{{\rm{2}}} = \frac{{{\rm{z}} + 1}}{5}\)

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

\(cosθ =\left|\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^{2}+b_1^{2}+c_1^{2} }\sqrt{a_2^{2}+b_2^{2}+c_2^{2} }} \right| \)

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

\(cosθ =\left|\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^{2}+b_1^{2}+c_1^{2} }\sqrt{a_2^{2}+b_2^{2}+c_2^{2} }} \right|\)

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

\(cosθ =\left|\frac{8-8}{\sqrt{9}\sqrt{36}} \right|\)

\(cosθ =\left|\frac{0}{3\times 6} \right|\)

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: \(\cos θ = \left| {\frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} \;\sqrt {a_2^2 + b_2^2 + c_2^2} }}} \right|\)

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: \(\cos θ = \left| {\frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} \;\sqrt {a_2^2 + b_2^2 + c_2^2} }}} \right|\)

⇒ \(\cos θ = \left| {\frac{{6 \cdot 3 + 3 \cdot 3 +6 \cdot 0}}{{\sqrt {6^2 + 3^2 + 6^2} \;\sqrt {3^2 + 3^2 + 0^2} }}} \right|\)

⇒ \(cos θ = \frac{1}{\sqrt 2}\)

⇒ θ = π/4

Hence, correct option is 2.

CONCEPT:

The angle between the lines with direction ratios \(\left\langle {{a_1},\;{b_1},\;{c_1}} \right\rangle \) and \(\left\langle {{a_2},\;{b_2},\;{c_2}} \right\rangle \) is given by: \(\cos {\bf{\theta }} = \;\frac{{{{\bf{a}}_1}{{\bf{a}}_2} + \;{{\bf{b}}_1}{{\bf{b}}_2} + {{\bf{c}}_1}{{\bf{c}}_2}\;}}{{\sqrt {{\bf{a}}_{1\;}^2 + \;{\bf{b}}_1^2 + \;{\bf{c}}_1^2} \sqrt {{\bf{a}}_{2\;}^2 + \;{\bf{b}}_2^2 + \;{\bf{c}}_2^2} }}\)

Concept:

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

cos θ = \(\rm \vec A\cdot\vec B\over|\vec A||\vec B|\)

Calculation:

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

Let the angle between them be θ 

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

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

cos θ = 0

∴ θ = \(\boldsymbol{\pi\over2}\)

Concept:

\(\rm\vec a.\vec b = |\vec a||\vec b|\cos \theta\)

\(\rm\text{If A = (x, y, z) and B = (x', y', z') then }\\ \vec {AB} = (x' - x)\hat i + (y' - y)\hat j+(z' - z)\hat k\\ |\vec {AB}|=\sqrt{(x' - x)^2 + (y' - y)^2+(z' - z)^2} \)

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)

\(\rm\vec {OA} = (a - 0)\hat i + (a - 0)\hat j+(a - 0)\hat k\\ =a\hat i + a \hat j+a \hat k\\ |\vec {OA}|=\sqrt{a^2 +a^2 +a^2 }=\sqrt{3} a\\ \rm\vec {BC} = (a - 0)\hat i + (a - 0)\hat j+(0 - a)\hat k\\ =a\hat i + a \hat j-a \hat k\\|\vec {BC}|=\sqrt{a^2 +a^2 +(-a)^2 }=\sqrt{3} a\\\)

Now,

 \(\rm\vec {OA} .\rm\vec {BC} =|\rm\vec {OA}||\rm\vec {BC} | \cos \theta \\ \Rightarrow a^2 +a^2 -a^2=\sqrt 3a\times \sqrt 3a \cos \theta \\ \Rightarrow a^2=3a^2\cos \theta \\ \Rightarrow 3\cos \theta=1 \)

Hence, option (4) is correct.