Angle between two lines MCQ Quiz - Objective Question with Answer for Angle between two lines - Download Free PDF
Last updated on May 3, 2025
Latest Angle between two lines MCQ Objective Questions
Angle between two lines Question 1:
If 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 mutually perpendicular, then k is equal to
Answer (Detailed Solution Below)
Angle between two lines Question 1 Detailed Solution
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 = -\(\frac{10}{7}\)
Angle between two lines Question 2:
The angle between two lines whose direction ratios are propotional to 1, 1, –2 and (√3 – 1), (−√3 – 1), –4 is
Answer (Detailed Solution Below)
Angle between two lines Question 2 Detailed Solution
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.
Angle between two lines Question 3:
Two equal sides of an isosceles triangle are along –x + 2y = 4 and x + y = 4. If m is the slope of its third side, then the sum, of all possible distinct values of m, is :
Answer (Detailed Solution Below)
Angle between two lines Question 3 Detailed Solution
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
Angle between two lines Question 4:
The angle between the lines whose direction ratios satisfy the equations l + m + n = 0 and l2 = m2 + n2 is
Answer (Detailed Solution Below)
Angle between two lines Question 4 Detailed Solution
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 \(\frac{\pi}{3}\).
Angle between two lines Question 5:
Find the angle between the lines \(\frac{x-3}{1} = \frac{y-2}{2}=\frac {z+1}{2}\) and \(\frac{x-0}{3} = \frac{y-5}{2}=\frac {z-2}{6}\)
Answer (Detailed Solution Below)
Angle between two lines Question 5 Detailed Solution
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 \(\frac{19}{21}\)
Top Angle between two lines MCQ Objective Questions
What is the acute angle between the lines x - 2 = 0 and √3x - y - 2 = 0?
Answer (Detailed Solution Below)
Angle between two lines Question 6 Detailed Solution
Download Solution PDFConcept:
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°
The angle between the straight lines \(\frac {x + 1}{2} = \frac {y - 2}{5} = \frac {z + 3}{4}\) and \(\frac {x - 1}{1} = \frac {y + 2}{2} = \frac {z - 3}{-3}\) is-
Answer (Detailed Solution Below)
Angle between two lines Question 7 Detailed Solution
Download Solution PDFConcept:
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°
Find the values of k so the line \(\frac{{2{\rm{x}} - 2}}{{2{\rm{k}}}} = \frac{{4 - {\rm{y}}}}{3} = \frac{{{\rm{z}} + 2}}{{ - 1}}\) and \(\frac{{{\rm{x}} - 5}}{1} = \frac{{\rm{y}}}{{\rm{k}}} = \frac{{{\rm{z}} + 6}}{4}\) are at right angles.
Answer (Detailed Solution Below)
Angle between two lines Question 8 Detailed Solution
Download Solution PDFConcept:
Let two lines having direction ratios a1, b1, c1, and a2, 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
Find the angle between the pair of lines
\(\frac{{x - 5}}{3} = \frac{{y + 2}}{5} = \frac{{z + 2}}{4}\)
And \(\frac{{x - 1}}{1} = \frac{{y - 3}}{1} = \frac{{z - 3}}{2}\)
Answer (Detailed Solution Below)
Angle between two lines Question 9 Detailed Solution
Download Solution PDFCONCEPT:
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: \(\cos {\bf{\theta }} = \;\frac{{{{\bf{a}}_1}{{\bf{a}}_2} + \;{{\bf{b}}_1}{{\bf{b}}_2} + {{\bf{c}}_1}{{\bf{c}}_2}\;}}{{\sqrt {{\bf{a}}_{1\;}^2 + \;{\bf{b}}_1^2 + \;{\bf{c}}_1^2} \sqrt {{\bf{a}}_{2\;}^2 + \;{\bf{b}}_2^2 + \;{\bf{c}}_2^2} }}\)
Find the values of k so the line \(\frac{{{\rm{x}} + 4}}{{2{\rm{}}}} = \frac{{4 - {\rm{y}}}}{-2} = \frac{{{\rm{2z}} - 4}}{{\rm 2k}}\) and \(\frac{{{\rm{x}} +3}}{\rm -k} = \frac{{\rm{y-3}}}{{\rm{2}}} = \frac{{{\rm{z}} + 1}}{5}\) are at right angles.
Answer (Detailed Solution Below)
Angle between two lines Question 10 Detailed Solution
Download Solution PDFConcept:
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
The angle between the lines \(\frac{{x - 4}}{2} = \frac{y}{1} = \frac{{z + 1}}{{ - 2}},\frac{{x - 1}}{4} = \frac{{y + 1}}{{ - 4}} = \frac{{z - 2}}{2} \) is:
Answer (Detailed Solution Below)
Angle between two lines Question 11 Detailed Solution
Download Solution PDFConcept:
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 = \(\frac{\pi}{2}\)
What is the angle between the two lines having direction ratios (6, 3, 6) and (3, 3, 0)?
Answer (Detailed Solution Below)
Angle between two lines Question 12 Detailed Solution
Download Solution PDFCONCEPT:
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.
Find the angle between the lines whose direction ratios are (1, 2, - 2) and (0, 3, -4) ?
Answer (Detailed Solution Below)
Angle between two lines Question 13 Detailed Solution
Download Solution PDFCONCEPT:
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: \(\cos {\bf{\theta }} = \;\frac{{{{\bf{a}}_1}{{\bf{a}}_2} + \;{{\bf{b}}_1}{{\bf{b}}_2} + {{\bf{c}}_1}{{\bf{c}}_2}\;}}{{\sqrt {{\bf{a}}_{1\;}^2 + \;{\bf{b}}_1^2 + \;{\bf{c}}_1^2} \sqrt {{\bf{a}}_{2\;}^2 + \;{\bf{b}}_2^2 + \;{\bf{c}}_2^2} }}\)
The angle between the vectors A = 3i + 5j - 4k and B = -5i + 11j + 10k is:
Answer (Detailed Solution Below)
Angle between two lines Question 14 Detailed Solution
Download Solution PDFConcept:
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 = \(\boldsymbol{\pi\over2}\)
If θ is the acute angle between the diagonals of a cube, then which one of the following is correct?
Answer (Detailed Solution Below)
Angle between two lines Question 15 Detailed Solution
Download Solution PDFConcept:
\(\rm\vec a.\vec b = |\vec a||\vec b|\cos \theta\)
\(\rm\text{If A = (x, y, z) and B = (x', y', z') then }\\ \vec {AB} = (x' - x)\hat i + (y' - y)\hat j+(z' - z)\hat k\\ |\vec {AB}|=\sqrt{(x' - x)^2 + (y' - y)^2+(z' - z)^2} \)
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.