Skew Lines MCQ Quiz in தமிழ் - Objective Question with Answer for Skew Lines - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

The shortest distance between two skew lines \(\frac{{x - x_1}}{l_1} = \frac{{y - y_1}}{m_1} = \frac{{z - z_1}}{n_1}\) and  \(\frac{{x - x_2}}{l_2} = \frac{{y - y_2}}{m_2} = \frac{{z - z_2}}{n_2}\)  is given by:

d = \(\frac{\begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\l_1&m_1&n_1\\l_2&m_2&n_2\end{vmatrix}}{\sqrt{(m_1n_2-m_2n_1)^2+(l_1n_2-n_1l_2)^2+(l_1m_2-m_1l_2)^2}}\)

Explanation:

Given,  \({L_1}:\frac{{x - a}}{l} = \frac{{y - 2}}{3} = \frac{{z - b}}{4}\) ---(1)

 and \({L_2}:\frac{{x - 2}}{3} = \frac{{y - 4}}{4} = \frac{{z - 5}}{5}\) ---(2)

The foot of the perpendicular from the point (4,3,8) on the line L₁ is (3,5,7).

So, point (3,5,7) satisfies the equation of the line L₁.

Substitute x=3, y=5, and z=7 in equation (1), we have

⇒ \(\frac{{3 - a}}{l} = \frac{{5 - 2}}{3} = \frac{{7 - b}}{4}\)

⇒ \(\frac{{3 - a}}{l} = \frac{{3}}{3} = \frac{{7 - b}}{4}\)

⇒\(\frac{{3 - a}}{l} = 1= \frac{{7 - b}}{4}\)

Now, \(\frac{3-a}{l}\) = 1

⇒ 3-a= l

⇒ a+l=3  ---(3)

And \( \frac{{7 - b}}{4}\) =1

⇒ 7-b =4

⇒ b=3

Direction cosines of the line joining points (4,3,8) and (3,5,7) are (3-4,5-3,7-8) i.e.(-1,2, -1).

The direction cosines of the line L₁ are (l,3,4).

Both lines are perpendicular.

∴ l× (-1)+2× 3+(-1)× 4 =0

⇒ -l +6-4=0

⇒-l+2=0

⇒ l=2

Substitute the value of l in equation (3), we have

⇒ a+2=3

⇒ a =1

Substitute the values of a,b and l in equation (1),

\({L_1}:\frac{{x - 1}}{2} = \frac{{y - 2}}{3} = \frac{{z - 3}}{4}\)

and  \({L_2}:\frac{{x - 2}}{3} = \frac{{y - 4}}{4} = \frac{{z - 5}}{5}\)

Direction cosines of the line L₁ and L₂ are (2,3,4) and (3,4,5) respectively.

Now, the shortest distance between line L₁ and L₂ is

d = \(\frac{\begin{vmatrix}2-1&4-2&5-3\\2&3&4\\3&4&5\end{vmatrix}}{\sqrt{(15-16)^2+(10-12)^2+(8-9)^2}}\)

=\(\frac{\begin{vmatrix}1&2&2\\2&3&4\\3&4&5\end{vmatrix}}{\sqrt{(15-16)^2+(10-12)^2+(8-9)^2}}\)

=\(\frac{1(15-16)-2(10-12)+2(8-9)}{\sqrt{(-1)^2+(-2)^2+(-1)^2}}\)

= \(\frac{-1+4-2}{\sqrt{1+4+1}}\)

=\(\frac{1}{\sqrt 6}\)

The shortest distance between lines L₁ and L₂ is \(\frac{1}{\sqrt 6}\).

Hence, the correct answer is option (1).

Concept:

Now, the shortest distance between two lines is given by 

\(= \left| {\frac{{\left( {{{\vec b}_1} \times {{\vec b}_2}} \right).\left( {{{\vec a}_2} - {{\vec a}_1}} \right)}}{{\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|}}} \right|\)

Calculation:

Given:  \( {{\vec b}_1} \times {{\vec b}_2} \) = 24î + 36ĵ + 72k̂ with its magnitude as 84

\(\Rightarrow | {{\vec b}_1} \times {{\vec b}_2} | = 84\)

\({\vec r}\) = 8î - 9ĵ + 10k̂ + λ (3î - 16ĵ + 7k̂)

⇒ \({{\vec a}_1}\)= 8î - 9ĵ + 10k̂ and \({{\vec b}_1}\)= (3î - 16ĵ + 7k̂) ...(i)

Also, \({\vec r}\)​ ​= 15î + 29ĵ + 5k̂ + μ(3î + 8ĵ - 5k)

\({{\vec a}_2}\) = 15î + 29ĵ + 5k̂ and \({{\vec b}_2}\) = (3î + 8ĵ - 5k̂) ....(ii)

Now, (\({{{\vec a}_2}}\) - \({{{\vec a}_1}}\) ) = (15 - 8)î + (29 + 9)ĵ + (5 - 10)k̂ 

= 7î + 38ĵ - 5k̂ 

∴ Shortest distance 

\( = \left| {\frac{{\left( {24\hat i + 36\hat j + 72\hat k} \right).\left( {7\hat i + 38\hat j - 5\hat k} \right)}}{{84}}} \right|\)

\( = \left| {\frac{{\left( {2\hat i + 3\hat j + 6\hat k} \right).\left( {7\hat i + 38\hat j - 5\hat k} \right)}}{7}} \right|\)

\(= \left| {\frac{{14 + 114 - 30}}{7}} \right| = 14\)

Hence, option 2 is the correct answer.

Concept:

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

\(SD = \frac{{\left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right|}}{{\sqrt {\left\{ {{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2} + {{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2}} \right\}} }}\)

Calculation:

Given: Equation of lines is \(\frac{{x + 1}}{2} = \frac{{y - 1}}{1} = \frac{{z - 9}}{{ - 3}}\;\;and\;\frac{{x - 3}}{2} = \frac{{y + 15}}{{ - 7}} = \frac{{z - 9}}{5}\)

By comparing the given equations with \(\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\;and\frac{{x - {x_2}}}{{{a_2}}} = \frac{{y - {y_2}}}{{{b_2}}} = \frac{{z - {z_2}}}{{{c_2}}}\), we get

⇒ x1 = -1, y1 = 1, z1 = 9, a1 = 2, b1 = 1 and c1 = - 3

Similarly, x2 = 3, y2 = -15, z2 = 9, a2 = 2, b2 = -7 and c2 = 5

So, \(\left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 4&{-16}&0\\ 2&1&{ - 3}\\ 2&{ - 7}&5 \end{array}} \right|\)

\(\left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = 192\)

And \({{\sqrt {\left\{ {{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2} + {{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2}} \right\}} }}=\sqrt{768}\)

As we know that shortest distance between two skew lines is given by:\(SD = \frac{{\left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right|}}{{\sqrt {\left\{ {{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2} + {{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2}} \right\}} }}\)

\(SD=\frac{192}{\sqrt{768}}=4 \sqrt3\)

⇒ \(SD = 4\sqrt{3} \ units\)

Hence, option D is the correct answer.

Calculation

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

Lines passed through the points

\(\vec{a}_{1}=(3,2,1) \text { and } \vec{a}_{2}=(-3,6,5)\),

\(\vec{b}_{1}=2 \hat{i}+3 \hat{j}-\hat{k}\)

\(\vec{b}_{1}=2 \hat{i}+\hat{j}-3 k, \vec{a}_{2}-\vec{a}_{1}=6 \hat{i}-4 f-4 \hat{k}\)

Shortest distance = \(\frac{\left|\left(\vec{a}_{2}-\vec{a}_{1}\right)\left(\vec{b}_{1} \times \vec{b}_{2}\right)\right|}{\left|\left(\vec{b}_{1} \times \vec{b}_{2}\right)\right|}\)

\(\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 2 & 1 & 3 \end{array}\right|=10 \hat{i}-8 \hat{j}-4 \hat{k}\)

\(\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)=60+32+16=108\)

\(\left|\vec{b}_{1} \times \vec{b}_{2}\right|=\sqrt{100+64+16}=\sqrt{180}\)

\(S . D=\frac{108}{\sqrt{180}}=\frac{108}{6 \sqrt{5}}=\frac{18}{\sqrt{5}}\)

Hence option 1 is correct

Concept:

The shortest distance between the 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:\(d= \frac{\begin{vmatrix} x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ a_{1}& b_{1} & c_{1}\\ a_{2}& b_{2} & c_{2} \end{vmatrix}}{\sqrt{(b_{1}c_{2}-b_{2}c_{1})^{2}+(c_{1}a_{2}-c_{2}a_{1})^{2}+(a_{1}b_{2}-a_{2}b_{1})^{2}}}\)

Calculation:

Here we have to find the shortest distance between the lines ​​\(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\) and \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\)

Let line L1 be represented by the equation \(\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}\) and line L2 be represented by the equation \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\)

⇒ x1 = 5, y1 = -2, z1 = 0  and a1 = 7, b1 = -5, c1 = 1.

⇒ x2 = 0, y2 = 0, z2 = 0  and a2 = 1, b2 = 2, c2 = 3.

∵ The shortest distance between the lines is given by:  \(d= \frac{\begin{vmatrix} x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ a_{1}& b_{1} & c_{1}\\ a_{2}& b_{2} & c_{2} \end{vmatrix}}{\sqrt{(b_{1}c_{2}-b_{2}c_{1})^{2}+(c_{1}a_{2}-c_{2}a_{1})^{2}+(a_{1}b_{2}-a_{2}b_{1})^{2}}}\)

⇒ \(d= \frac{\begin{vmatrix} 0-5 &0+2&0-0 \\ 7& -5 & 1\\ 1& 2 & 3 \end{vmatrix}}{\sqrt{(-15-2)^{2}+(1-21)^{2}+(14+5)^{2}}}\)

⇒ \(d= \frac{\begin{vmatrix} -5 &2&0 \\ 7& -5 & 1\\ 1& 2 & 3 \end{vmatrix}}{\sqrt{(-17)^{2}+(-20)^{2}+(19)^{2}}}\)

⇒ \(d= \frac{-5(-15-2)-2(21-1)}{\sqrt{289+400+361}}\)

⇒ \(d= \frac{85-40}{\sqrt{1050}} = \frac{9}{\sqrt {42}}\)

Hence, option 3 is correct.

Concept:

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

\(SD = \frac{{\left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right|}}{{\sqrt {\left\{ {{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2} + {{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2}} \right\}} }}\)

Calculation:

Given: Equation of lines is \(\frac{{x - 3}}{{ - 1}} = \frac{{y - 4}}{2} = \frac{{z + 2}}{1}\;\;and\;\frac{{x - 1}}{1} = \frac{{y + 7}}{3} = \frac{{z + 2}}{2}\)

By comparing the given equations with \(\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\;and\frac{{x - {x_2}}}{{{a_2}}} = \frac{{y - {y_2}}}{{{b_2}}} = \frac{{z - {z_2}}}{{{c_2}}}\), we get

⇒ x1 = 3, y1 = 4, z1 = - 2, a1 = -1, b1 = 2 and c1 = 1

Similarly, x2 = 1, y2 = - 7, z2 = -2, a2 = 1, b2 = 3 and c2 = 2

So, \(\left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} { - 2}&{ - 11}&0\\ { - 1}&2&1\\ 1&3&2 \end{array}} \right|\)

As we know that shortest distance between two skew lines is given by:\(SD = \frac{{\left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right|}}{{\sqrt {\left\{ {{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2} + {{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2}} \right\}} }}\)

⇒ \(SD = \sqrt{35} \ units\)

Hence, option C is the correct answer.

Calculation:

Distance between skew lines \(\vec{r}=\vec{a_1}+\lambda\vec{b_1}\) and \(\vec{r}=\vec{a_2}+\mu\vec{b_2}\) is given by:

d = \(\rm \frac{[(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1 \times\vec{b}_2)]}{|\vec{b}_1 \times \vec{b}_2|}\)

Calculation:

Given, (x – λ)/1 = (y – 1/2)/(1/2) = z/(-1/2)

(x – λ)/2 = (y-1/2)/1 = z/(-1) …(1) Point on line = (λ, 1/2, 0)

x/1 = (y + 2λ)/1 = (z – λ)/1 …(2) Point on line = (0, -2λ, λ)

∴ Distance between skew lines = \(\rm \frac{[(\vec{a}_2-\vec{a}_1)\cdot(\vec{b}_1 \times\vec{b}_2)]}{|\vec{b}_1 \times \vec{b}_2|}\)

= \(\frac{\left|\begin{array}{ccc} \lambda & \frac{1}{2}+2 \lambda & -\lambda \\ 2 & 1 & -1 \\ 1 & 1 & 1 \end{array}\right|}{\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 1 & -1 \\ 1 & 1 & 1 \end{array}\right|}\)

= |-5λ – 3/2|/\(\sqrt{14}\)

= √7/(2√2) (Given)

⇒ |10λ + 3| = 7

⇒ 10λ + 3 = ± 7

⇒ λ = - 1 [∵ λ is an integer]

⇒ |λ| = 1

∴ The value of |λ| is 1. 

Concept:

The shortest distance between the lines \(\frac{x-x_0}{a_0}=\frac{y-y_0}{b_0}=\frac{z-z_0}{c_0} \) and \(\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1} \) is given by d = \(\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|\)

Calculation:

Given, \(\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5} \) and \(\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}\)

∴ a₁ = \(3\hat{i} -15\hat{j}+ 9\hat{k}\) and b₁ = \(2\hat{i} -7\hat{j}+5\hat{k}\)

a₂ = \(-1\hat{i}+\hat{j}+9\hat{k}\) and b₂ = \(2\hat{i}+\hat{j}-3\hat{k}\)

⇒ a₂ – a₁ = \(-4\hat{i}+16\hat{j}+0\hat{k}\)

∴ \(\overline{\mathrm{b}}_1 \times \overline{\mathrm{b}}_2=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{array}\right|=\hat{\mathrm{i}}(16)-\hat{\mathrm{j}}(-16)+\hat{\mathrm{k}}(16)\)

= \(16(\hat{i}+\hat{j}+\hat{k})\)

⇒ \(\left|\bar{b}_1 \times \bar{b}_2\right|=16 \sqrt{3}\)

∴ \(\left(\overline{\mathrm{a}}_2-\overline{\mathrm{a}}_1\right) \cdot\left(\overline{\mathrm{b}}_1-\overline{\mathrm{b}}_2\right)\) = 16[-4 + 16] = (16)(12)

⇒ d = \(\frac{(16)(12)}{16 \sqrt{3}}\) = 4√3

∴ The shortest distance is 4√3.

The correct answer is Option 2.

Calculation

\(L_2=\frac{x+4}{3}=\frac{y-4}{2}=\frac{z-3}{0}\)

\(\therefore \text { S.D }=\frac{\left|\begin{array}{ccc} \mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{array}\right|}{\left|\overrightarrow{\mathrm{n}_1} \times \overrightarrow{\mathrm{n}_2}\right|}\)

⇒ \(\frac{\left|\begin{array}{ccc} 5 & -5 & -7 \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{array}\right|}{\left|\overrightarrow{n_1} \times \overrightarrow{n_2}\right|}\)

⇒ \(\frac{141}{|-4 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+13 \hat{k}|}\)

⇒ \(\frac{141}{\sqrt{16+36+169}}\)

⇒ \(\frac{141}{\sqrt{221}}\)

Hence, Option (3) is correct

Calculation

Given

\(\mathrm{L}_1: \frac{\mathrm{x}+1}{1}=\frac{\mathrm{y}}{1 / 2}=\frac{\mathrm{z}}{-1 / 12}, \mathrm{~L}_2: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}+2}{1}=\frac{\mathrm{z}-1}{\frac{1}{6}}\)

d₁ = shortest distance between L₁ & L₂ 

⇒ d₁= \(\left|\frac{\left(\overrightarrow{\mathrm{a}}_2-\overrightarrow{\mathrm{a}}_1\right) \cdot\left(\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2\right)}{\left|\left(\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2\right)\right|}\right|\)

⇒ d₁ = 2

\(\mathrm{L}_3: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+8}{-7}=\frac{\mathrm{z}-4}{5}, \mathrm{~L}_4: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-6}{-3}\)

d₂ = shortest distance between L₃ & L₄

⇒ \(\mathrm{d}_2=\frac{12}{\sqrt{3}}\) 

Hence

\(\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}=\frac{32 \sqrt{3} \times 2}{\frac{12}{\sqrt{3}}}\) = 16