Shortest Distance MCQ Quiz in मराठी - Objective Question with Answer for Shortest Distance - मोफत PDF डाउनलोड करा

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

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 \(\frac{{x - 12}}{{ - 9}} = \frac{{y - 1}}{4} = \frac{{z - 5}}{2}\) and \(\frac{{x - 23}}{{ - 6}} = \frac{{y - 19}}{{ - 4}} = \frac{{z - 25}}{3}\)

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 = 12, y1 = 1, z1 = 5, a1 = -9, b1 = 4 and c1 = 2

Similarly, x2 = 23, y2 = 19, z2 = 25, a2 = -6, b2 = -4 and c2 = 3

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

\(= \left| {\begin{array}{*{20}{c}} {11}&{18}&{20}\\ { - 9}&4&2\\ { - 6}&{ - 4}&3 \end{array}} \right|\)

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 = 26 units

Hence, option A 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: The equation of lines is \(\frac{{x - 3}}{1} = \frac{{y - 5}}{{ - 2}} = \frac{{z - 7}}{1}\;\;and\;\frac{{x + 1}}{7} = \frac{{y + 1}}{{ - 6}} = \frac{{z + 1}}{1}\)

By comparing the given equations, we get

⇒ x1 = 3, y1 = 5, z1 = 7, a1 = 1, b1 = - 2 and c1 = 1

Similarly, x2 = - 1, y2 = -1, z2 = -1, a2 = 7, b2 = - 6 and c2 = 1

So, \(\left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} { - 4}&{ - 6}&{ - 8}\\ 1&{ - 2}&1\\ 7&{ - 6}&1 \end{array}} \right|\)

Similarly, 

\(\sqrt{(a_1b_2- a_2b_1)^2 +( b_1c_2 - b_2c_1)^2+ (c_1a_2 - c_2a_1)^2}\)

= 2√29

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 = \(2\sqrt{29}\)

Calculation

\(\dfrac{x-3}{1}=\dfrac{y+2}{-1}=\dfrac{z+\lambda}{-2}=k\)

Let P be any point on the line,

\(P=(k+3, -k-2, -2k-\lambda)\)

P lies on the plane \(2x-4y+3z=2\)

⇒ \(2(k+3)-4(-k-2)+3(-2k-\lambda)=2\)

⇒ \(2k+6+4k+8-6k-3\lambda =2\)

⇒ \(14-3\lambda =2\)

⇒ \(3\lambda =12\)

⇒ \(\lambda =4\)

\(\therefore\) Line \(1\) is \(\dfrac{x-3}{1}=\dfrac{y+2}{-1}=\dfrac{z+4}{-2}\)

Another line is \(\dfrac{x-1}{12}=\dfrac{y}{9}=\dfrac{z}{4}\)(line \(2\)) Shortest distance be d.

⇒ \(a^2=\begin{vmatrix} x_1-x_2 & y_1-y_2 & z_1-z_2\\\ l_1 & m_1 & n_1\\\ l_2 & m_2 & n_2\end{vmatrix}\) where \(l_1m_1n\) are DRS of lines

\(⇒ \begin{vmatrix} 3-1 & -2-0 & -4-0\ \\1 & -1 & -2 \ \\12 & 9 & 4\end{vmatrix}\)

⇒ \(\begin{vmatrix} 2 & -2 & -4 \\\ 1 & -1 & -2 \ \\12 & 9 & 4\end{vmatrix}=|2(14)+2(28)-4(21)|=|28+56-84|=0\)

\(\therefore d^2=0\)

\(\Rightarrow d=0\)

Hence option 1 is correct

Calculation

Given: Shortest distance(S.D) = 1

Passing points of lines L₁ & L₂ are

(λ, 2, 1) & (√3, 1, 2)

\(\text { S.D }=\frac{\left|\begin{array}{ccc} √{3}-\lambda & -1 & 1 \\ -2 & 1 & 1 \\ 1 & -2 & 1 \end{array}\right|}{\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 1 \\ 1 & -2 & 1 \end{array}\right|}\)

⇒ \(1=\left|\frac{√{3}-\lambda}{√{3}}\right|\)

⇒ λ = 0, λ = 2√3

∴ The sum of all possible values of λ is \(2\sqrt3\)

Explanation -

\(\left.\begin{array}{c} \bar{r}_1=(λ \hat{i}+4 \hat{j}+3 \hat{k})+\alpha(2 \hat{i}+3 \hat{j}+4 \hat{k}) \\ \bar{r}_2=(2 \hat{i}+4 \hat{j}+7 \hat{k})+\beta(2 \hat{i}+3 \hat{j}+4 \hat{k}) \end{array}\right\} \begin{aligned} & \overline{\mathrm{b}}=2 \hat{i}+3 \mathrm{j}+4 \hat{k} \\ & \bar{a}_1+λ \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & \mathrm{a}_2=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{k} \end{aligned}\)

Shortest dist. = \(\frac{\left|\overline{\mathrm{b}} \times\left(\overline{\mathrm{a}}_2-\overline{\mathrm{a}}_1\right)\right|}{|\mathrm{b}|}=\frac{13}{\sqrt{29}}\)

\(\frac{|(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times((2-λ) \hat{\mathrm{i}}+4 \hat{\mathrm{k}})|}{\sqrt{29}}=\frac{13}{\sqrt{29}}\)

\(|-8 \hat{\mathrm{j}}-3(2-λ) \hat{\mathrm{k}}+12 \hat{\mathrm{i}}+4(2-λ) \hat{\mathrm{j}}|=13\)

\(|12 \hat{i}-4 λ \hat{j}+(3 λ-6) \hat{k}|=13\)

144 + 16λ² + (3λ – 6)² = 169

16λ² + (3λ – 6)² = 25  ⇒ λ = 1 

Hence Option (3) is correct.