Skew Lines MCQ Quiz - Objective Question with Answer for Skew Lines - Download Free PDF

Calculation

\(\vec{a}=(2,1,-3)\)

\(\overrightarrow{\mathrm{b}}=(-1,-3,-5)\)

\(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{array}\right|\)

= \(2 \hat{i}-\hat{\mathrm{j}}\)

\(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}=-3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)

\(S_{d}=\frac{|(\vec{b}-\vec{a}) \cdot(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})|}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}\)

= \(\frac{2}{\sqrt{5}}\)

\(\left(\mathrm{S}_{\mathrm{d}}\right)^{2}=\frac{4}{5}\)

m = 4, n = 5 ⇒ m + n = 9 

Hence option 2 is correct

Calculation

P(2λ + 1, 3λ + 2, 4λ + 3) on L₁

Q(3µ + 2, 4µ + 4, 5µ + 5) on L₂

Dr’s of PQ = 3µ – 2λ + 1, 4µ – 3λ + 2, 5µ – 4λ + 2

PQ ⊥ L₁ 

⇒ (3µ – 2λ + 1)² + (4µ – 3λ + 2)3 + (5µ – 4λ + 2)4 = 0

38µ – 29λ + 16 = 0 …(1)

PQ ⊥ L₂ 

⇒ (3µ – 2λ + 1)3 + (4µ – 3λ + 2)4 + (5µ – 4λ + 2)5 = 0

50µ – 38λ + 21 = 0 …(2)

By (1) & (2) 

\(λ=\frac{1}{3} ; \mu=\frac{-1}{6}\)

∴ \(\mathrm{P}\left(\frac{5}{3}, 3, \frac{13}{3}\right) \& \ \mathrm{Q}\left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right)\)

Line PQ 

\(\begin{array}{ccc} \frac{x-\frac{5}{3}}{\frac{1}{6}} & \frac{y-3}{\frac{-1}{3}} & \frac{z-\frac{13}{3}}{\frac{1}{6}} \end{array}\)

\(\frac{x-\frac{5}{3}}{1}=\frac{y-3}{-2}=\frac{z-\frac{13}{3}}{1}\)

\(\text { Point }\left(\frac{14}{3},-3, \frac{22}{3}\right)\)

lies on the line PQ 

Hence option 4 is correct

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

Concept Used:

The shortest distance between two skew lines is given by:

\(d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}\)

where \(\vec{a_1}\) and \(\vec{a_2}\) are points on the two lines, and \(\vec{b_1}\) and \(\vec{b_2}\) are the direction vectors of the two lines.

Calculation:

Given:

Skew lines:

\(\vec{r} = (-\hat{i} - 2\hat{j} - 3\hat{k}) + t(3\hat{i} - 2\hat{j} - 2\hat{k})\)

\(\vec{r} = (7\hat{i} + 4\hat{k}) + s(\hat{i} - 2\hat{j} + 2\hat{k})\)

Here, \(\vec{a_1} = -\hat{i} - 2\hat{j} - 3\hat{k}\), \(\vec{b_1} = 3\hat{i} - 2\hat{j} - 2\hat{k}\)

and \(\vec{a_2} = 7\hat{i} + 4\hat{k}\), \(\vec{b_2} = \hat{i} - 2\hat{j} + 2\hat{k}\)

Then \(\vec{a_2} - \vec{a_1} = (7 - (-1))\hat{i} + (0 - (-2))\hat{j} + (4 - (-3))\hat{k} = 8\hat{i} + 2\hat{j} + 7\hat{k}\)

Now, \(\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & -2 \\ 1 & -2 & 2 \end{vmatrix} = (-4 - 4)\hat{i} - (6 + 2)\hat{j} + (-6 + 2)\hat{k} = -8\hat{i} - 8\hat{j} - 4\hat{k}\)

So, \(|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})| = |(8\hat{i} + 2\hat{j} + 7\hat{k}) \cdot (-8\hat{i} - 8\hat{j} - 4\hat{k})| = |-64 - 16 - 28| = |-108| = 108\)

And \(|\vec{b_1} \times \vec{b_2}| = \sqrt{(-8)^2 + (-8)^2 + (-4)^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12\)

Therefore, \(d = \frac{108}{12} = 9\)

Hence option 3 is correct

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 magnitude of the shortest distance between the lines \( \vec{r_1} = \vec a_1 + \lambda \vec b_1\) and \(\vec{r_2} = \vec a_2 + \mu\vec b_2\) is 

\(d = \left | \frac{(\vec b_1\times\vec b_2).(\vec a_2 - \vec a_1)}{|\vec b_1\times\vec b_2|} \right|\)

Given:  

The lines \(\frac{x-0}{2} = \frac{y-0}{-3}=\frac {z-0}{1} \) and \(\frac{x-2}{3} = \frac{y-1}{-5}=\frac {z+2}{2} \).

Rewriting the given equations,

\( \vec{r_1} = \lambda(2\vec i-3\vec j+\vec k) \) and \( \vec{r_2} = (2\vec i+\vec j-2\vec k) + \mu(3\vec i-5\vec j+2\vec k) \)

⇒ \(\vec a_1=0\) ,  \(\vec b_1=2\vec i-3\vec j+\vec k\) and  \(\vec a_2=2\vec i+\vec j-2\vec k\),  \(\vec b_2=3\vec i-5\vec j+2\vec k\)

Therefore, the magnitude of the shortest distance between the given lines is

\(d = \left | \frac{(\vec b_1\times\vec b_2).(\vec a_2 - \vec a_1)}{|\vec b_1\times\vec b_2|} \right|\)

\(d = \left | \frac{[(2\vec i-3\vec j+\vec k)\times(3\vec i-5\vec j+2\vec k)].[(2\vec i+\vec j-2\vec k)-0]}{|(2\vec i-3\vec j+\vec k)\times(3\vec i-5\vec j+2\vec k)|} \right|\)

\(d = \left | \frac{(-\vec i-\vec j-\vec k).(2\vec i+\vec j-2\vec k)]}{|-\vec i-\vec j-\vec k|} \right|\)

\(d = \frac{1}{\sqrt{3}}\)

Therefore, the magnitude of the shortest distance between the given lines is \(\frac{1}{\sqrt3}\).

Concept:

• If two lines are parallel, then the distance between them is fixed.
• The distance between two parallel lines \(\rm \vec{r}=\vec{a}_1 +\lambda \vec{b}\) and \(\rm r=\vec{a}_2 + \mu\vec{b}\) is given by the formula: \(\rm d=\left|\dfrac{\vec{b}\times (\vec{a}_2-\vec{a}_1)}{|\vec{b}|}\right|\).

Calculation:

Using the formula for the distance between two parallel lines, we can say that the distance is \(\rm d=\left|\dfrac{\vec{b}\times (\vec{a}_2-\vec{a}_1)}{|\vec{b}|}\right|\).

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 \(\frac{{x - 8}}{3} = \frac{{y + 9}}{{ - 16}} = \frac{{z - 10}}{7}\;\;and\;\frac{{x - 15}}{3} = \frac{{y - 29}}{8} = \frac{{z - 5}}{{ - 5}}\)

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 = 8, y1 = - 9, z1 = 10, a1 = 3, b1 = -16 and c1 = 7

Similarly, x2 = 15, y2 = 29, z2 = 5, a2 = 3, b2 = 8 and c2 = -5

So, \(\left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 7&{38}&{ - 5}\\ 3&{ - 16}&7\\ 3&8&{ - 5} \end{array}} \right|\)

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 = 14 units

Hence, option B is the correct answer.

Concept:

The shortest distance between parallel lines \(\vec{r}= \vec{a_{1}}+ \lambda \vec{b} \) and \(\vec{r}= \vec{a_{2}}+ \mu \vec{b} \) is given by: \(d= \left | \frac{\vec{b}\times (\vec{a_{2}}-\vec{a_{1}})}{|b|} \right | \)

Calculation:

L₁: \(\vec{r}=(3s+2) \hat{i}-3\hat{j}+(4s+4)\hat{k}\) can be written as \(\vec{r}=2 \hat{i}-3\hat{j}+4\hat{k}+s(3 \hat{i}+4\hat{k})\).

L₂: \(\vec{r}=(3t+2) \hat{i}-3\hat{j}+(4t)\hat{k}\) can be written as \(\vec{r}=2 \hat{i}-3\hat{j}+t(3 \hat{i}+4\hat{k})\).

Here, we see both lines are parallel and \(\vec{a_{1}}=2\hat{i}-3\hat{j}+4\hat{k}\) , \(\vec{a_{2}}= 2\hat{i}-3\hat{j}\) and \(\vec{b}= 3\hat{i}+4\hat{k}\).

\(\therefore \) The shortest distance between parallel lines L₁ and L₂: 

\(d= \left | \frac{(3\hat{i}+4\hat{k})\times [(2\hat{i}-3\hat{j})-(2\hat{i}-3\hat{j}+4\hat{k})]}{|3\hat{i}+4\hat{k}|} \right | \)

⇒ \(d= \left | \frac{(3\hat{i}+4\hat{k})\times (-4\hat{k})}{\sqrt{9+16}} \right | \)

⇒ \(d= \left | \frac{12\hat{j}}{5} \right | \) ⇒ \(d= \frac{12}{5} =2.4\) unit.

Hence, option 1 is correct.

Concept:

The shortest distance between parallel lines \(\vec{r}= \vec{a_{1}}+ \lambda \vec{b} \) and \(\vec{r}= \vec{a_{2}}+ \mu \vec{b} \) is given by: \(d= \left | \frac{\vec{b}\times (\vec{a_{2}}-\vec{a_{1}})}{|b|} \right | \)

Calculation:

L₁: \(\vec{r}=(3s+2) \hat{i}-3\hat{j}+(4s+4)\hat{k}\) can be written as \(\vec{r}=2 \hat{i}-3\hat{j}+4\hat{k}+s(3 \hat{i}+4\hat{k})\).

L₂: \(\vec{r}=(3t+2) \hat{i}-3\hat{j}+(4t)\hat{k}\) can be written as \(\vec{r}=2 \hat{i}-3\hat{j}+t(3 \hat{i}+4\hat{k})\).

Here, we see both lines are parallel and \(\vec{a_{1}}=2\hat{i}-3\hat{j}+4\hat{k}\) , \(\vec{a_{2}}= 2\hat{i}-3\hat{j}\) and \(\vec{b}= 3\hat{i}+4\hat{k}\).

\(\therefore \) The shortest distance between parallel lines L₁ and L₂: 

\(d= \left | \frac{(3\hat{i}+4\hat{k})\times [(2\hat{i}-3\hat{j})-(2\hat{i}-3\hat{j}+4\hat{k})]}{|3\hat{i}+4\hat{k}|} \right | \)

⇒ \(d= \left | \frac{(3\hat{i}+4\hat{k})\times (-4\hat{k})}{\sqrt{9+16}} \right | \)

⇒ \(d= \left | \frac{12\hat{j}}{5} \right | \) ⇒ \(d= \frac{12}{5} =2.4\) unit.

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 ​​\(\frac{x}{-1}=\frac{y-2}{0}=\frac{z}{1}\) and \(\frac{x+2}{1}=\frac{y}{1}=\frac{z}{0}\)

Let line L₁ be represented by the equation \(\frac{x}{-1}=\frac{y-2}{0}=\frac{z}{1}\) and line L₂ be represented by the equation \(\frac{x+2}{1}=\frac{y}{1}=\frac{z}{0}\)

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

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

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

⇒ \(d= \frac{\begin{vmatrix} -2-0 &0-2&0-0 \\ -1& 0 & 1\\ 1& 1 & 0 \end{vmatrix}}{\sqrt{(0-1)^{2}+(0-1)^{2}+(-1-0)^{2}}}\)    

⇒ \(d= \frac{\begin{vmatrix} -2 &-2&0 \\ -1& 0 & 1\\ 1& 1 & 0 \end{vmatrix}}{\sqrt{1+1+1}}\)

⇒ d = 0

Hence, option 4 is correct.

Concept:

The shortest distance between parallel lines \(\vec{r}= \vec{a_{1}}+ \lambda \vec{b} \) and \(\vec{r}= \vec{a_{2}}+ \mu \vec{b} \) is given by: \(d= \left | \frac{\vec{b}\times (\vec{a_{2}}-\vec{a_{1}})}{|b|} \right | \)

Calculation:

Given: Equation of lines \(\vec{r}= \hat{i}+\hat{2k}+ \lambda ( \hat{i}+2\hat{j}+3\hat{k} )\)  and \(\vec{r}= \hat{i}+2\hat{j}+\hat{2k}+ \lambda ( \hat{i}+2\hat{j}+3\hat{k} )\).

So, by comparing the above equations with \(\vec{r}= \vec{a_{1}}+ \lambda \vec{b} \) and \(\vec{r}= \vec{a_{2}}+ \mu \vec{b} \) we get

⇒ \(\vec{a_{1}}= \hat{i}+2\hat{k}\) , \(\vec{a_{2}}= \hat{i}+2\hat{j}+2\hat{k}\)  and \(\vec{b}= \hat{i}+2\hat{j}+3\hat{k}\).

\(\therefore \) The shortest distance between parallel lines \(\vec{r}= \vec{a_{1}}+ \lambda \vec{b} \) and \(\vec{r}= \vec{a_{2}}+ \mu \vec{b} \) is given by: \(d= \left | \frac{\vec{b}\times (\vec{a_{2}}-\vec{a_{1}})}{|b|} \right | \)

\(⇒ d= \left | \frac{(\hat{i}+2\hat{j}+3\hat{k})\times [(\hat{i}+2\hat{j}+2\hat{k})-(\hat{i}+2\hat{k})]}{|\hat{i}+2\hat{j}+3\hat{k}|} \right | \)

⇒ \(d= \left | \frac{(\hat{i}+2\hat{j}+3\hat{k})\times 2\hat{j}}{|\sqrt{1+4+9}|} \right | \)

⇒ \(d= \left | \frac{2\hat{k}-6\hat{i}}{\sqrt{14}} \right | \)

⇒ \(d = \sqrt \frac{{40}}{{14}}\)

\(\Rightarrow d = \sqrt \frac{{40}}{{14}} \)  \(\Rightarrow d = \sqrt \frac{{20}}{{7}} \)

⇒ k = 20

Hence, option 4 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 \(\frac{{x + 3}}{{ - \;4}} = \frac{{y - 6}}{3} = \frac{z}{2}\;and \ \frac{{x + 2}}{{ - \;4}} = \frac{y}{1} = \frac{{z - 7}}{1}\)

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 = 6, z₁ = 0, a₁ = - 4, b₁ = 3 and c₁ = 2

Similarly, x₂ = - 2, y₂ = 0, z₂ = 7, a₂ = - 4, b₂ = 1 and c₂ = 1

So, \(\left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 1&{ - \;6}&7\\ { - \;4}&3&2\\ { - \;4}&1&1 \end{array}} \right|\)

Similarly, \(\sqrt {\left\{ {{{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2} + {{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2}} \right\}} = \sqrt {81} = 9\)

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

\(\Rightarrow SD = \frac{{\left| {\begin{array}{*{20}{c}} 1&{ - \;6}&7\\ { - \;4}&3&2\\ { - \;4}&1&1 \end{array}} \right|}}{9} = \frac{{81}}{9} = 9\)

Concept -

Shortest distance between two lines is:

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

Explanation -

The given lines are :

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

So, \(\vec{b}_1=4 \hat{i}+5 \hat{j}+3 \hat{k}\)

\(\vec{b}_2=3 \hat{i}+4 \hat{j}+2 \hat{k}\)

\(\vec{a}_1=4 \hat{i}-2 \hat{j}-3 \hat{k}\), \( \vec{a}_2=\hat{i}+3 \hat{j}+4 \hat{k}\)

∴ \(\vec{b}_1 \times \vec{b}_2=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{array}\right|\)

= \((10-12) \hat{i}-(8-9) \hat{j}+(16-15) \hat{k}\)

= \(-2 \hat{i}+\hat{j}+\hat{k}\)

Shortest distance, \(d=\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|\)

= \(\left|\frac{(3 \hat{i}-5 \hat{j}-7 \hat{k}) \cdot(-2 \hat{i}+\hat{j}+\hat{k})}{\sqrt{4+1+1}}\right|\)

= \(\left|\frac{-6-5-7}{\sqrt{6}}\right|=\frac{18}{\sqrt{6}}=3 \sqrt{6}\) units

Hence Option (2) 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.