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

Calculation: 

\(\rm \vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k}) \quad \vec{r}=\vec{a}+\lambda \vec{p} \)

⇒ \(\rm \vec{r}=(+\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}-\hat{j}) \quad \vec{r}=\vec{b}+\mu \vec{q} \)

⇒ \(\rm \vec{p} × \vec{q}=\left|\begin{array}{ccc} \rm \hat{i} &\rm \hat{j} &\rm \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 0 \end{array}\right|=\hat{i}+2 \hat{j}-3 \hat{k} \)

⇒ \(\rm d=\left|\frac{(\vec{b}-\vec{a}) \cdot(\vec{p} × \vec{q})}{|\vec{p} × \vec{q}|}\right| \)

⇒ \(\rm d=\left|\frac{(-3 \hat{j}-\hat{k}) \cdot(\hat{i}+2 \hat{j}-3 \hat{k})}{\sqrt{14}}\right| \)

\(=\left|\frac{-6+3}{\sqrt{14}}\right|=\frac{3}{\sqrt{14}} \)

⇒ \(α=\frac{3}{\sqrt{14}}\)

Now, 28α² = 28× \(\frac{9}{14} = 18\)

Hence, the correct answer is 18. 

Calculation: 

\(\rm \frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{-1}{12}}\) and \(\rm \frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{\frac{1}{6}} \)

\(\rm \Rightarrow \text { Shortest distance }=\frac{(\vec{b}-\vec{a}) \cdot(\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \)

\(\rm \text { S.D. }=(-\hat{i}+2 \hat{j}-\hat{k}) \cdot \frac{(\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|}\)

\(\left\{\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}} \equiv\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & \frac{1}{2} & \frac{-1}{12} \\ 1 & 1 & \frac{1}{6} \end{array}\right|=\frac{1}{6} \hat{\mathrm{i}}-\frac{1}{4} \hat{\mathrm{j}}+\frac{1}{2} \hat{\mathrm{k}} \text { or } 2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}\right\}\)

\(\rm \text { S.D. }=\frac{(-\hat{i}+2 \hat{j}-\hat{k}) \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})}{\sqrt{2^{2}+3^{2}+6^{2}}}=\left|\frac{-14}{7}\right|=2\)

Hence, the correct answer is Option 1.

Calculation: 

\(\rm \frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5} \quad \vec{a}=\hat{i}-8 \hat{j}+4 \hat{k}\)

\(\rm \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3} \quad \vec{b}=\hat{i}+2 \hat{j}+6 \hat{k}\)

⇒ \(\overrightarrow{\mathrm{p}}=2 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}, \overrightarrow{\mathrm{q}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}}\)

⇒ \(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{array}\right|\)

= \(\hat{\mathrm{i}}(16)-\hat{\mathrm{j}}(-16)+\hat{\mathrm{k}}(16)\)

= \(16(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)

⇒ d = \(\left|\frac{(\mathrm{a}-\mathrm{b}) \cdot(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}\right|=\left|\frac{(-10 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot 16(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})}{16 \sqrt{3}}\right|\)

= \(\left|\frac{-12}{\sqrt{3}}\right|=4 \sqrt{3}\)

Hence, the correct answer is Option 2.

Calculation:

We have two Lines

⇒ \(L_1: \frac{x-1}{0} = \frac{y-2}{0} = \frac{z-3}{1}, \qquad\)

⇒  \(L_2: \frac{x-\lambda}{0} = \frac{y-5}{1} = \frac{z-6}{0}.\)

To find their shortest distance, form the determinant

⇒ \(\det \begin{pmatrix} \lambda-1 & 3 & 3\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix} = (\lambda-1)\det\!\begin{pmatrix}0&1\\1&0\end{pmatrix} -3\det\!\begin{pmatrix}0&1\\0&0\end{pmatrix} +3\det\!\begin{pmatrix}0&0\\0&1\end{pmatrix} \) 

\(= -(\lambda-1). \)

⇒ \(\mathbf{d}_1\times\mathbf{d}_2 = \det \begin{pmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 0&0&1\\ 0&1&0 \end{pmatrix} = -\,\mathbf{i}, \quad \bigl\|-\mathbf{i}\bigr\| = 1.\)

\(SD = \bigl|-(\lambda-1)\bigr| = |\lambda-1| = 3 \quad\Longrightarrow\quad \lambda = 1\pm 3 \quad\Longrightarrow\quad \lambda = 4 \text{ or } -2. \)

Foot of the perpendicular from P  to L₁

Take \(P(\lambda_1,\lambda_2,7) = (4,-2,7)\), A general point on L₁ ay be written as Q ( \((1,2,\,3+t).\)

Then, \(\overrightarrow{PQ} = (1-4,\;2-(-2),\;(3+t)-7) = (-3,\,4,\,t-4).\)

Perpendicularity to L₁ direction d₁ = (0, 0, 1) gives 

⇒ \(\overrightarrow{PQ}\cdot\mathbf{d}_1 = 0 \;\Longrightarrow\; (-3,4,t-4)\cdot(0,0,1)=0 \;\Longrightarrow\; t-4=0 \;\Longrightarrow\; t=4.\)

Thus, \(Q = (1,2,\,3+4) = (1,2,7).\)

⇒ \(PQ^2 = \|P-Q\|^2 = (4-1)^2 + (-2-2)^2 + (7-7)^2 = 3^2 + (-4)^2 + 0^2 \)

\(= 9 + 16 = 25. \)

Hence, the correct answer is Option 3.

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

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.