Advertisements
Advertisements
Question
The shortest distance between the lines \[\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} \text{ and }, \frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}\]
Options
\[\sqrt{30}\]
\[2\sqrt{30}\]
\[5\sqrt{30}\]
\[3\sqrt{30}\]
Advertisements
Solution
\[3\sqrt{30}\]
We have ,
\[\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} . . . (1) \]
\[\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4} . . . (2)\]
We know that line (1) passes through the point (3, 8, 3) and has direction ratios proportional to 3,-1 , 1 .
Its vector equation is ,
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} , \text{ where }\overrightarrow{a_1} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} \text{ and } \overrightarrow{b_1} = 3 \hat{i} - \hat{j} + \hat{k} .\]
Also, line (2) passes through the point ( -3 , -7 , 6 ) and has direction ratios proportional to -3, 2 4 .
Its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} , \text{ where } \overrightarrow{a_2} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} \text{ and } \overrightarrow{b_2} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k} .\]
Now,
\[\overrightarrow{a_2} - \overrightarrow{a_1} = - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \]
\[ \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & - 1 & 1 \\ - 3 & 2 & 4\end{vmatrix}\]
\[ = - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{\left( - 6 \right)^2 + \left( - 15 \right)^2 + 3^2}\]
\[ = \sqrt{36 + 225 + 9}\]
\[ = \sqrt{270}\]
\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \right) . \left( - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \right)\]
\[ = 36 + 225 + 9\]
\[ = 270\]
The shortest distance between the lines
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\] is given by
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ = \left| \frac{270}{\sqrt{270}} \right|\]
\[ = \sqrt{270}\]
\[ = 3\sqrt{30}\]
APPEARS IN
RELATED QUESTIONS
A line passes through (2, −1, 3) and is perpendicular to the lines `vecr=(hati+hatj-hatk)+lambda(2hati-2hatj+hatk) and vecr=(2hati-hatj-3hatk)+mu(hati+2hatj+2hatk)` . Obtain its equation in vector and Cartesian from.
Show that the line through the points (1, −1, 2) (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5).
Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).
Find the vector equation of the line passing through the points (−1, 0, 2) and (3, 4, 6).
Find the vector equation of a line passing through (2, −1, 1) and parallel to the line whose equations are \[\frac{x - 3}{2} = \frac{y + 1}{7} = \frac{z - 2}{- 3} .\]
The cartesian equations of a line are \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} .\] Find a vector equation for the line.
Find the cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line \[\frac{- x - 2}{1} = \frac{y + 3}{7} = \frac{2z - 6}{3} .\]
Show that the line through the points (1, −1, 2) and (3, 4, −2) is perpendicular to the through the points (0, 3, 2) and (3, 5, 6).
Find the equation of a line parallel to x-axis and passing through the origin.
Find the angle between the following pair of line:
\[\overrightarrow{r} = \lambda\left( \hat{i} + \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = 2 \hat{j} + \mu\left\{ \left( \sqrt{3} - 1 \right) \hat{i} - \left( \sqrt{3} + 1 \right) \hat{j} + 4 \hat{k} \right\}\]
Find the angle between the following pair of line:
\[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3} \text { and } \frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}\]
Find the angle between the following pair of line:
\[\frac{x - 5}{1} = \frac{2y + 6}{- 2} = \frac{z - 3}{1} \text{ and } \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}\]
Find the angle between the pairs of lines with direction ratios proportional to 5, −12, 13 and −3, 4, 5
Find the angle between the pairs of lines with direction ratios proportional to 2, 2, 1 and 4, 1, 8 .
Find the angle between the pairs of lines with direction ratios proportional to a, b, c and b − c, c − a, a − b.
Find the equations of the line passing through the point (−1, 2, 1) and parallel to the line \[\frac{2x - 1}{4} = \frac{3y + 5}{2} = \frac{2 - z}{3} .\]
Determine the equations of the line passing through the point (1, 2, −4) and perpendicular to the two lines \[\frac{x - 8}{8} = \frac{y + 9}{- 16} = \frac{z - 10}{7} \text{ and } \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}\]
Show that the lines \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} \text{ and } \frac{x}{1} = \frac{y}{2} = \frac{z}{3}\] are perpendicular to each other.
If the lines \[\frac{x - 1}{- 3} = \frac{y - 2}{2 \lambda} = \frac{z - 3}{2} \text{ and } \frac{x - 1}{3\lambda} = \frac{y - 1}{1} = \frac{z - 6}{- 5}\] are perpendicular, find the value of λ.
If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.
Find the foot of the perpendicular drawn from the point A (1, 0, 3) to the joint of the points B (4, 7, 1) and C (3, 5, 3).
Find the foot of perpendicular from the point (2, 3, 4) to the line \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .\] Also, find the perpendicular distance from the given point to the line.
Find the foot of the perpendicular drawn from the point \[\hat{i} + 6 \hat{j} + 3 \hat{k} \] to the line \[\overrightarrow{r} = \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) .\] Also, find the length of the perpendicular
Find the distance of the point (2, 4, −1) from the line \[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9}\]
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} + 4 \hat{j} + 5 \hat{k} \right) + \mu\left( 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \right)\]
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 8 + 3\lambda \right) \hat{i} - \left( 9 + 16\lambda \right) \hat{j} + \left( 10 + 7\lambda \right) \hat{k} \]\[\overrightarrow{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + \mu\left( 3 \hat{i} + 8 \hat{j} - 5 \hat{k} \right)\]
Find the shortest distance between the following pairs of lines whose cartesian equations are: \[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} and \frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 5}{5}\]
By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2\]
Write the coordinate axis to which the line \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 1}{0}\] is perpendicular.
The lines `x/1 = y/2 = z/3 and (x - 1)/-2 = (y - 2)/-4 = (z - 3)/-6` are
The perpendicular distance of the point P (1, 2, 3) from the line \[\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}\] is
If a line makes angles α, β and γ with the axes respectively, then cos 2 α + cos 2 β + cos 2 γ =
If the direction ratios of a line are proportional to 1, −3, 2, then its direction cosines are
Find the value of λ for which the following lines are perpendicular to each other:
`(x - 5)/(5 lambda + 2 ) = ( 2 - y )/5 = (1 - z ) /-1 ; x /1 = ( y + 1/2)/(2 lambda ) = ( z -1 ) / 3`
Choose correct alternatives:
The difference between the slopes of the lines represented by 3x2 - 4xy + y2 = 0 is 2
Find the position vector of a point A in space such that `vec"OA"` is inclined at 60º to OX and at 45° to OY and `|vec"OA"|` = 10 units.
Equation of a line passing through (1, 1, 1) and parallel to z-axis is ______.
