Advertisements
Advertisements
Question
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 1 - t \right) \hat{i} + \left( t - 2 \right) \hat{j} + \left( 3 - t \right) \hat{k} \text{ and } \overrightarrow{r} = \left( s + 1 \right) \hat{i} + \left( 2s - 1 \right) \hat{j} - \left( 2s + 1 \right) \hat{k} \]
Advertisements
Solution
\[\overrightarrow{r} = \left( 1 - t \right) \hat{i} + \left( t - 2 \right) \hat{j} + \left( 3 - t \right) \hat{k} \text{ and } \vec{r} = \left( s + 1 \right) \hat{i} + \left( 2s - 1 \right) \hat{j} - \left( 2s + 1 \right) \hat{k} \]
The vector equations of the given lines can be re-written as
\[\overrightarrow{r} = \hat{i} - 2 \hat{j} + 3 \hat{k} + t\left( - \hat{i} + \hat{j} - \hat{k}\right) \text{ and } \overrightarrow{r} = \hat{i} - \hat{j} - \hat{k} + s\left( \hat{i} + 2 \hat{j} - 2 \hat{k} \right)\]
Comparing the given equations with the equations \[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\]
we get ,
\[\overrightarrow{a_1} = \hat{i} - 2 \hat{j} + 3 \hat{k} \]
\[ \overrightarrow{a_2} = \hat{i} - \hat{j} - \hat{k} \]
\[ \overrightarrow{b_1} = - \hat{i} + \hat{j} - \hat{k} \]
\[ \overrightarrow{b_2} = \hat{i} + 2 \hat{j} - 2 \hat{k} \]
\[ \therefore \overrightarrow{a_2} - \overrightarrow{a_1} = \hat{j} - 4 \hat{k} \]
\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ - 1 & 1 & - 1 \\ 1 & 2 & - 2\end{vmatrix}\]
\[ = - 3 \hat{j} - 3 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{\left( - 3 \right)^2 + \left( - 3 \right)^2}\]
\[ = \sqrt{9 + 9}\]
\[ = 3\sqrt{2}\]
\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( \hat{j} - 4 \hat{k} \right) . \left( - 3 \hat{j} - 3 \hat{k} \right)\]
\[ = - 3 + 12\]
\[ = 9\]
The shortest distance between the line
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\] is given by
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ = \left| \frac{9}{3\sqrt{2}} \right|\]
\[ = \frac{3}{\sqrt{2}}\]
APPEARS IN
RELATED QUESTIONS
If a line drawn from the point A( 1, 2, 1) is perpendicular to the line joining P(1, 4, 6) and Q(5, 4, 4) then find the co-ordinates of the foot of the perpendicular.
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 three lines with direction cosines `12/13, (-3)/13, (-4)/13; 4/13, 12/13, 3/13; 3/13, (-4)/13, 12/13 ` are mutually perpendicular.
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector `3hati+2hatj-2hatk`.
Find the Cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by `(x+3)/3 = (y-4)/5 = (z+8)/6`.
The Cartesian equation of a line is `(x-5)/3 = (y+4)/7 = (z-6)/2` Write its vector form.
Find the equation of a line parallel to x-axis and passing through the origin.
Find the vector and Cartesian equations of a line passing through (1, 2, –4) and perpendicular to the two lines `(x - 8)/3 = (y + 19)/(-16) = (z - 10)/7` and `(x - 15)/3 = (y - 29)/8 = (z - 5)/(-5)`
Find the vector and cartesian equations of the line through the point (5, 2, −4) and which is parallel to the vector \[3 \hat{i} + 2 \hat{j} - 8 \hat{k} .\]
A line passes through the point with position vector \[2 \hat{i} - 3 \hat{j} + 4 \hat{k} \] and is in the direction of \[3 \hat{i} + 4 \hat{j} - 5 \hat{k} .\] Find equations of the line in vector and cartesian form.
The cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form.
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 three lines with direction cosines \[\frac{12}{13}, \frac{- 3}{13}, \frac{- 4}{13}; \frac{4}{13}, \frac{12}{13}, \frac{3}{13}; \frac{3}{13}, \frac{- 4}{13}, \frac{12}{13}\] are mutually perpendicular.
Find the angle between the following pair of line:
\[\overrightarrow{r} = \left( 3 \hat{i} + 2 \hat{j} - 4 \hat{k} \right) + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 5 \hat{j} - 2 \hat{k} \right) + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\]
Find the angle between the following pair of line:
\[\frac{5 - x}{- 2} = \frac{y + 3}{1} = \frac{1 - z}{3} \text{ and } \frac{x}{3} = \frac{1 - y}{- 2} = \frac{z + 5}{- 1}\]
Find the equation of the line passing through the point (1, −1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, −1, 0) and (1, 2, −1), (2, 1, 1).
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.
Show that the lines \[\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3} \text{ and } \frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}\] intersect and find their point of intersection.
Prove that the lines through A (0, −1, −1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (−4, 4, 4). Also, find their point of intersection.
Determine whether the following pair of lines intersect or not:
\[\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{- 5} and \frac{x - 8}{7} = \frac{y - 4}{1} = \frac{3 - 5}{3}\]
A (1, 0, 4), B (0, −11, 3), C (2, −3, 1) are three points and D is the foot of perpendicular from A on BC. Find the coordinates of D.
Find the equation of line passing through the points A (0, 6, −9) and B (−3, −6, 3). If D is the foot of perpendicular drawn from a point C (7, 4, −1) on the line AB, then find the coordinates of the point D and the equation of line CD.
Find the shortest distance between the following pairs of lines whose vector are: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} \right) + \lambda\left( 2 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } , \overrightarrow{r} = 2 \hat{i} + \hat{j} - \hat{k} + \mu\left( 3 \hat{i} - 5 \hat{j} + 2 \hat{k} \right)\]
Find the shortest distance between the following pairs of lines whose cartesian equations are : \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{3} = \frac{y - 2}{1}; z = 2\]
By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\frac{x - 5}{4} = \frac{y - 7}{- 5} = \frac{z + 3}{- 5} \text{ and } \frac{x - 8}{7} = \frac{y - 7}{1} = \frac{z - 5}{3}\]
Find the shortest distance between the lines \[\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} \text{ and } \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}\]
Write the cartesian and vector equations of X-axis.
Write the value of λ for which the lines \[\frac{x - 3}{- 3} = \frac{y + 2}{2\lambda} = \frac{z + 4}{2} \text{ and } \frac{x + 1}{3\lambda} = \frac{y - 2}{1} = \frac{z + 6}{- 5}\] are perpendicular to each other.
Find the angle between the lines 2x=3y=-z and 6x =-y=-4z.
If a line makes angles α, β and γ with the axes respectively, then cos 2 α + cos 2 β + cos 2 γ =
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`
If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k.
If the lines represented by kx2 − 3xy + 6y2 = 0 are perpendicular to each other, then
The equation 4x2 + 4xy + y2 = 0 represents two ______
If 2x + y = 0 is one of the line represented by 3x2 + kxy + 2y2 = 0 then k = ______
Find the cartesian equation of the line which passes ·through the point (– 2, 4, – 5) and parallel to the line given by.
`(x + 3)/3 = (y - 4)/5 = (z + 8)/6`
Find the vector equation of a line passing through a point with position vector `2hati - hatj + hatk` and parallel to the line joining the points `-hati + 4hatj + hatk` and `-hati + 2hatj + 2hatk`.
Which parametric form corresponds to a line through \[A(x_1,y_1,z_1)\] with direction ratios \[a,b,c\]?
For the point \[(5,2,-4)\], what is its position vector \[\vec{a}\]?
