Advertisements
Advertisements
प्रश्न
Find the equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines
(1, 3, 0) and (0, 3, 0)
Advertisements
उत्तर
The equation of the line passing through the points (1, 3, 0) and (0, 3, 0) is
\[\frac{x - 1}{0 - 1} = \frac{y - 3}{3 - 3} = \frac{z - 0}{0 - 0}\]
\[ = \frac{x - 1}{- 1} = \frac{y - 3}{0} = \frac{z}{0}\]
Since the first line passes through the point (0, 0, 0) and has direction ratios proportional to 1, 0, 2, its vector equation is \[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} . . . (1) \]
\[\text{ Here }, \]
\[ \overrightarrow{a_1} = 0 \hat{i} + 0 \hat{j} + 0 \hat{k} \]
\[ \overrightarrow{b_1} = \hat{i} + 0 \hat{j}+ 2 \hat{k} \]
Also, the second line passes through the point (1, 3, 0) and has direction ratios proportional to -1, 0 , 0 .
Its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} . . . (2) \]
\[\text{ Here }, \]
\[ \overrightarrow{a_2} = \hat{i} + 3 \hat{j} + 0 \hat{k} \]
\[ \overrightarrow{b_2} = - \hat{i} + 0 \hat{j} + 0 \hat{k} \]
Now,
\[\overrightarrow{a_2} - \overrightarrow{a_1} = \hat{i} + 3 \hat{j} + 0 \hat{k} \]
\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 2 \\ - 1 & 0 & 0\end{vmatrix} = 0 \hat{i} - 2 \hat{j} + 0 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{0^2 + \left( - 2 \right)^2 + 0^2}\]
\[ = \sqrt{0 + 4 + 0}\]
\[ = 2\]
\[\text{ and } \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( \hat{i} + 3 \hat{j} + 0 \hat{k} \right) . \left( 0 \hat{i} - 2 \hat{j} + 0 \hat{k} \right)\]
\[ = - 6\]
The shortest distance between the lines
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\]
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ = \left| \frac{- 6}{2} \right|\]
\[ = 3 \]
\[ \therefore d = \text{ 3 } units\]
APPEARS IN
संबंधित प्रश्न
Find the separate equations of the lines represented by the equation 3x2 – 10xy – 8y2 = 0.
The Cartestation equation of line is `(x-6)/2=(y+4)/7=(z-5)/3` find its vector equation.
If the Cartesian equations of a line are ` (3-x)/5=(y+4)/7=(2z-6)/4` , write the vector equation for the line.
Find the value of p, so that the lines `l_1:(1-x)/3=(7y-14)/p=(z-3)/2 and l_2=(7-7x)/3p=(y-5)/1=(6-z)/5 ` are perpendicular to each other. Also find the equations of a line passing through a point (3, 2, – 4) and parallel to line l1.
Let `A(bara)` and `B(barb)` be any two points in the space and `R(barr)` be a point on the line segment AB dividing it internally in the ratio m : n, then prove that `bar r=(mbarb+nbara)/(m+n)`. Hence find the position vector of R which divides the line segment joining the points A(1, –2, 1) and B(1, 4, –2) internally in the ratio 2 : 1.
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.
The Cartesian equation of a line is `(x-5)/3 = (y+4)/7 = (z-6)/2` Write its vector form.
Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).
Find the vector equation of a line which is parallel to the vector \[2 \hat{i} - \hat{j} + 3 \hat{k}\] and which passes through the point (5, −2, 4). Also, reduce it to cartesian form.
Find the angle between two lines, one of which has direction ratios 2, 2, 1 while the other one is obtained by joining the points (3, 1, 4) and (7, 2, 12).
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.
Find the perpendicular distance of the point (1, 0, 0) from the line \[\frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 10}{8}.\] Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular.
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 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 equation of the perpendicular drawn from the point P (2, 4, −1) to the line \[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9} .\] Also, write down the coordinates of the foot of the perpendicular from P.
Find the length of the perpendicular drawn from the point (5, 4, −1) to the line \[\overrightarrow{r} = \hat{i} + \lambda\left( 2 \hat{i} + 9 \hat{j} + 5 \hat{k} \right) .\]
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 foot of the perpendicular from (1, 2, −3) to the line \[\frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1} .\]
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 cartesian equations are : \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{3} = \frac{y - 2}{1}; z = 2\]
Find the shortest distance between the lines \[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + \hat{k} \right) + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) \text{ and } , \overrightarrow{r} = 2 \hat{i} - \hat{j} - \hat{k} + \mu\left( 2 \hat{i} + \hat{j} + 2 \hat{k} \right)\]
Find the shortest distance between the lines \[\overrightarrow{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + \lambda\left( \hat{i} - 3 \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} + \mu\left( 2 \hat{i} + 3 \hat{j} + \hat{k} \right)\]
Write the direction cosines of the line whose cartesian equations are 6x − 2 = 3y + 1 = 2z − 4.
Write the angle between the lines \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z - 2}{1} \text{ and } \frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{3} .\]
Write the vector equation of a line given by \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} .\]
Find the angle between the lines
\[\vec{r} = \left( 2 \hat{i} - 5 \hat{j} + \hat{k} \right) + \lambda\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\] and \[\vec{r} = 7 \hat{i} - 6 \hat{k} + \mu\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right)\]
Find the angle between the lines 2x=3y=-z and 6x =-y=-4z.
The direction ratios of the line perpendicular to the lines \[\frac{x - 7}{2} = \frac{y + 17}{- 3} = \frac{z - 6}{1} \text{ and }, \frac{x + 5}{1} = \frac{y + 3}{2} = \frac{z - 4}{- 2}\] are proportional to
The angle between the lines
The direction ratios of the line x − y + z − 5 = 0 = x − 3y − 6 are proportional to
Find the equation of a plane which passes through the point (3, 2, 0) and contains the line \[\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}\].
Find the value of p for which the following lines are perpendicular :
`(1-x)/3 = (2y-14)/(2p) = (z-3)/2 ; (1-x)/(3p) = (y-5)/1 = (6-z)/5`
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`
Auxillary equation of 2x2 + 3xy − 9y2 = 0 is ______
Find the separate equations of the lines given by x2 + 2xy tan α − y2 = 0
