Advertisements
Advertisements
प्रश्न
Find the shortest distance between the following pairs of lines whose cartesian equations are: \[\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} \text{ and } \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}\]
Advertisements
उत्तर
\[\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} . . . (1) \]
\[\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} . . . (2)\]
Since line (1) passes through the point (3, 5, 7) and has direction ratios proportional to 1, -2,1 its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \]
\[Here, \]
\[ \overrightarrow{a_1} = 3 \hat{i} + 5 \hat{j} + 7 \hat{k} \]
\[ \overrightarrow{b_1} = \hat{i} - 2 \hat{j} + \hat{k}\]
Also, line (2) passes through the point ( -1,-1,-1) and has direction ratios proportional to 7,-6,1 Its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} \]
\[Here, \]
\[ \overrightarrow{a_2} = - \hat{i} - \hat{j} - \hat{k} \]
\[ \overrightarrow{b_2} = 7 \hat{i} - 6 \hat{j} + \hat{k} \]
Now,
\[\overrightarrow{a_2} - \overrightarrow{a_1} = - 4 \hat{i} - 6 \hat{j} - 8 \hat{k} \]
\[\text{ and }\overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 1 \\ 7 & - 6 & 1\end{vmatrix}\]
\[ = 4 \hat{i} + 6 \hat{j} + 8 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{4^2 + 6^2 + 8^2}\]
\[ = \sqrt{16 + 36 + 64}\]
\[ = \sqrt{116}\]
\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \vec{b_1} \times \overrightarrow{b_2} \right) = \left( - 4 \hat{i} - 6 \hat{j} - 8 \hat{k} \right) . \left( 4 \hat{i} + 6 \hat{j} + 8 \hat{k } \right)\]
\[ = - 16 - 36 - 64\]
\[ = - 116\]
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
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ = \left| \frac{- 116}{\sqrt{116}} \right|\]
\[ = \sqrt{116}\]
\[ = 2\sqrt{29}\]
APPEARS IN
संबंधित प्रश्न
Find the Cartesian equation of the line which passes through the point (−2, 4, −5) and is parallel to the line `(x+3)/3=(4-y)/5=(z+8)/6`
Find the separate equations of the lines represented by the equation 3x2 – 10xy – 8y2 = 0.
Find the vector and cartesian equations of the line passing through the point (2, 1, 3) and perpendicular to the lines
`(x-1)/1=(y-2)/2=(z-3)/3 and x/(-3)=y/2=z/5`
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.
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).
Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).
Show that the lines `(x-5)/7 = (y + 2)/(-5) = z/1` and `x/1 = y/2 = z/3` are perpendicular to each other.
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 direction cosines of the line \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .\] Also, reduce it to vector form.
Show that the points whose position vectors are \[- 2 \hat{i} + 3 \hat{j} , \hat{i} + 2 \hat{j} + 3 \hat{k} \text{ and } 7 \text{ i} - \text{ k} \] are collinear.
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 (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (−1, −2, 1) and, (1, 2, 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.
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:
\[\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} \text{ and } \frac{x + 2}{- 1} = \frac{2y - 8}{4} = \frac{z - 5}{4}\]
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).
Find the equations of the line passing through the point (−1, 2, 1) and parallel to the line \[\frac{2x - 1}{4} = \frac{3y + 5}{2} = \frac{2 - z}{3} .\]
Find the equation of the line passing through the point (2, −1, 3) and parallel to the line \[\overrightarrow{r} = \left( \hat{i} - 2 \hat{j} + \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} - 5 \hat{k} \right) .\]
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 value of λ so that the following lines are perpendicular to each other. \[\frac{x - 5}{5\lambda + 2} = \frac{2 - y}{5} = \frac{1 - z}{- 1}, \frac{x}{1} = \frac{2y + 1}{4\lambda} = \frac{1 - z}{- 3}\]
Find the perpendicular distance of the point (3, −1, 11) from the line \[\frac{x}{2} = \frac{y - 2}{- 3} = \frac{z - 3}{4} .\]
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 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: \[\overrightarrow{r} = \left( 3 \hat{i} + 5 \hat{j} + 7 \hat{k} \right) + \lambda\left( \hat{i} - 2 \hat{j} + 7 \hat{k} \right) \text{ and } \overrightarrow{r} = - \hat{i} - \hat{j} - \hat{k} + \mu\left( 7 \hat{i} - 6 \hat{j} + \hat{k} \right)\]
Find the equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines
(i) (0, 0, 0) and (1, 0, 2)
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.
Write the condition for the lines \[\vec{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\] to be intersecting.
The cartesian equations of a line AB are \[\frac{2x - 1}{\sqrt{3}} = \frac{y + 2}{2} = \frac{z - 3}{3} .\] Find the direction cosines of a line parallel to AB.
Find the Cartesian equations of the line which passes through the point (−2, 4 , −5) and is parallel to the line \[\frac{x + 3}{3} = \frac{4 - y}{5} = \frac{z + 8}{6} .\]
Find the angle between the lines 2x=3y=-z and 6x =-y=-4z.
If the direction ratios of a line are proportional to 1, −3, 2, then its direction cosines are
The lines \[\frac{x}{1} = \frac{y}{2} = \frac{z}{3} \text { and } \frac{x - 1}{- 2} = \frac{y - 2}{- 4} = \frac{z - 3}{- 6}\]
The straight line \[\frac{x - 3}{3} = \frac{y - 2}{1} = \frac{z - 1}{0}\] is
If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k.
If slopes of lines represented by kx2 - 4xy + y2 = 0 differ by 2, then k = ______
