Advertisements
Advertisements
प्रश्न
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 8 + 3\lambda \right) \hat{i} - \left( 9 + 16\lambda \right) \hat{j} + \left( 10 + 7\lambda \right) \hat{k} \]\[\overrightarrow{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + \mu\left( 3 \hat{i} + 8 \hat{j} - 5 \hat{k} \right)\]
Advertisements
उत्तर
The vector equations of the given lines can be re-written as
\[\overrightarrow{r} = 8 \hat{i} - 9 \hat{j} + 10 \hat{k} + \lambda\left( 3 \hat{i} - 16 \hat{j} + 7 \hat{k} \right)\] and \[\overrightarrow{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + \mu\left( 3 \hat{i} + 8 \hat{j} - 5 \hat{k} \right)\]
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 ,
\[\overrightarrow{a_1} = 8 \hat{i} - 9 \hat{j} + 10 \hat{k}\]
\[\overrightarrow{b_1} = 3 \hat{i} - 16 \hat{j} + 7 \hat{k}\]
\[\overrightarrow{a_2} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k}\]
\[\overrightarrow{b_2} = 3 \hat{i} + 8 \hat{j} - 5 \hat{k}\]
\[\therefore \overrightarrow{a_2} - \overrightarrow{a_1} = \left( 15 \hat{i} + 29 \hat{j} + 5 \hat{k} \right) - \left( 8 \hat{i} - 9 \hat{j} + 10 \hat{k} \right) = 7 \hat{i} + 38 \hat{j} - 5 \hat{k} \]
\[\overrightarrow{b_1} \times \vec{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & - 16 & 7 \\ 3 & 8 & - 5\end{vmatrix} = 24 \hat{i} + 36 \hat{j} + 72 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{{24}^2 + {36}^2 + {72}^2} = \sqrt{576 + 1296 + 5184} = \sqrt{7056} = 84\]
Also,
\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)\]
\[ = \left( 7 \hat{i} + 38 \hat{j} - 5 \hat{k} \right) . \left( 24 \hat{i} + 36 \hat{j} + 72 \hat{k} \right)\]
\[ = 7 \times 24 + 38 \times 36 + \left( - 5 \right) \times 72\]
\[ = 168 + 1368 - 360\]
\[ = 1176\]
We know that 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
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
∴ Required shortest distance between the given pairs of lines,
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ = \left| \frac{1176}{84} \right|\]
\[ = 14\]
APPEARS IN
संबंधित प्रश्न
Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, – 1), (4, 3, – 1).
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.
Find in vector form as well as in cartesian form, the equation of the line passing through the points A (1, 2, −1) and B (2, 1, 1).
Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector \[\hat{i} - 2 \hat{j} + 3 \hat{k} .\] Reduce the corresponding equation in cartesian from.
The cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form.
The cartesian equation of a line are 3x + 1 = 6y − 2 = 1 − z. Find the fixed point through which it passes, its direction ratios and also its vector equation.
Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z.
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).
Find the cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by \[\frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6} .\]
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 pairs of lines with direction ratios proportional to 1, 2, −2 and −2, 2, 1 .
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 \[\frac{x - 8}{8} = \frac{y + 9}{- 16} = \frac{z - 10}{7} \text{ and } \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}\]
Find the value of λ so that the following lines are perpendicular to each other. \[\frac{x - 5}{5\lambda + 2} = \frac{2 - y}{5} = \frac{1 - z}{- 1}, \frac{x}{1} = \frac{2y + 1}{4\lambda} = \frac{1 - z}{- 3}\]
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.
Prove that the line \[\vec{r} = \left( \hat{i }+ \hat{j }- \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and } \vec{r} = \left( 4 \hat{i} - \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{k} \right)\] intersect and find their point of intersection.
Show that the lines \[\vec{r} = 3 \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \vec{r} = 5 \hat{i} - 2 \hat{j} + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\] are intersecting. Hence, find their point of intersection.
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 shortest distance between the following pairs of lines whose vector are: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} \right) + \lambda\left( 2 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } , \overrightarrow{r} = 2 \hat{i} + \hat{j} - \hat{k} + \mu\left( 3 \hat{i} - 5 \hat{j} + 2 \hat{k} \right)\]
By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} - \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and } \overrightarrow{r} = \left( 4 \hat{i} - \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{k} \right)\]
By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2\]
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 vector equations of the following lines and hence determine the distance between them \[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6} \text{ and } \frac{x - 3}{4} = \frac{y - 3}{6} = \frac{z + 5}{12}\]
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 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
If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k.
If 2x + y = 0 is one of the line represented by 3x2 + kxy + 2y2 = 0 then k = ______
Find the separate equations of the lines given by x2 + 2xy tan α − y2 = 0
Find the vector equation of the lines passing through the point having position vector `(-hati - hatj + 2hatk)` and parallel to the line `vecr = (hati + 2hatj + 3hatk) + λ(3hati + 2hatj + hatk)`.
The lines `(x - 1)/2 = (y + 1)/2 = (z - 1)/4` and `(x - 3)/1 = (y - k)/2 = z/1` intersect each other at point
