Advertisements
Advertisements
प्रश्न
Find the shortest distance between the lines \[\overrightarrow{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + \lambda\left( \hat{i} - 3 \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} + \mu\left( 2 \hat{i} + 3 \hat{j} + \hat{k} \right)\]
Advertisements
उत्तर
\[\overrightarrow{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + \lambda\left( \hat{i} - 3 \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} + \mu\left( 2 \hat{i} + 3 \hat{j} + \hat{k} \right)\]
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 ,
\[\overrightarrow{a_1} = \hat{i} + 2 \hat{j} + 3 \hat{k} \]
\[ \overrightarrow{a_2} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} \]
\[ \overrightarrow{b_1} = \hat{i} - 3 \hat{j} + 2 \hat{k} \]
\[ \vec{b_2} = 2 \hat{i} + 3 \hat{j} + \hat{k} \]
\[ \therefore \overrightarrow{a_2} - \overrightarrow{a_1} = 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \]
\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & - 3 & 2 \\ 2 & 3 & 1\end{vmatrix}\]
\[ = - 9 \hat{i} + 3 \hat{j} + 9 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{\left( - 9 \right)^2 + 3^2 + 9^2}\]
\[ = \sqrt{81 + 9 + 81}\]
\[ = \sqrt{171}\]
\[\text{ and } \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( 3 \hat{i} + 3 \hat{j} + 3 \hat{k} \right) . \left( - 9 \hat{i} + 3 \hat{j} + 9 \hat{k} \right)\]
\[ = - 27 + 9 + 27\]
\[ = 9\]
The shortest distance between the lines
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\]
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ \Rightarrow d = \left| \frac{9}{\sqrt{171}} \right|\]
\[ = \frac{3}{\sqrt{19}}\]
APPEARS IN
संबंधित प्रश्न
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 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).
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 the cartesian equation of a line passing through (1, −1, 2) and parallel to the line whose equations are \[\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z + 1}{- 2}\] Also, reduce the equation obtained in vector form.
Find the direction cosines of the line \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .\] Also, reduce it to vector form.
Find the points on the line \[\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2}\] at a distance of 5 units from the point P (1, 3, 3).
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 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:
\[\frac{5 - x}{- 2} = \frac{y + 3}{1} = \frac{1 - z}{3} \text{ and } \frac{x}{3} = \frac{1 - y}{- 2} = \frac{z + 5}{- 1}\]
Find the angle between the following pair of line:
\[\frac{x - 2}{3} = \frac{y + 3}{- 2}, z = 5 \text{ and } \frac{x + 1}{1} = \frac{2y - 3}{3} = \frac{z - 5}{2}\]
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 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 equations of the line passing through the point (2, 1, 3) and perpendicular to the lines \[\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} \text{ and } \frac{x}{- 3} = \frac{y}{2} = \frac{z}{5}\]
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.
Find the equation of the perpendicular drawn from the point P (−1, 3, 2) to the line \[\overrightarrow{r} = \left( 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) .\] Also, find the coordinates of the foot of the perpendicular from P.
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 shortest distance between the following pairs of lines whose cartesian equations are : \[\frac{x - 1}{- 1} = \frac{y + 2}{1} = \frac{z - 3}{- 2} \text{ and } \frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z + 1}{- 2}\]
By computing the shortest distance determine whether the following pairs of lines intersect or not : \[\overrightarrow{r} = \left( \hat{i} - \hat{j} \right) + \lambda\left( 2 \hat{i} + \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} - \hat{j} \right) + \mu\left( \hat{i} + \hat{j} - \hat{k} \right)\]
Find the shortest distance between the following pairs of parallel lines whose equations are: \[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} - \hat{j} - \hat{k} \right) + \mu\left( - \hat{i} + \hat{j} - \hat{k} \right)\]
Find the equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines
(1, 3, 0) and (0, 3, 0)
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 coordinate axis to which the line \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 1}{0}\] is perpendicular.
Write the direction cosines of the line whose cartesian equations are 2x = 3y = −z.
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.
Write the vector equation of a line given by \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} .\]
The direction ratios of the line x − y + z − 5 = 0 = x − 3y − 6 are proportional to
The projections of a line segment on X, Y and Z axes are 12, 4 and 3 respectively. The length and direction cosines of the line segment are
The lines \[\frac{x}{1} = \frac{y}{2} = \frac{z}{3} \text { and } \frac{x - 1}{- 2} = \frac{y - 2}{- 4} = \frac{z - 3}{- 6}\]
If the lines represented by kx2 − 3xy + 6y2 = 0 are perpendicular to each other, then
The equation 4x2 + 4xy + y2 = 0 represents two ______
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`.
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
Find the equations of the diagonals of the parallelogram PQRS whose vertices are P(4, 2, – 6), Q(5, – 3, 1), R(12, 4, 5) and S(11, 9, – 2). Use these equations to find the point of intersection of diagonals.
