Advertisements
Advertisements
प्रश्न
Find the shortest distance between the following pairs of lines whose vector equations are: \[\vec{r} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} + \lambda\left( 3 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } \vec{r} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} + \mu\left( - 3 \hat{i} + 2 \hat{j} + 4 \hat{k} \right)\]
Advertisements
उत्तर
\[\vec{r} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} + \lambda\left( 3 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } \vec{r} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} + \mu\left( - 3 \hat{i} + 2 \hat{j} + 4 \hat{k} \right)\] Comparing the given equations with the equations
\[\vec{r} = \vec{a_1} + \lambda \vec{b_1} \text{ and } \vec{r} = \vec{a_2} + \mu \vec{b_2}\]
\[\vec{a_1} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} \]
\[ \vec{a_2} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} \]
\[ \vec{b_1} = 3 \hat{i} - \hat{j} + \hat{k} \]
\[ \vec{b_2} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k} \]
\[ \therefore \vec{a_2} - \vec{a_1} = - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \]
\[and \vec{b_1} \times \vec{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & - 1 & 1 \\ - 3 & 2 & 4\end{vmatrix}\]
\[ = - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \]
\[ \Rightarrow \left| \vec{b_1} \times \vec{b_2} \right| = \sqrt{\left( - 6 \right)^2 + \left( - 15 \right)^2 + 3^2}\]
\[ = \sqrt{36 + 225 + 9}\]
\[ = \sqrt{270}\]
\[\left( \vec{a_2} - \vec{a_1} \right) . \left( \vec{b_1} \times \vec{b_2} \right) = \left( - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \right) . \left( - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \right)\]
\[ = 36 + 225 + 9\]
\[ = 270\]
The shortest distance between the lines
\[\vec{r} = \vec{a_1} + \lambda \vec{b_1} \text{ and } \vec{r} = \vec{a_2} + \mu \vec{b_2}\] is given by
\[d = \left| \frac{\left( \vec{a_2} - \vec{a_1} \right) . \left( \vec{b_1} \times \vec{b_2} \right)}{\left| \vec{b_1} \times \vec{b_2} \right|} \right|\]
\[ = \frac{270}{\sqrt{270}}\]
\[ = \sqrt{270}\]
APPEARS IN
संबंधित प्रश्न
The Cartesian equations of line are 3x+1=6y-2=1-z find its equation in vector form.
Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5).
Show that the lines `(x-5)/7 = (y + 2)/(-5) = z/1` and `x/1 = y/2 = z/3` are perpendicular to each other.
ABCD is a parallelogram. The position vectors of the points A, B and C are respectively, \[4 \hat{ i} + 5 \hat{j} -10 \hat{k} , 2 \hat{i} - 3 \hat{j} + 4 \hat{k} \text{ and } - \hat{i} + 2 \hat{j} + \hat{k} .\] Find the vector equation of the line BD. Also, reduce it to 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 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.
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.
Show that the line through the points (1, −1, 2) and (3, 4, −2) is perpendicular to the through the points (0, 3, 2) and (3, 5, 6).
Find the equation of a line parallel to x-axis and passing through the origin.
Find the angle between the following pair of line:
\[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3} \text { and } \frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}\]
Find the angle between the following pair of line:
\[\frac{x - 5}{1} = \frac{2y + 6}{- 2} = \frac{z - 3}{1} \text{ and } \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}\]
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 the pairs of lines with direction ratios proportional to 5, −12, 13 and −3, 4, 5
Find the value of λ so that the following lines are perpendicular to each other. \[\frac{x - 5}{5\lambda + 2} = \frac{2 - y}{5} = \frac{1 - z}{- 1}, \frac{x}{1} = \frac{2y + 1}{4\lambda} = \frac{1 - z}{- 3}\]
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.
Determine whether the following pair of lines intersect or not:
\[\frac{x - 1}{3} = \frac{y - 1}{- 1} = \frac{z + 1}{0} and \frac{x - 4}{2} = \frac{y - 0}{0} = \frac{z + 1}{3}\]
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 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 distance of the point (2, 4, −1) from the line \[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9}\]
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 vector equations are: \[\overrightarrow{r} = \left( 2 \hat{i} - \hat{j} - \hat{k} \right) + \lambda\left( 2 \hat{i} - 5 \hat{j} + 2 \hat{k} \right) \text{ and }, \overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + \hat{k} \right) + \mu\left( \hat{i} - \hat{j} + \hat{k} \right)\]
Write the cartesian and vector equations of X-axis.
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
If a line makes angles α, β and γ with the axes respectively, then cos 2 α + cos 2 β + cos 2 γ =
The lines \[\frac{x}{1} = \frac{y}{2} = \frac{z}{3} \text { and } \frac{x - 1}{- 2} = \frac{y - 2}{- 4} = \frac{z - 3}{- 6}\]
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`
If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k.
The separate equations of the lines represented by `3x^2 - 2sqrt(3)xy - 3y^2` = 0 are ______
Auxillary equation of 2x2 + 3xy − 9y2 = 0 is ______
P is a point on the line joining the points A(0, 5, −2) and B(3, −1, 2). If the x-coordinate of P is 6, then its z-coordinate is ______.
Equation of a line passing through (1, 1, 1) and parallel to z-axis is ______.
In the vector equation \[\vec{r}=\vec{a}+\lambda\vec{b}\], what does \[\vec{r}\] represent?
A line passes through \[A(x_1,y_1,z_1)\] and has direction ratios \[a,b,c\]. Which is its Cartesian equation?
