Advertisements
Advertisements
प्रश्न
Find the shortest distance between the lines \[\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} \text{ and } \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}\]
Advertisements
उत्तर
\[\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} and \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}\]
Since the first line passes through the point (-1,-1,-1) and has direction ratios proportional to 7, -6 , 1 , its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \]
\[\text{ Here }, \]
\[ \overrightarrow{a_1} = - \hat{i} - \hat{j} - \hat{k} \]
\[ \overrightarrow{b_1} = 7 \hat{i} - 6 \hat{j} + \hat{k} \]
Also, the second line passing through the point (3, 5, 7) has direction ratios proportional to 1, -2,1.Its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} \]
\[\text{ Here }, \]
\[ \overrightarrow{a_2} = 3 \hat{i} + 5 \hat{j} + 7 \hat{k} \]
\[ \overrightarrow{b_2} = \hat{i} - 2 \hat{j} + \hat{k} \]
Now,
\[\overrightarrow{a_2} - \overrightarrow{a_1} = 4 \hat{i} + 6 \hat{j} + 8 \hat{k} \]
\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 7 & - 6 & 1 \\ 1 & - 2 & 1\end{vmatrix}\]
\[ = - 4 \hat{i} - 6 \hat{j} - 8 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{\left( - 4 \right)^2 + \left( - 6 \right)^2 + \left( - 8 \right)^2}\]
\[ = \sqrt{16 + 36 + 64}\]
\[ = \sqrt{116}\]
\[\text{ and } \left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( 4 \hat{i} + 6 \hat{j} + 8 \hat{k} \right) . \left( - 4 \hat{i} - 6 \hat{j} - 8 \hat{k} \right)\]
\[ = - 16 - 36 - 64\]
\[ = - 116\]
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
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ \Rightarrow d = \left| \frac{- 116}{\sqrt{116}} \right|\]
\[ = \sqrt{116}\]
\[ = 2\sqrt{29}\]
APPEARS IN
संबंधित प्रश्न
The Cartesian equations of line are 3x+1=6y-2=1-z find its equation in vector form.
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.
A line passes through (2, −1, 3) and is perpendicular to the lines `vecr=(hati+hatj-hatk)+lambda(2hati-2hatj+hatk) and vecr=(2hati-hatj-3hatk)+mu(hati+2hatj+2hatk)` . Obtain its equation in vector and Cartesian from.
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).
The Cartesian equation of a line is `(x-5)/3 = (y+4)/7 = (z-6)/2` Write its vector form.
Find the vector and cartesian equations of the line through the point (5, 2, −4) and which is parallel to the vector \[3 \hat{i} + 2 \hat{j} - 8 \hat{k} .\]
Find the vector equation of the line passing through the points (−1, 0, 2) and (3, 4, 6).
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 (1, −1, 2) and (3, 4, −2) is perpendicular to the through the points (0, 3, 2) and (3, 5, 6).
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:
\[\overrightarrow{r} = \lambda\left( \hat{i} + \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = 2 \hat{j} + \mu\left\{ \left( \sqrt{3} - 1 \right) \hat{i} - \left( \sqrt{3} + 1 \right) \hat{j} + 4 \hat{k} \right\}\]
Find the equation of the line passing through the point (1, −1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, −1, 0) and (1, 2, −1), (2, 1, 1).
Find the shortest distance between the following pairs of lines whose cartesian equations are: \[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} and \frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 5}{5}\]
Find the shortest distance between the following pairs of lines whose cartesian equations are: \[\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} \text{ and } \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}\]
Find the distance between the lines l1 and l2 given by \[\overrightarrow{r} = \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right) \text{ and } , \overrightarrow{r} = 3 \hat{i} + 3 \hat{j} - 5 \hat{k} + \mu\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right)\]
Write the cartesian and vector equations of X-axis.
Write the cartesian and vector equations of Y-axis.
Write the coordinate axis to which the line \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 1}{0}\] is perpendicular.
Write the angle between the lines 2x = 3y = −z and 6x = −y = −4z.
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 \[\frac{x + 3}{3} = \frac{4 - y}{5} = \frac{z + 8}{6} .\]
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 direction ratios of the line x − y + z − 5 = 0 = x − 3y − 6 are proportional to
The perpendicular distance of the point P (1, 2, 3) from the line \[\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}\] is
If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k.
Find the value of λ, so that the lines `(1-"x")/(3) = (7"y" -14)/(λ) = (z -3)/(2) and (7 -7"x")/(3λ) = ("y" - 5)/(1) = (6 -z)/(5)` are at right angles. Also, find whether the lines are intersecting or not.
If the lines represented by kx2 − 3xy + 6y2 = 0 are perpendicular to each other, then
Choose correct alternatives:
The difference between the slopes of the lines represented by 3x2 - 4xy + y2 = 0 is 2
Auxillary equation of 2x2 + 3xy − 9y2 = 0 is ______
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
The equation of line passing through (3, -1, 2) and perpendicular to the lines `overline("r")=(hat"i"+hat"j"-hat"k")+lambda(2hat"i"-2hat"j"+hat"k")` and `overline("r")=(2hat"i"+hat"j"-3hat"k")+mu(hat"i"-2hat"j"+2hat"k")` is ______.
The distance of the point (4, 3, 8) from the Y-axis is ______.
Find the cartesian equation of the line which passes ·through the point (– 2, 4, – 5) and parallel to the line given by.
`(x + 3)/3 = (y - 4)/5 = (z + 8)/6`
