Advertisements
Advertisements
Question
Find the shortest distance between the lines \[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + \hat{k} \right) + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) \text{ and } , \overrightarrow{r} = 2 \hat{i} - \hat{j} - \hat{k} + \mu\left( 2 \hat{i} + \hat{j} + 2 \hat{k} \right)\]
Solution
\[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + \hat{k} \right) + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) \text{ and } , \overrightarrow{r} = 2 \hat{i} - \hat{j} - \hat{k} + \mu\left( 2 \hat{i} + \hat{j} + 2 \hat{k} \right)\]
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 ,
\[\overrightarrow{a_1} = \hat{i} + 2 \hat{j} + \hat{k} \]
\[ \overrightarrow{a_2} = 2 \hat{i} - \hat{j} - \hat{k} \]
\[ \overrightarrow{b_1} = \hat{i} - \hat{j} + \hat{k} \]
\[ \overrightarrow{b_2} = 2 \hat{i} + \hat{j} + 2 \hat{k} \]
\[ \therefore \overrightarrow{a_2} - \overrightarrow{a_1} = \hat{i} - 3 \hat{j} - 2 \hat{k} \]
\[\text{ and } \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 1 \\ 2 & 1 & 2\end{vmatrix}\]
\[ = - 3 \hat{i} + 3 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{\left( - 3 \right)^2 + 3^2}\]
\[ = \sqrt{9 + 9}\]
\[ = 3\sqrt{2}\]
\[\text{ and }\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( \hat{i} - 3 \hat{j} - 2 \hat{k} \right) . \left( - 3 \hat{i} + 3 \hat{k} \right)\]
\[ = - 3 - 6\]
\[ = - 9\]
The shortest distance between the lines
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\] is given by
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ \Rightarrow d = \left| \frac{- 9}{3\sqrt{2}} \right|\]
\[ = \frac{3}{\sqrt{2}}\]
APPEARS IN
RELATED QUESTIONS
The Cartestation equation of line is `(x-6)/2=(y+4)/7=(z-5)/3` find its vector equation.
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).
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).
Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).
Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).
Find the equation of a line parallel to x-axis and passing through the origin.
Find the vector equation of the lines which passes through the point with position vector `4hati - hatj +2hatk` and is in the direction of `-2hati + hatj + hatk`
Find the vector equation of a line passing through (2, −1, 1) and parallel to the line whose equations are \[\frac{x - 3}{2} = \frac{y + 1}{7} = \frac{z - 2}{- 3} .\]
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) and (4, 3, −1).
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:
\[\frac{5 - x}{- 2} = \frac{y + 3}{1} = \frac{1 - z}{3} \text{ and } \frac{x}{3} = \frac{1 - y}{- 2} = \frac{z + 5}{- 1}\]
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 pairs of lines with direction ratios proportional to 2, 2, 1 and 4, 1, 8 .
Find the angle between the pairs of lines with direction ratios proportional to 1, 2, −2 and −2, 2, 1 .
Find the angle between the pairs of lines with direction ratios proportional to a, b, c and b − c, c − a, a − b.
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 vector equation of the line passing through the point (2, −1, −1) which is parallel to the line 6x − 2 = 3y + 1 = 2z − 2.
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} + 4 \hat{j} + 5 \hat{k} \right) + \mu\left( 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \right)\]
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 8 + 3\lambda \right) \hat{i} - \left( 9 + 16\lambda \right) \hat{j} + \left( 10 + 7\lambda \right) \hat{k} \]\[\overrightarrow{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + \mu\left( 3 \hat{i} + 8 \hat{j} - 5 \hat{k} \right)\]
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 \[\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} \text{ and } \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}\]
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 angle between the lines 2x = 3y = −z and 6x = −y = −4z.
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.
The equations of a line are given by \[\frac{4 - x}{3} = \frac{y + 3}{3} = \frac{z + 2}{6} .\] Write the direction cosines of a line parallel to this line.
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 angle between the straight lines \[\frac{x + 1}{2} = \frac{y - 2}{5} = \frac{z + 3}{4} and \frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 3}{- 3}\] is
If a line makes angles α, β and γ with the axes respectively, then cos 2 α + cos 2 β + cos 2 γ =
If the direction ratios of a line are proportional to 1, −3, 2, then its direction cosines are
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
If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k.
The equation 4x2 + 4xy + y2 = 0 represents two ______
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 ______.
