Advertisements
Advertisements
प्रश्न
Find the angle between the pairs of lines with direction ratios proportional to 5, −12, 13 and −3, 4, 5
उत्तर
5, −12, 13 and −3, 4, 5
\[\text{ Let } \overrightarrow{m_1} \text{ and } \overrightarrow{m_2} \text{ be vectors parallel to the two given lines } . \]
\[\text { Then, the angle between the two given lines is same as the angle between } \overrightarrow{m_1} \text{ and } \overrightarrow{m_2} . \]
\[\text{ Now }, \]
\[ \overrightarrow{m_1} = \text{ Vector parallel to the line having direction ratios proportional to } 5, - 12, 13\]
\[ \overrightarrow{m_2} = \text{ Vector parallel to the line having direction ratios proportional to } - 3, 4, 5\]
\[ \therefore \overrightarrow{m_1} = 5 \hat{i} - 12 \hat{j} + 13 \hat{k} \]
\[ \overrightarrow{m_2} = - 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \]
\[\text{ Let } \theta \text{ be the angle between the lines } \].
\[\text{ Now } , \]
\[\cos \theta = \frac{\overrightarrow{m_1} . \overrightarrow{m_2}}{\left| \overrightarrow{m_1} \right| \left| \overrightarrow{m_2} \right|}\]
\[ = \frac{\left( 5 \hat{i} - 12 \hat{j} + 13 \hat{k} \right) . \left( - 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \right)}{\sqrt{5^2 + \left( - 12 \right)^2 + {13}^2} \sqrt{\left( - 3 \right)^2 + 4^2 + 5^2}}\]
\[ = \frac{- 15 - 48 + 65}{13\sqrt{2} \times 5\sqrt{2}}\]
\[ = \frac{1}{65}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{1}{65} \right)\]
APPEARS IN
संबंधित प्रश्न
Find the separate equations of the lines represented by the equation 3x2 – 10xy – 8y2 = 0.
Find the vector and Cartesian equations of the line through the point (1, 2, −4) and perpendicular to the two lines.
`vecr=(8hati-19hatj+10hatk)+lambda(3hati-16hatj+7hatk) " and "vecr=(15hati+29hatj+5hatk)+mu(3hati+8hatj-5hatk)`
Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).
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 equation of the lines which passes through the point with position vector `4hati - hatj +2hatk` and is in the direction of `-2hati + hatj + hatk`
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.
Show that the three lines with direction cosines \[\frac{12}{13}, \frac{- 3}{13}, \frac{- 4}{13}; \frac{4}{13}, \frac{12}{13}, \frac{3}{13}; \frac{3}{13}, \frac{- 4}{13}, \frac{12}{13}\] are mutually perpendicular.
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 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 pairs of lines with direction ratios proportional to a, b, c and b − c, c − a, a − b.
Find the equations of the line passing through the point (−1, 2, 1) and parallel to the line \[\frac{2x - 1}{4} = \frac{3y + 5}{2} = \frac{2 - z}{3} .\]
If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.
Show that the lines \[\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} \text{ and } \frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}\] intersect. Find their point of intersection.
Find the foot of perpendicular from the point (2, 3, 4) to the line \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .\] Also, find the perpendicular distance from the given point to the line.
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: \[\overrightarrow{r} = \left( 8 + 3\lambda \right) \hat{i} - \left( 9 + 16\lambda \right) \hat{j} + \left( 10 + 7\lambda \right) \hat{k} \]\[\overrightarrow{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + \mu\left( 3 \hat{i} + 8 \hat{j} - 5 \hat{k} \right)\]
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 : \[\frac{x - 1}{- 1} = \frac{y + 2}{1} = \frac{z - 3}{- 2} \text{ and } \frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z + 1}{- 2}\]
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)
Write the cartesian and vector equations of Z-axis.
Write the direction cosines of the line whose cartesian equations are 6x − 2 = 3y + 1 = 2z − 4.
Write the coordinate axis to which the line \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 1}{0}\] is perpendicular.
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.
If the equations of a line AB are
\[\frac{3 - x}{1} = \frac{y + 2}{- 2} = \frac{z - 5}{4},\] write the direction ratios of a line parallel to AB.
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 the direction ratios of a line are proportional to 1, −3, 2, then its direction cosines are
Choose correct alternatives:
If the equation 4x2 + hxy + y2 = 0 represents two coincident lines, then h = _______
The separate equations of the lines represented by `3x^2 - 2sqrt(3)xy - 3y^2` = 0 are ______
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
A line passes through the point (2, – 1, 3) and is perpendicular to the lines `vecr = (hati + hatj - hatk) + λ(2hati - 2hatj + hatk)` and `vecr = (2hati - hatj - 3hatk) + μ(hati + 2hatj + 2hatk)` obtain its equation.
