Advertisements
Advertisements
प्रश्न
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).
उत्तर
Let OA be the line joining the origin, O (0, 0, 0), and the point, A (2, 1, 1).
Also, let BC be the line joining the points, B (3, 5, −1) and C (4, 3, −1).
The direction ratios of OA are 2, 1, and 1 and of BC are (4 − 3) = 1, (3 − 5) = −2, and (−1 + 1) = 0
OA is perpendicular to BC, if a1a2 + b1b2 + c1c2 = 0
∴ a1a2 + b1b2 + c1c2 = 2 × 1 + 1 (−2) + 1 ×0 = 2 − 2 = 0
Thus, OA is perpendicular to BC.
APPEARS IN
संबंधित प्रश्न
The Cartesian equations of line are 3x -1 = 6y + 2 = 1 - z. Find the vector equation of line.
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 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)`
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.
Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).
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 vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector \[\hat{i} - 2 \hat{j} + 3 \hat{k} .\] Reduce the corresponding equation in cartesian from.
The cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form.
Find the cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line \[\frac{- x - 2}{1} = \frac{y + 3}{7} = \frac{2z - 6}{3} .\]
Find the angle between the following pair of line:
\[\overrightarrow{r} = \left( 3 \hat{i} + 2 \hat{j} - 4 \hat{k} \right) + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 5 \hat{j} - 2 \hat{k} \right) + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\]
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 pairs of lines with direction ratios proportional to 2, 2, 1 and 4, 1, 8 .
Find the angle between two lines, one of which has direction ratios 2, 2, 1 while the other one is obtained by joining the points (3, 1, 4) and (7, 2, 12).
Find the equations of the line passing through the point (−1, 2, 1) and parallel to the line \[\frac{2x - 1}{4} = \frac{3y + 5}{2} = \frac{2 - z}{3} .\]
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 \[\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3} \text{ and } \frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}\] intersect and find their point of intersection.
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.
Prove that the line \[\vec{r} = \left( \hat{i }+ \hat{j }- \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and } \vec{r} = \left( 4 \hat{i} - \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{k} \right)\] intersect and find their point of intersection.
Determine whether the following pair of lines intersect or not:
\[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2\]
Determine whether the following pair of lines intersect or not:
\[\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{- 5} and \frac{x - 8}{7} = \frac{y - 4}{1} = \frac{3 - 5}{3}\]
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).
A (1, 0, 4), B (0, −11, 3), C (2, −3, 1) are three points and D is the foot of perpendicular from A on BC. Find the coordinates of D.
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( \lambda - 1 \right) \hat{i} + \left( \lambda + 1 \right) \hat{j} - \left( 1 + \lambda \right) \hat{k} \text{ and } \overrightarrow{r} = \left( 1 - \mu \right) \hat{i} + \left( 2\mu - 1 \right) \hat{j} + \left( \mu + 2 \right) \hat{k} \]
Write the cartesian and vector equations of X-axis.
Write the cartesian and vector equations of Y-axis.
Write the direction cosines of the line whose cartesian equations are 6x − 2 = 3y + 1 = 2z − 4.
Write the direction cosines of the line whose cartesian equations are 2x = 3y = −z.
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.
Find the angle between the lines 2x=3y=-z and 6x =-y=-4z.
The direction ratios of the line x − y + z − 5 = 0 = x − 3y − 6 are proportional to
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 \[\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}\].
Show that the lines \[\frac{5 - x}{- 4} = \frac{y - 7}{4} = \frac{z + 3}{- 5} \text { and } \frac{x - 8}{7} = \frac{2y - 8}{2} = \frac{z - 5}{3}\] are coplanar.
Auxillary equation of 2x2 + 3xy − 9y2 = 0 is ______
If slopes of lines represented by kx2 - 4xy + y2 = 0 differ by 2, then k = ______
Find the position vector of a point A in space such that `vec"OA"` is inclined at 60º to OX and at 45° to OY and `|vec"OA"|` = 10 units.