Advertisements
Advertisements
प्रश्न
Equation of a line passing through (1, 1, 1) and parallel to z-axis is ______.
पर्याय
`x/1 = y/1 = z/1`
`(x - 1)/1 = (y - 1)/1 = (z - 1)/1`
`x/0 = y/0 = (z - 1)/1`
`(x - 1)/0 = (y - 1)/0 = (z - 1)/1`
Advertisements
उत्तर
Equation of a line passing through (1, 1, 1) and parallel to z-axis is `underlinebb((x - 1)/0 = (y - 1)/0 = (z - 1)/1)`.
Explanation:
Required equation of line is given by
`(x - a)/l = (y - b)/m = (z - c)/n`
Here, a = 1, b = 1, c = 1
And line is parallel to z-axis. Then
l = 0, m = 0, n = 1
∴ `(x - 1)/0 = (y - 1)/0 = (z - 1)/1`.
APPEARS IN
संबंधित प्रश्न
If a line drawn from the point A( 1, 2, 1) is perpendicular to the line joining P(1, 4, 6) and Q(5, 4, 4) then find the co-ordinates of the foot of the perpendicular.
Let `A(bara)` and `B(barb)` be any two points in the space and `R(barr)` be a point on the line segment AB dividing it internally in the ratio m : n, then prove that `bar r=(mbarb+nbara)/(m+n)`. Hence find the position vector of R which divides the line segment joining the points A(1, –2, 1) and B(1, 4, –2) internally in the ratio 2 : 1.
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 and Cartesian equations of a line passing through (1, 2, –4) and perpendicular to the two lines `(x - 8)/3 = (y + 19)/(-16) = (z - 10)/7` and `(x - 15)/3 = (y - 29)/8 = (z - 5)/(-5)`
Find the vector equation of the line passing through the points (−1, 0, 2) and (3, 4, 6).
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.
Find the angle between the following pair of line:
\[\frac{x + 4}{3} = \frac{y - 1}{5} = \frac{z + 3}{4} \text { and } \frac{x + 1}{1} = \frac{y - 4}{1} = \frac{z - 5}{2}\]
Show that the lines \[\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 1}{5} \text{ and } \frac{x + 2}{4} = \frac{y - 1}{3} = \frac{z + 1}{- 2}\] do not intersect.
Determine whether the following pair of lines intersect or not:
\[\overrightarrow{r} = \left( \hat{i} - \hat{j} \right) + \lambda\left( 2 \hat{i} + \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} - \hat{j} \right) + \mu\left( \hat{i} + \hat{j} - \hat{k} \right)\]
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:
\[\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{- 5} and \frac{x - 8}{7} = \frac{y - 4}{1} = \frac{3 - 5}{3}\]
Show that the lines \[\vec{r} = 3 \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \vec{r} = 5 \hat{i} - 2 \hat{j} + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\] are intersecting. Hence, find their point of intersection.
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 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 cartesian equations are: \[\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} \text{ and } \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}\]
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 vector equation of a line passing through a point having position vector \[\overrightarrow{\alpha}\] and parallel to vector \[\overrightarrow{\beta}\] .
Cartesian equations of a line AB are \[\frac{2x - 1}{2} = \frac{4 - y}{7} = \frac{z + 1}{2} .\] Write the direction ratios of a line parallel to AB.
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
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.
Auxillary equation of 2x2 + 3xy − 9y2 = 0 is ______
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 \[A(x_1,y_1,z_1)\] and has direction ratios \[a,b,c\]. Which is its Cartesian equation?
For the point \[(5,2,-4)\], what is its position vector \[\vec{a}\]?
A line passes through \[(5,2,-4)\] and is parallel to \[3\hat{i}+2\hat{j}-8\hat{k}\]. Which is its vector equation?
What is the parametric form of the line through \[(5,2,-4)\] with direction ratios \[3,2,-8\]?
