Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
he cartesian equation of the given line is
3x + 1 = 6y − 2 = 1 − z
It can be re-written as
\[\frac{x + \frac{1}{3}}{\frac{1}{3}} = \frac{y - \frac{1}{3}}{\frac{1}{6}} = \frac{z - 1}{- 1}\]
\[ = \frac{x - \left( - \frac{1}{3} \right)}{2} = \frac{y - \frac{1}{3}}{1} = \frac{z - 1}{- 6}\]
Thus, the given line passes through the point
\[\left( - \frac{1}{3}, \frac{1}{3}, 1 \right)\] and its direction ratios are proportional to 2, 1,-6. It is parallel to the vector \[\overrightarrow{b} = 2 \hat{i} + \hat{j} - 6 \hat{k} \]
We know that the vector equation of a line passing through a point with position vector
\[\overrightarrow{a}\]and parallel to the vector \[\overrightarrow{b}\] is \[\vec{r} = \vec{a} + \lambda \vec{b}\]
Vector equation of the required line is
\[\overrightarrow{r} = \left( - \frac{1}{3} \hat{i} + \frac{1}{3} \hat{j} + \hat{k} \right) + \lambda \left( 2 \hat{i} + \hat{j} - 6 \hat{k} \right)\]
\[\text{ Here }, \lambda \text { is a parameter } . \]
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`
The Cartestation equation of line is `(x-6)/2=(y+4)/7=(z-5)/3` find its vector equation.
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.
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.
Find the equation of the line in vector and in Cartesian form that passes through the point with position vector `2hati -hatj+4hatk` and is in the direction `hati + 2hatj - hatk`.
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).
The cartesian equations of a line are \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} .\] Find a vector equation for the line.
The cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form.
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 \[\frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6} .\]
Show that the lines \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} \text{ and } \frac{x}{1} = \frac{y}{2} = \frac{z}{3}\] are perpendicular to each other.
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.
Find the value of λ so that the following lines are perpendicular to each other. \[\frac{x - 5}{5\lambda + 2} = \frac{2 - y}{5} = \frac{1 - z}{- 1}, \frac{x}{1} = \frac{2y + 1}{4\lambda} = \frac{1 - z}{- 3}\]
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.
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 length of the perpendicular drawn from the point (5, 4, −1) to the line \[\overrightarrow{r} = \hat{i} + \lambda\left( 2 \hat{i} + 9 \hat{j} + 5 \hat{k} \right) .\]
Find the coordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, −1, 3) and C(2, −3, −1).
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 : \[\frac{x - 1}{- 1} = \frac{y + 2}{1} = \frac{z - 3}{- 2} \text{ and } \frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z + 1}{- 2}\]
By computing the shortest distance determine whether the following pairs of lines intersect or not : \[\overrightarrow{r} = \left( \hat{i} - \hat{j} \right) + \lambda\left( 2 \hat{i} + \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} - \hat{j} \right) + \mu\left( \hat{i} + \hat{j} - \hat{k} \right)\]
Find the equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines
(i) (0, 0, 0) and (1, 0, 2)
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 coordinate axis to which the line \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 1}{0}\] is perpendicular.
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
If the direction ratios of a line are proportional to 1, −3, 2, then its direction cosines are
If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k.
Choose correct alternatives:
If the equation 4x2 + hxy + y2 = 0 represents two coincident lines, then h = _______
Find the separate equations of the lines given by x2 + 2xy tan α − y2 = 0
If slopes of lines represented by kx2 - 4xy + y2 = 0 differ by 2, then k = ______
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`
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
