Advertisements
Advertisements
प्रश्न
Find the cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line \[\frac{- x - 2}{1} = \frac{y + 3}{7} = \frac{2z - 6}{3} .\]
Advertisements
उत्तर
We have \[\frac{- x - 2}{1} = \frac{y + 3}{7} = \frac{2z - 6}{3} .\]
It can be re-written as
\[\frac{x + 2}{- 1} = \frac{y + 3}{7} = \frac{z - 3}{\left( \frac{3}{2} \right)}\]
\[ = \frac{x + 2}{- 2} = \frac{y + 3}{14} = \frac{z - 3}{3}\]
This shows that the given line passes through the point (-2, -3,3) and its direction ratios are proportional to -2,14,3
Thus, the parallel vector is \[\overrightarrow{b} = - 2 \hat{i} + 14 \hat{j} + 3 \hat{k} \]
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 \[\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b}\]
Here,
\[\overrightarrow{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} \]
\[ \overrightarrow{b} = - 2 \hat{i} + 14 \hat{j} + 3 \hat{k}\]
Vector equation of the required line is
\[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \lambda \left( - 2 \hat{i} + 14 \hat{j} + 3 \hat{k} \right) . . . (1)\]
\[\text{ Here } , \lambda \text { is a parameter } . \]
Reducing (1) to cartesian form, we get
\[x \hat{i} + y \hat{j} + z \hat{k} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \lambda \left( - 2 \hat{i} + 14 \hat{j} + 3 \hat{k} \right) [\text { Putting } \overrightarrow{r} = x \hat{i} + y \hat{j} + z k \text{ in } (1)]\]
\[ \Rightarrow x \hat{ i }+ y \hat{j} + z \hat{k} = \left( 1 - 2\lambda \right) \hat{i} + \left( 2 + 14 \lambda \right) \hat{j} + \left( 3 + 3\lambda \right) \hat{k} \]
\[\text{ Comparing the coefficients of } \hat{i} , \hat{j} \text{ and } \hat{k} , \text{we get } \]
\[x = 1 - 2\lambda, y = 2 + 14 \lambda, z = 3 + 3\lambda\]
\[ \Rightarrow \frac{x - 1}{- 2} = \lambda, \frac{y - 2}{14} = \lambda, \frac{z - 3}{3} = \lambda\]
\[ \Rightarrow \frac{x - 1}{- 2} = \frac{y - 2}{14} = \frac{z - 3}{3} = \lambda\]
\[\text{ Hence, the cartesian form of (1) is } \]
\[\frac{x - 1}{- 2} = \frac{y - 2}{14} = \frac{z - 3}{3}\]
APPEARS IN
संबंधित प्रश्न
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.
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 equation of the line which passes through the point (1, 2, 3) and is parallel to the vector `3hati+2hatj-2hatk`.
Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).
Show that the lines `(x-5)/7 = (y + 2)/(-5) = z/1` and `x/1 = y/2 = z/3` are perpendicular to each other.
Find the vector equation of a line passing through the point with position vector \[\hat{i} - 2 \hat{j} - 3 \hat{k}\] and parallel to the line joining the points with position vectors \[\hat{i} - \hat{j} + 4 \hat{k} \text{ and } 2 \hat{i} + \hat{j} + 2 \hat{k} .\] Also, find the cartesian equivalent of this equation.
Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z.
Find the equation of a line parallel to x-axis and passing through the origin.
Find the angle between the following pair of line:
\[\frac{x + 4}{3} = \frac{y - 1}{5} = \frac{z + 3}{4} \text { and } \frac{x + 1}{1} = \frac{y - 4}{1} = \frac{z - 5}{2}\]
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 \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} \text{ and } \frac{x}{1} = \frac{y}{2} = \frac{z}{3}\] are perpendicular to each other.
Find the value of λ so that the following lines are perpendicular to each other. \[\frac{x - 5}{5\lambda + 2} = \frac{2 - y}{5} = \frac{1 - z}{- 1}, \frac{x}{1} = \frac{2y + 1}{4\lambda} = \frac{1 - z}{- 3}\]
Determine whether the following pair of lines intersect or not:
\[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2\]
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 foot of the perpendicular drawn from the point \[\hat{i} + 6 \hat{j} + 3 \hat{k} \] to the line \[\overrightarrow{r} = \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) .\] Also, find the length of the perpendicular
Find the foot of the perpendicular from (1, 2, −3) to the line \[\frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1} .\]
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 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 : \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{3} = \frac{y - 2}{1}; z = 2\]
By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} - \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and } \overrightarrow{r} = \left( 4 \hat{i} - \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{k} \right)\]
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 X-axis.
Write the direction cosines of the line whose cartesian equations are 6x − 2 = 3y + 1 = 2z − 4.
Write the angle between the lines 2x = 3y = −z and 6x = −y = −4z.
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
The angle between the lines
The straight line \[\frac{x - 3}{3} = \frac{y - 2}{1} = \frac{z - 1}{0}\] is
The separate equations of the lines represented by `3x^2 - 2sqrt(3)xy - 3y^2` = 0 are ______
If 2x + y = 0 is one of the line represented by 3x2 + kxy + 2y2 = 0 then k = ______
Find the separate equations of the lines given by x2 + 2xy tan α − y2 = 0
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 equations of the diagonals of the parallelogram PQRS whose vertices are P(4, 2, – 6), Q(5, – 3, 1), R(12, 4, 5) and S(11, 9, – 2). Use these equations to find the point of intersection of diagonals.
In the vector equation \[\vec{r}=\vec{a}+\lambda\vec{b}\], what does \[\vec{r}\] represent?
For the point \[(5,2,-4)\], what is its position vector \[\vec{a}\]?
