हिंदी

Find the Value of λ So that the Following Lines Are Perpendicular to Each Other. X − 5 5 λ + 2 = 2 − Y 5 = 1 − Z − 1 , X 1 = 2 Y + 1 4 λ = 1 − Z − 3 - Mathematics

Advertisements
Advertisements

प्रश्न

Find the value of λ so that the following lines are perpendicular to each other. \[\frac{x - 5}{5\lambda + 2} = \frac{2 - y}{5} = \frac{1 - z}{- 1}, \frac{x}{1} = \frac{2y + 1}{4\lambda} = \frac{1 - z}{- 3}\]

योग

उत्तर

The equations of the given line \[\frac{x - 5}{5\lambda + 2} = \frac{2 - y}{5} = \frac{1 - z}{- 1} \text{    and    } \frac{x}{1} = \frac{2y + 1}{4\lambda} = \frac{1 - z}{- 3}\]s   

can be re-written as

\[\frac{x - 5}{5\lambda + 2} = \frac{y - 2}{- 5} = \frac{z - 1}{1} \text{            and                     } \frac{x}{1} = \frac{y + \frac{1}{2}}{2\lambda} = \frac{z - 1}{3}\] Since the given lines are perpendicular to each other,

we have 

\[\left( 5\lambda + 2 \right)1 - 5\left( 2\lambda \right) + 1\left( 3 \right) = 0\]

\[ \Rightarrow 5\lambda = 5\]

\[ \Rightarrow \lambda = 1\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 28: Straight Line in Space - Exercise 28.2 [पृष्ठ १७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 28 Straight Line in Space
Exercise 28.2 | Q 23 | पृष्ठ १७

वीडियो ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्न

The Cartesian equations of line are 3x -1 = 6y + 2 = 1 - z. Find the vector equation of line.


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 passing through the point (2, 1, 3) and perpendicular to the lines

`(x-1)/1=(y-2)/2=(z-3)/3 and x/(-3)=y/2=z/5`


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 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 equation of a line parallel to x-axis and passing through the origin.


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 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 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 cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by  \[\frac{x + 3}{3} = \frac{y - 4}{5} = \frac{z + 8}{6} .\]


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 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 equation of the line passing through the point  \[\hat{i}  + \hat{j}  - 3 \hat{k} \] and perpendicular to the lines  \[\overrightarrow{r} = \hat{i}  + \lambda\left( 2 \hat{i} + \hat{j}  - 3 \hat{k}  \right) \text { and }  \overrightarrow{r} = \left( 2 \hat{i}  + \hat{j}  - \hat{ k}  \right) + \mu\left( \hat{i}  + \hat{j}  + \hat{k}  \right) .\]

  

 

 

 


Find the equation of the line passing through the point (1, −1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, −1, 0) and (1, 2, −1), (2, 1, 1).


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 equation of the perpendicular drawn from the point P (−1, 3, 2) to the line  \[\overrightarrow{r} = \left( 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + \hat{j} + 3 \hat{k}  \right) .\]  Also, find the coordinates of the foot of the perpendicular from P.


Find the foot of the perpendicular from (0, 2, 7) on the line \[\frac{x + 2}{- 1} = \frac{y - 1}{3} = \frac{z - 3}{- 2} .\]


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: \[\vec{r} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k}  + \lambda\left( 3 \hat{i}  - \hat{j}  + \hat{k}  \right) \text{ and }  \vec{r} = - 3 \hat{i}  - 7 \hat{j}  + 6 \hat{k}  + \mu\left( - 3 \hat{i}  + 2 \hat{j}  + 4 \hat{k} \right)\]


Find the shortest distance between the following pairs of lines whose cartesian equations are:  \[\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} \text{ and } \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}\]


By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2\]


By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\frac{x - 5}{4} = \frac{y - 7}{- 5} = \frac{z + 3}{- 5} \text{ and } \frac{x - 8}{7} = \frac{y - 7}{1} = \frac{z - 5}{3}\]


Write the vector equations of the following lines and hence determine the distance between them  \[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6} \text{ and } \frac{x - 3}{4} = \frac{y - 3}{6} = \frac{z + 5}{12}\]


Write the vector equation of a line passing through a point having position vector  \[\overrightarrow{\alpha}\] and parallel to vector \[\overrightarrow{\beta}\] .


The equations of a line are given by \[\frac{4 - x}{3} = \frac{y + 3}{3} = \frac{z + 2}{6} .\]  Write the direction cosines of a line parallel to this line.


The lines `x/1 = y/2 = z/3 and (x - 1)/-2 = (y - 2)/-4 = (z - 3)/-6` are


The equation of the line passing through the points \[a_1 \hat{i}  + a_2 \hat{j}  + a_3 \hat{k}  \text{ and }  b_1 \hat{i} + b_2 \hat{j}  + b_3 \hat{k} \]  is 


If a line makes angles α, β and γ with the axes respectively, then cos 2 α + cos 2 β + cos 2 γ =


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.


Find the value of p for which the following lines are perpendicular : 

`(1-x)/3 = (2y-14)/(2p) = (z-3)/2 ; (1-x)/(3p) = (y-5)/1 = (6-z)/5`


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.


The separate equations of the lines represented by `3x^2 - 2sqrt(3)xy - 3y^2` = 0 are ______ 


Auxillary equation of 2x2 + 3xy − 9y2 = 0 is ______ 


Find the joint equation of pair of lines through the origin which is perpendicular to the lines represented by 5x2 + 2xy - 3y2 = 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 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`.


Equation of a line passing through (1, 1, 1) and parallel to z-axis is ______.


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×