English

Find the Equation of the Perpendicular Drawn from Point P (−1, 3, 2) to Line → R = ( 2 ^ J + 3 ^ K ) + λ ( 2 ^ I + ^ J + 3 ^ K ) . Also, Find the Coordinates of Foot of the Perpendicular from P. - Mathematics

Advertisements
Advertisements

Question

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.

Sum
Advertisements

Solution

Let L be the foot of the perpendicular drawn from the point P ( -1, 3, 2) to the line \[\overrightarrow{r} = \left( 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + \hat{j} + 3 \hat{k}  \right) .\]

Let the position vector L be \[\overrightarrow{r} = \left( 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) = 2\lambda \hat{i}  + \left( 2 + \lambda \right) \hat{j}  + \left( 3 + 3\lambda \right) \hat{k} \]  ...........(1)

Now, 

\[\overrightarrow{PL} = \text{ Position vector of L - Position vector of P } \]

\[ \Rightarrow \overrightarrow{PL} = \left\{ 2\lambda \hat{i} + \left( 2 + \lambda \right) \hat{j} + \left( 3 + 3\lambda \right) \hat{k} \right\} - \left( - \hat{i} + 3 \hat{j} + 2 \hat{k} \right)\]

\[ \Rightarrow \overrightarrow{PL} = \left( 2\lambda + 1 \right) \hat{i} + \left( \lambda - 1 \right) \hat{j} + \left( 3\lambda + 1 \right) \hat{k} . . . (2)\]

Since  \[\overrightarrow{PL}\]  is perpendicular to the given line, which is parallel to \[\overrightarrow{b} = 2 \hat{i} + \hat{j} + 3 \hat{k} \]

 we have , 

\[\overrightarrow{PL} . \overrightarrow{b} = 0\]

\[ \Rightarrow \left\{ \left( 2\lambda + 1 \right) \hat{i} + \left( \lambda - 1 \right) \hat{j} + \left( 3\lambda + 1 \right) \hat{k} \right\} . \left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) = 0 \]

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

\[ \Rightarrow \lambda = - \frac{2}{7}\]

Substituting  \[ \Rightarrow \lambda = - \frac{2}{7}\]   in (1),

we get the position vector of as \[- \frac{4}{7} \hat{i} + \frac{12}{7} \hat{j} + \frac{15}{7} \hat{k} \] So, the coordinates of the foot of the perpendicular from P to the given line is L \[\left( - \frac{4}{7}, \frac{12}{7}, \frac{15}{7} \right)\]

Substituting  \[\lambda = - \frac{2}{7}\]  in (2), we get  \[\overrightarrow{PL} = \frac{3}{7} \hat{i} - \frac{9}{7} \hat{j} + \frac{1}{7} \hat{k} \]

Equation of the perpendicular drawn from P to the given line is 

\[\overrightarrow{r} = \text{ Position vector of P } + \lambda\left( \vec{PL} \right)\]

\[ = \left( - \hat{i} + 3 \hat{j} + 2 \hat{k}  \right) + \lambda\left( 3 \hat{i} - 9 \hat{j}  + \hat{k} \right)\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 28: Straight Line in Space - Exercise 28.4 [Page 30]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 28 Straight Line in Space
Exercise 28.4 | Q 9 | Page 30

RELATED QUESTIONS

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


The Cartesian equations of line are 3x+1=6y-2=1-z find its equation in vector form.

 


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 value of p, so that the lines `l_1:(1-x)/3=(7y-14)/p=(z-3)/2 and l_2=(7-7x)/3p=(y-5)/1=(6-z)/5 ` are perpendicular to each other. Also find the equations of a line passing through a point (3, 2, – 4) and parallel to line l1.

 

Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5).


Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).

 


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 and cartesian equations of the line through the point (5, 2, −4) and which is parallel to the vector  \[3 \hat{i} + 2 \hat{j} - 8 \hat{k} .\]


Find in vector form as well as in cartesian form, the equation of the line passing through the points A (1, 2, −1) and B (2, 1, 1).


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.


Find the cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line  \[\frac{- x - 2}{1} = \frac{y + 3}{7} = \frac{2z - 6}{3} .\] 


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) and (4, 3, −1). 


Find the angle between the following pair of line: 

\[\overrightarrow{r} = \left( 4 \hat{i} - \hat{j} \right) + \lambda\left( \hat{i} + 2 \hat{j} - 2 \hat{k} \right) \text{ and }\overrightarrow{r} = \hat{i} - \hat{j} + 2 \hat{k} - \mu\left( 2 \hat{i} + 4 \hat{j} - 4 \hat{k} \right)\]


Find the angle between the following pair of line:

\[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3} \text { and } \frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}\]


Find the angle between the following pair of line:

\[\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} \text{  and  } \frac{x + 2}{- 1} = \frac{2y - 8}{4} = \frac{z - 5}{4}\]


Find the angle between the pairs of lines with direction ratios proportional to  1, 2, −2 and −2, 2, 1 .


Find the angle between the pairs of lines with direction ratios proportional to   abc and b − cc − aa − b.


Find the equations of the line passing through the point (2, 1, 3) and perpendicular to the lines  \[\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3} \text{  and  } \frac{x}{- 3} = \frac{y}{2} = \frac{z}{5}\]


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

  

 

 

 


If the lines \[\frac{x - 1}{- 3} = \frac{y - 2}{2 \lambda} = \frac{z - 3}{2} \text{     and     } \frac{x - 1}{3\lambda} = \frac{y - 1}{1} = \frac{z - 6}{- 5}\]  are perpendicular, find the value of λ.


If the coordinates of the points ABCD 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 perpendicular distance of the point (1, 0, 0) from the line  \[\frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 10}{8}.\] Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular.


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 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 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 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 cartesian and vector equations of Y-axis.

 

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.


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


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 angle \[\frac{\pi}{3} \text{ and } \frac{\pi}{4}\]  with x-axis and y-axis respectively, then the angle made by the line with z-axis is


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.


 The equation of a line is 2x -2 = 3y +1 = 6z -2 find the direction ratios and also find the vector equation of the line. 


If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k. 


The equation 4x2 + 4xy + y2 = 0 represents two ______ 


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 


The distance of the point (4, 3, 8) from the Y-axis is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×