English

Show that the three lines with direction cosines 1213,-313,-413; 413,1213,313; 313,-413,1213 are mutually perpendicular. - Mathematics

Advertisements
Advertisements

Question

Show that the three lines with direction cosines `12/13, (-3)/13, (-4)/13;  4/13, 12/13, 3/13;  3/13, (-4)/13, 12/13 ` are mutually perpendicular.

Sum
Advertisements

Solution

Two lines whose direction cosines are l1, m1, n1 and l2, m2, n2 respectively will be mutually perpendicular.

If,  l1l2 + m1m2 + n1n2 = 0

(i) Lines with direction cosines `12/13, (-3)/13, (-4)/13; 4/13, 12/13, 3/13`

Here the values ​​of l1, m1, n are `2/13, (-3)/13, (-4)/13` respectively

And the values ​​of l2, m2, n are `4/13, 12/13, 3/13` respectively.

Now, l1l2 + m1m2 + n1n2

= `12/13 xx 4/13 + ((-3)/13) (12/13) + ((-4)/13) (3/13)`

`= 48/169 - 36/169 - 12/169`

= 0

⇒ Hence, both the lines are perpendicular to each other.

(ii)  Lines with direction cosines `4/13, 12/13, 3/13; 3/13, (-4)/13, 12/13`

Now, l1l2 + m1m2 + n1n2

= `4/13 xx 3/13 + 12/13 ((-4)/13) + 3/13 xx (12/13)`

`= 12/169 - 48/169 + 36/169`

= `(12 - 48 + 36)/169`

= 0

⇒ These two lines are perpendicular to each other.

(iii) Lines with direction cosines `3/13, (-4)/13, 12/13; 12/13, (-3)/13, (-4)/13`

Now, l1l2 + m1m2 + n1n2

= `3/13 xx 12/13 + ((-4)/13) ((-3)/13) + 12/13 xx ((-4)/13)`

`= 36/169 + 12/169 - 48/169`

= `(36 + 12 - 48)/169`

= 0

⇒ Both lines are perpendicular to each other.

Therefore, the given lines are perpendicular to each other.

shaalaa.com
  Is there an error in this question or solution?
Chapter 11: Three Dimensional Geometry - Exercise 11.2 [Page 477]

APPEARS IN

NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 11 Three Dimensional Geometry
Exercise 11.2 | Q 1 | Page 477

RELATED QUESTIONS

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.


 

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. 

 

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`


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 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 the vector equation of a line passing through (2, −1, 1) and parallel to the line whose equations are \[\frac{x - 3}{2} = \frac{y + 1}{7} = \frac{z - 2}{- 3} .\]


Find the cartesian equation of a line passing through (1, −1, 2) and parallel to the line whose equations are  \[\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z + 1}{- 2}\]  Also, reduce the equation obtained in vector form.


Show that the three lines with direction cosines \[\frac{12}{13}, \frac{- 3}{13}, \frac{- 4}{13}; \frac{4}{13}, \frac{12}{13}, \frac{3}{13}; \frac{3}{13}, \frac{- 4}{13}, \frac{12}{13}\] are mutually perpendicular. 


Find the angle between the following pair of line: 

\[\overrightarrow{r} = \left( 3 \hat{i} + 2 \hat{j} - 4 \hat{k} \right) + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 5 \hat{j} - 2 \hat{k}  \right) + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\]


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:

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


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


Find the equation of the line passing through the point (2, −1, 3) and parallel to the line  \[\overrightarrow{r} = \left( \hat{i} - 2 \hat{j} + \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} - 5 \hat{k} \right) .\]


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. \[\frac{x - 5}{5\lambda + 2} = \frac{2 - y}{5} = \frac{1 - z}{- 1}, \frac{x}{1} = \frac{2y + 1}{4\lambda} = \frac{1 - z}{- 3}\]


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.


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


Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( \lambda - 1 \right) \hat{i} + \left( \lambda + 1 \right) \hat{j}  - \left( 1 + \lambda \right) \hat{k}  \text{ and }  \overrightarrow{r} = \left( 1 - \mu \right) \hat{i}  + \left( 2\mu - 1 \right) \hat{j}  + \left( \mu + 2 \right) \hat{k} \]


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


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 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 direction cosines of the line \[\frac{x - 2}{2} = \frac{2y - 5}{- 3}, z = 2 .\]


Write the vector equation of a line given by \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} .\]

 


The angle between the lines

\[\frac{x - 1}{1} = \frac{y - 1}{1} = \frac{z - 1}{2} \text{ and }, \frac{x - 1}{- \sqrt{3} - 1} = \frac{y - 1}{\sqrt{3} - 1} = \frac{z - 1}{4}\] is 

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


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. 


If the lines represented by kx2 − 3xy + 6y2 = 0 are perpendicular to each other, then


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


Find the separate equations of the lines given by x2 + 2xy tan α − y2 = 0 


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


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×