हिंदी

Show that the Points Whose Position Vectors Are − 2 ^ I + 3 ^ J , ^ I + 2 ^ J + 3 ^ K and 7 I − K Are Collinear. - Mathematics

Advertisements
Advertisements

प्रश्न

Show that the points whose position vectors are  \[- 2 \hat{i} + 3 \hat{j} , \hat{i} + 2 \hat{j} + 3 \hat{k}  \text{ and }  7 \text{ i}  - \text{ k} \]  are collinear.

योग
Advertisements

उत्तर

Let the given points be PQ and R and let their position vectors be  \[\overrightarrow{a} , \overrightarrow{b} \text { and } \overrightarrow{c} , \text{ respectively } . \] 

\[\overrightarrow{a} = - 2 \hat{i} + 3 \hat{j} \]

\[ \overrightarrow{b} = \hat{i} + 2 \hat{j} + 3 \hat{k}  \]

\[ \overrightarrow{c} = 7 \hat{i} + 9 \hat{k} \] 

Vector equation of line passing through P and Q is 

\[\overrightarrow{r} = \overrightarrow{a} + \lambda\left( \overrightarrow{b} - \overrightarrow{a} \right)\]

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

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

If points P, Q and R are collinear, then point R must satisfy (1). 

\[\text{ Replacing } \overrightarrow{r} \text{ by } \overrightarrow{c} = 7 \hat{i} + 9 \hat{k}  \text{ in }  (1), \text { we get } \]

\[7 \hat{i} + 9 \hat{k}  = \left( - 2 \hat{i} + 3 \hat{j}  \right) + \lambda\left( 3 \hat{i}  - \hat{j}  + 3 \hat{k}  \right)\] 

Comparing the coefficients of  \[\hat{i}  , \hat{j}  \text{ and }  \hat{k}\] 

we get

\[7 = - 2 + 3\lambda, 0 = 3 - \lambda, 9 = 3\lambda\] 

∴ \[\lambda\] = 3 

These three equations are consistent, i.e. they give the same value of  \[\lambda\]  Hence, the given three points are collinear. 

Disclaimer: The question given in the book has a minor error. The third vectors should be    \[7 \hat{i }+ 9 \hat{k} \]  The solution here is created accordingly. 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 28: Straight Line in Space - Exercise 28.1 [पृष्ठ १०]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 28 Straight Line in Space
Exercise 28.1 | Q 15 | पृष्ठ १०

वीडियो ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्न

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 equation of the line which passes through the point (1, 2, 3) and is parallel to the vector `3hati+2hatj-2hatk`.


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 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 the lines which passes through the point with position vector `4hati - hatj +2hatk` and is in the direction of `-2hati + hatj + hatk`


Find the vector equation of the line passing through the points (−1, 0, 2) and (3, 4, 6).


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.


The cartesian equations of a line are x = ay + bz = cy + d. Find its direction ratios and reduce it to vector form. 


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.


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


Find the angle between the following pair of line: 

\[\overrightarrow{r} = \lambda\left( \hat{i} + \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = 2 \hat{j} + \mu\left\{ \left( \sqrt{3} - 1 \right) \hat{i} - \left( \sqrt{3} + 1 \right) \hat{j} + 4 \hat{k} \right\}\]

 


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


Determine the equations of the line passing through the point (1, 2, −4) and perpendicular to the two lines \[\frac{x - 8}{8} = \frac{y + 9}{- 16} = \frac{z - 10}{7} \text{    and    } \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}\]


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


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


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


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


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


Write the direction cosines of the line whose cartesian equations are 2x = 3y = −z.

 

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 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

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

The shortest distance between the lines  \[\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} \text{ and }, \frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}\] 

 

 

 

 


Choose correct alternatives:

If the equation 4x2 + hxy + y2 = 0 represents two coincident lines, then h = _______


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


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×