मराठी

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

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.

बेरीज
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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 27: Straight Line in Space - Exercise 28.1 [पृष्ठ १०]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 27 Straight Line in Space
Exercise 28.1 | Q 15 | पृष्ठ १०

व्हिडिओ ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्‍न

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.


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

 


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`


Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XZ plane. Also find the angle which this line makes with the XZ plane.


Show that the line through the points (1, −1, 2) (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).


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


The cartesian equations of a line are \[\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2} .\]  Find a vector equation for the line.


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.


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 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: 

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

 


Find the 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:

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


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


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 direction cosines of the line 

\[\frac{x + 2}{2} = \frac{2y - 7}{6} = \frac{5 - z}{6}\]  Also, find the vector equation of the line through the point A(−1, 2, 3) and parallel to the given line.  


Determine whether the following pair of lines intersect or not:  

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


Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 3 \hat{i} + 5 \hat{j} + 7 \hat{k} \right) + \lambda\left( \hat{i} - 2 \hat{j} + 7 \hat{k} \right) \text{ and } \overrightarrow{r} = - \hat{i} - \hat{j} - \hat{k}  + \mu\left( 7 \hat{i}  - 6 \hat{j}  + \hat{k}  \right)\]


By computing the shortest distance determine whether the following pairs of lines intersect or not  : \[\overrightarrow{r} = \left( \hat{i} - \hat{j} \right) + \lambda\left( 2 \hat{i} + \hat{k}  \right) \text{ and }  \overrightarrow{r} = \left( 2 \hat{i} - \hat{j}  \right) + \mu\left( \hat{i} + \hat{j} - \hat{k} \right)\]


Find the shortest distance between the following pairs of parallel lines whose equations are: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} \right) + \lambda\left( 2 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} + \hat{j} - \hat{k} \right) + \mu\left( 4 \hat{i} - 2 \hat{j} + 2 \hat{k} \right)\]


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)


Find the shortest distance between the lines \[\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} \text{ and }  \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}\]


Write the cartesian and vector equations of X-axis.

 

Write the cartesian and vector equations of Z-axis.

 

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 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 lines  \[\frac{x}{1} = \frac{y}{2} = \frac{z}{3} \text { and } \frac{x - 1}{- 2} = \frac{y - 2}{- 4} = \frac{z - 3}{- 6}\] 

 


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


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`.


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.


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 \[\vec{a}\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×