English

Find the position vector of a point A in space such that OAOA→ is inclined at 60º to OX and at 45° to OY and OA|OA→| = 10 units. - Mathematics

Advertisements
Advertisements

Question

Find the position vector of a point A in space such that `vec"OA"` is inclined at 60º to OX and at 45° to OY and `|vec"OA"|` = 10 units.

Sum
Advertisements

Solution

Let a = 60°, b = 45° and the angle inclined to OZ axis be g

We know that cos2a + cos2b + cos2ϒ = 1

⇒ cos2 60° + cos2 45° + cos2ϒ = 1

⇒ `(1/2)^2 + (1/sqrt(2))^2 + cos^2ϒ` = 1

⇒ `1/4 + 1/2 + cos^2ϒ` = 1

⇒ `3/4 + cos^2ϒ` = 1

⇒ cos2ϒ = `1 - 3/4 = 1/4`

∴ cos ϒ = `+- 1/2`

⇒ cos ϒ = `1/2`  .....(Rejecting cos ϒ = `- 1/2`, since ϒ < 90°)

∴ `vec"OA" = |vec"OA"| (1/2hat"i" + 1/sqrt(2)hat"j" + 1/2hat"k")`

= `10(1/2hat"i" + 1/sqrt(2)hat"j" + 1/2hat"k")`

= `5hat"i" + 5sqrt(2)hat"j" + 5hat"k"`

Hence, the position vector of A is `(5hat"i" + 5sqrt(2)hat"j" + 5hat"k")`.

shaalaa.com
  Is there an error in this question or solution?
Chapter 11: Three Dimensional Geometry - Exercise [Page 235]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 11 Three Dimensional Geometry
Exercise | Q 1 | Page 235

RELATED QUESTIONS

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

 


Find the separate equations of the lines represented by the equation 3x2 – 10xy – 8y2 = 0.


If the Cartesian equations of a line are ` (3-x)/5=(y+4)/7=(2z-6)/4` , write the vector equation for the line.


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`


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.


The Cartesian equation of a line is `(x-5)/3 = (y+4)/7 = (z-6)/2` Write its vector form.


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


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`


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.


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.


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 angle between the pairs of lines with direction ratios proportional to  2, 2, 1 and 4, 1, 8 .

 


Find the angle between two lines, one of which has direction ratios 2, 2, 1 while the  other one is obtained by joining the points (3, 1, 4) and (7, 2, 12). 


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


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 (1, −1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, −1, 0) and (1, 2, −1), (2, 1, 1).


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 vector equation of the line passing through the point (2, −1, −1) which is parallel to the line 6x − 2 = 3y + 1 = 2z − 2. 


Show that the lines  \[\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3} \text{          and         } \frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}\] intersect and find their point of intersection. 


Find the equation of the perpendicular drawn from the point P (2, 4, −1) to the line  \[\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{- 9} .\]  Also, write down 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 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 vector equations are: \[\overrightarrow{r} = \left( 8 + 3\lambda \right) \hat{i} - \left( 9 + 16\lambda \right) \hat{j} + \left( 10 + 7\lambda \right) \hat{k} \]\[\overrightarrow{r} = 15 \hat{i} + 29 \hat{j} + 5 \hat{k} + \mu\left( 3 \hat{i}  + 8 \hat{j} - 5 \hat{k}  \right)\]


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


Write the coordinate axis to which the line \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 1}{0}\]  is  perpendicular.


Write the value of λ for which the lines  \[\frac{x - 3}{- 3} = \frac{y + 2}{2\lambda} = \frac{z + 4}{2} \text{ and } \frac{x + 1}{3\lambda} = \frac{y - 2}{1} = \frac{z + 6}{- 5}\]  are perpendicular to each other.


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 direction ratios of the line perpendicular to the lines \[\frac{x - 7}{2} = \frac{y + 17}{- 3} = \frac{z - 6}{1} \text{ and }, \frac{x + 5}{1} = \frac{y + 3}{2} = \frac{z - 4}{- 2}\] are proportional to


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


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

 


The straight line \[\frac{x - 3}{3} = \frac{y - 2}{1} = \frac{z - 1}{0}\] is


Find the value of  λ for which the following lines are perpendicular to each other: 

`(x - 5)/(5 lambda + 2 ) = ( 2 - y )/5 = (1 - z ) /-1 ; x /1 = ( y + 1/2)/(2 lambda ) = ( z -1 ) / 3`


Choose correct alternatives:

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


Find the joint equation of pair of lines through the origin which is perpendicular to the lines represented by 5x2 + 2xy - 3y2 = 0 


The lines `(x - 1)/2 = (y + 1)/2 = (z - 1)/4` and `(x - 3)/1 = (y - k)/2 = z/1` intersect each other at point


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×