English

ABijkAB→=3i^-j^+k^ and CDijkCD→=-3i^+2j^+4k^ are two vectors. The position vectors of the points A and C are ijk6i^+7j^+4k^ and jk-9j^+2k^, respectively. Find the position vector of a point P on the - Mathematics

Advertisements
Advertisements

Question

`vec"AB" = 3hat"i" - hat"j" + hat"k"` and `vec"CD" = -3hat"i" + 2hat"j" + 4hat"k"` are two vectors. The position vectors of the points A and C are `6hat"i" + 7hat"j" + 4hat"k"` and `-9hat"j" + 2hat"k"`, respectively. Find the position vector of a point P on the line AB and a point Q on the line Cd such that `vec"PQ"` is perpendicular to `vec"AB"` and `vec"CD"` both.

Sum
Advertisements

Solution

Position vector of A is `6hat"i" + 7hat"j" + 4hat"k"` and `vec"AB" = 3hat"i" - hat"j" + hat"k"`

So, equation of any line passing through A and parallel to `vec"AB"`

`vec"r" = (6hat"i" + 7hat"j" + 4hat"k") + lambda(3hat"i" - hat"j" + hat"k")`  .....(i)

Now any point P on `vec"AB" = (6 + 3lambda, 7 - lambda, 4 + lambda)`

Similarly, position vector of C is `-9hat"j" + 2hat"k"`

And `vec"CD" = -3hat"i" + 2hat"j" + 4hat"k"`

So, equation of any line passing through C and parallel to CD is

`vec"r" = (-9hat"j" + 2hat"k") + mu(-3hat"i" + 2hat"j" + 4hat"k")`  .....(ii)

Any point Q on `vec"CD" = (-3mu, -9 + 2mu, 2 + 4mu)`

d’ratios of PQ are `(6 + 3lambda + 3mu, 7 - lambda + 9 - 2mu, 4 + lambda - 2 - 4mu)`

⇒ `(6 + 3lambda + 3mu), (16 - lambda - 2mu), (2 + lambda - 4mu)`

Now `vec"PQ"` is ⊥ to equation (i), then

3(6 + 3λ + 3m) – 1(16 – λ – 2m) + 1(2 + λ – 4m) = 0

⇒ 18 + 9λ + 9m – 16 + λ + 2m + 2 + λ – 4m = 0

⇒ 11λ + 7m + 4 = 0  .....(iii)

`vec"PQ"` is also ⊥ to equation (ii), then

`-3(6 + 3lambda + 3mu) + 2(16 - lambda - 2mu) + 4(2 + lambda - 4mu)` = 0

⇒ – 18 – 9λ – 9m + 32 – 2λ – 4m + 8 + 4λ – 16m = 0

⇒ – 7λ – 29m + 22 = 0

⇒ 7λ + 29m – 22 = 0  ......(iv)

Solving equation (iii) and (iv) we get

77λ +   49μ +  28 = 0
77λ + 319μ – 242 = 0
(–)       (–)    (+)            
       – 270μ  + 270 = 0

∴ μ = 1

Now using μ = 1 in equation (iv) we get

7λ + 29 – 22 = 0

⇒ λ = – 1

∴ Position vector of P = [6 + 3(– 1), 7 + 1, 4 – 1]

= (3, 8, 3)

And position vector of Q = [– 3(1), –9 + 2(1), 2 + 4(1)]

= (– 3, –7, 6)

Hence, the position vectors of P = `3hat"i" + 8hat"j" + 3hat"k"` and Q = `-3hat"i" - 7hat"j" + 6hat"k"`

shaalaa.com
  Is there an error in this question or solution?
Chapter 11: Three Dimensional Geometry - Exercise [Page 237]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 11 Three Dimensional Geometry
Exercise | Q 26 | Page 237

RELATED QUESTIONS

Find the equations of the planes that passes through three points.

(1, 1, −1), (6, 4, −5), (−4, −2, 3)


Find the equations of the planes that passes through three points.

(1, 1, 0), (1, 2, 1), (−2, 2, −1)


Find the equation of the plane passing through (abc) and parallel to the plane `vecr.(hati + hatj + hatk) = 2`


Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes `vecr = (hati - hatj + 2hatk)  = 5`and `vecr.(3hati + hatj + hatk) = 6`


Find the vector equation of the line passing through the point (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 Cartesian form of the equation of a plane whose vector equation is 

  \[\vec{r} \cdot \left( - \hat{i} + \hat{j}  + 2 \hat{k} \right) = 9\]

 


Find the vector and Cartesian equations of a plane passing through the point (1, −1, 1) and normal to the line joining the points (1, 2, 5) and (−1, 3, 1).

 

Show that the normals to the following pairs of planes are perpendicular to each other.

\[\vec{r} \cdot \left( 2 \hat{i}  - \hat{j}  + 3 \hat{k}  \right) = 5 \text{ and }  \vec{r} \cdot \left( 2 \hat{i}  - 2 \hat{j}  - 2 \hat{k}  \right) = 5\]

find the equation of the plane passing through the point (1, 2, 1) and perpendicular to the line joining the points (1, 4, 2) and (2, 3, 5). Find also the perpendicular distance of the origin from this plane


Find the vector equation of the plane passing through the points P (2, 5, −3), Q (−2, −3, 5) and R (5, 3, −3).


Find the vector equation of the plane passing through the points \[3 \hat{i}  + 4 \hat{j}  + 2 \hat{k} , 2 \hat{i} - 2 \hat{j} - \hat{k}  \text{ and }  7 \hat{i}  + 6 \hat{k}  .\]

 

Determine the value of λ for which the following planes are perpendicular to each ot

 2x − 4y + 3z = 5 and x + 2y + λz = 5


Find the equation of a plane passing through the point (−1, −1, 2) and perpendicular to the planes 3x + 2y − 3z = 1 and 5x − 4y + z = 5.

 

Find the equation of the plane passing through the points (1, −1, 2) and (2, −2, 2) and which is perpendicular to the plane 6x − 2y + 2z = 9.

 

Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

 

Find the vector equation of the plane through the points (2, 1, −1) and (−1, 3, 4) and perpendicular to the plane x − 2y + 4z = 10 


Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \[\vec{r} \cdot \left( \hat{i}  + 2 \hat{j}  - 5 \hat{k}  \right) + 9 = 0 .\]

 

If the lines  \[\frac{x - 1}{- 3} = \frac{y - 2}{- 2k} = \frac{z - 3}{2} \text{ and }\frac{x - 1}{k} = \frac{y - 2}{1} = \frac{z - 3}{5}\] are perpendicular, find the value of and, hence, find the equation of the plane containing these lines.


Find the image of the point (0, 0, 0) in the plane 3x + 4y − 6z + 1 = 0.

 

Find the coordinates of the foot of the perpendicular from the point (1, 1, 2) to the plane 2x − 2y + 4z + 5 = 0. Also, find the length of the perpendicular.

 

Find the direction cosines of the unit vector perpendicular to the plane  \[\vec{r} \cdot \left( 6 \hat{i}  - 3 \hat{j} - 2 \hat{k} \right) + 1 = 0\] passing through the origin.

 

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x − 3y + 4z − 6 = 0.


Write the equation of the plane passing through points (a, 0, 0), (0, b, 0) and (0, 0, c).

 

Write the distance between the parallel planes 2x − y + 3z = 4 and 2x − y + 3z = 18.  


Write the distance of the plane  \[\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + 2 \hat{k} \right) = 12\] from the origin.

  

Write the intercept cut off by the plane 2x + y − z = 5 on x-axis.

 

Write the equation of a plane which is at a distance of \[5\sqrt{3}\] units from origin and the normal to which is equally inclined to coordinate axes.

 

If the line drawn from (4, −1, 2) meets a plane at right angles at the point (−10, 5, 4), find the equation of the plane.


Find the equation of the plane which bisects the line segment joining the points (−1, 2, 3) and (3, −5, 6) at right angles.


Find the vector equation of the plane which contains the line of intersection of the planes `vec("r").(hat"i"+2hat"j"+3hat"k"),-4=0, vec("r").(2hat"i"+hat"j"-hat"k")+5=0`and which is perpendicular to the plane`vec("r").(5hat"i"+3hat"j"-6hat"k"),+8=0`


Find the co-ordinates 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).


Two systems of rectangular axis have the same origin. If a plane cuts them at distances a, b, c and a′, b′, c′, respectively, from the origin, prove that `1/"a"^2 + 1/"b"^2 + 1/"c"^2 = 1/"a'"^2 + 1/"b'"^2 + 1/"c'"^2`


Find the foot of perpendicular from the point (2, 3, –8) to the line `(4 - x)/2 = y/6 = (1 - z)/3`. Also, find the perpendicular distance from the given point to the line.


Find the equations of the line passing through the point (3, 0, 1) and parallel to the planes x + 2y = 0 and 3y – z = 0.


The locus represented by xy + yz = 0 is ______.


The equation of a line, which is parallel to `2hat"i" + hat"j" + 3hat"k"` and which passes through the point (5, –2, 4), is `(x - 5)/2 = (y + 2)/(-1) = (z - 4)/3`.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×