हिंदी

Prove that the lines x = py + q, z = ry + s and x = p′y + q′, z = r′y + s′ are perpendicular if pp′ + rr′ + 1 = 0. - Mathematics

Advertisements
Advertisements

प्रश्न

Prove that the lines x = py + q, z = ry + s and x = p′y + q′, z = r′y + s′ are perpendicular if pp′ + rr′ + 1 = 0.

योग
Advertisements

उत्तर

Given that, x = py + q

⇒ y = `(x - "q")/"p"`

And z = ry + s

⇒ y = `(z - "s")/"r"`

∴ The equation becomes `(x - "q")/"p" = y/1 = (z - "s")/"r"` in which d’ratios are a1 = p, b1 = 1, c1 = r

Similarly x = p'y + q' 

⇒ y = `(x - "q'")/"p'"`

And z = r'y + s' 

⇒ y = `(z - "s'")/"r'"`

∴ The equation becomes `(x - "q'")/"p'" = y/1 = (z - "s'")/"r'"` in which a2 = p', b2 = 1, c2 = r'

If the lines are perpendicular to each other, then

a1a2 + b1b2 + c1c2 = 0

pp' + 1.1 + rr' = 0

Hence, pp' + rr' + 1 = 0 is the required condition.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Three Dimensional Geometry - Exercise [पृष्ठ २३५]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 11 Three Dimensional Geometry
Exercise | Q 6 | पृष्ठ २३५

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector.`3hati + 5hatj - 6hatk`


If the points (1, 1, p) and (−3, 0, 1) be equidistant from the plane `vecr.(3hati + 4hatj - 12hatk)+ 13 = 0`, then find the value of p.


\[\vec{n}\] is a vector of magnitude \[\sqrt{3}\] and is equally inclined to an acute angle with the coordinate axes. Find the vector and Cartesian forms of the equation of a plane which passes through (2, 1, −1) and is normal to \[\vec{n}\] .

 


Find the equation of the plane that bisects the line segment joining the points (1, 2, 3) and (3, 4, 5) and is at right angle to it.

 

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

x − y + z − 2 = 0 and 3x + 2y − z + 4 = 0 


Find the vector equation of a plane which is at a distance of 3 units from the origin and has \[\hat{k}\] as the unit vector normal to it.


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 points A (a, 0, 0), B (0, b, 0) and C(0, 0, c). Reduce it to normal form. If plane ABC is at a distance p from the origin, prove that \[\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} .\]

 


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

\[\vec{r} \cdot \left( \hat{i}  + 2 \hat{j} + 3 \hat{k} \right) = 7 \text{ and }  \vec{r} \cdot \left( \lambda \hat{i} + 2 \hat{j}  - 7 \hat{k}  \right) = 26\]

 


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

 2x − 4y + 3z = 5 and x + 2y + λz = 5


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

 3x − 6y − 2z = 7 and 2x + y − λz = 5

 

Find the equation of the plane passing through the origin and perpendicular to each of the planes x + 2y − z = 1 and 3x − 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 points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1.

 

Find the equation of the plane that contains the point (1, −1, 2) and is perpendicular to each of the planes 2x + 3y − 2z = 5 and x + 2y − 3z = 8.


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 equation of the plane through (2, 3, −4) and (1, −1, 3) and parallel to x-axis.

 

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

 

Find the equation of the plane passing through the intersection of the planes x − 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios proportional to 1, 2, 1. Also, find the perpendicular distance of (1, 1, 1) from this plane


Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the   yz - plane .


Find the coordinates of the point where the line through (3, −4, −5) and (2, −3, 1) crosses the plane 2x + y + z = 7.

 

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 coordinates of the foot of the perpendicular drawn from the point (5, 4, 2) to the line \[\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1} .\]

 Hence, or otherwise, deduce the length of the perpendicular.

 
 

Find the image of the point (1, 3, 4) in the plane 2x − y + z + 3 = 0.

 

Write the equation of the plane parallel to the YOZ- plane and passing through (−4, 1, 0).

 

Write the equation of the plane passing through (2, −1, 1) and parallel to the plane 3x + 2y −z = 7.


Write the position vector of the point where the line \[\vec{r} = \vec{a} + \lambda \vec{b}\] meets the plane  \[\vec{r} . \vec{n} = 0 .\]


Find the length of the perpendicular drawn from the origin to the plane 2x − 3y + 6z + 21 = 0.

 

The equation of the plane parallel to the lines x − 1 = 2y − 5 = 2z and 3x = 4y − 11 = 3z − 4 and passing through the point (2, 3, 3) is


Find the coordinates of the foot of the perpendicular Q  drawn from P(3, 2, 1) to the plane 2x − y + z + 1 = 0. Also, find the distance PQ and the image of the point P treating this plane as a mirror

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.


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


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`.


The point at which the normal to the curve y = `"x" + 1/"x", "x" > 0` is perpendicular to the line 3x – 4y – 7 = 0 is:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×