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Three Dimensional Geometry

Class 12th Mathematics Part Ii CBSE Solution
Exercise 11.1
  1. If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively,…
  2. Find the direction cosines of a line which makes equal angles with the…
  3. If a line has the direction ratios -18, 12, -4, then what are its direction…
  4. Show that the points (2, 3, 4), (-1, -2, 1), (5, 8, 7) are collinear.…
  5. Find the direction cosines of the sides of the triangle whose vertices are (3,…
Exercise 11.2
  1. Show that the three lines with direction cosines 12/13 , -3/13 , -4/13 4/13 ,…
  2. Show that the line through the points (1, -1, 2), (3, 4, -2) is perpendicular…
  3. Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the…
  4. Find the equation of the line which passes through the point (1, 2, 3) and is…
  5. Find the equation of the line in vector and in cartesian form that passes…
  6. Find the cartesian equation of the line which passes through the point (-2, 4,…
  7. The Cartesian equation of a line is x-5/3 = y+4/7 = z-6/2 . Write its vector…
  8. Find the vector and the cartesian equations of the lines that passes through…
  9. . Find the vector and the cartesian equations of the line that passes through…
  10. Find the angle between the following pairs of lines:
  11. . Find the angle between the following pair of lines: x-2/2 = y-1/5 - z+3/-3…
  12. Find the values of p so that the lines 1-x/3 = 7y-14/2p = z-3/2 7-7x/3p =…
  13. Show that the lines x-5/7 = y+2/-5 = z/1 x/1 = y/2 = z/3 are perpendicular to…
  14. vector r = (i+2 j + k) + lambda (i - j + k) vector r = 2 i - j - k + μ (2 i +…
  15. x+1/7 = y+1/-6 = z+1/1 x-3/1 = y-5/-2 = z-7/1 Find the shortest distance…
  16. Find the shortest distance between the lines whose vector equations are…
  17. vector r = (1-t) i + (t-2) j + (3-2t) k vector r = (s+1) i + (2s-1) j - (2s+1)…
Exercise 11.3
  1. z = 2 In each of the following cases, determine the direction cosines of the…
  2. x + y + z = 1 In each of the following cases, determine the direction cosines…
  3. 2x + 3y - z = 5 In each of the following cases, determine the direction…
  4. 5y + 8 = 0 In each of the following cases, determine the direction cosines of…
  5. Find the vector equation of a plane which is at a distance of 7 units from the…
  6. vector t (i + j - k) = 2 Find the Cartesian equation of the following planes:…
  7. Find the Cartesian equation of the following planes:
  8. Find the Cartesian equation of the following planes:
  9. 2x + 3y + 4z - 12 = 0 In the following cases, find the coordinates of the foot…
  10. 2x + 3y + 4z - 12 = 0 In the following cases, find the coordinates of the foot…
  11. 3y + 4z - 6 = 0 In the following cases, find the coordinates of the foot of…
  12. x + y + z = 1 In the following cases, find the coordinates of the foot of the…
  13. that passes through the point (1, 0, -2) and the normal to the plane is i + j…
  14. that passes through the point (1,4, 6) and the normal vector to the plane is…
  15. (1, 1, -1), (6, 4, -5), (-4, -2, 3) Find the equations of the planes that…
  16. (1, 1, 0), (1, 2, 1), (-2, 2, -1) Find the equations of the planes that passes…
  17. Find the intercepts cut off by the plane 2x + y - z = 5.
  18. Find the equation of the plane with intercept 3 on the y-axis and parallel to…
  19. Find the equation of the plane through the intersection of the planes 3x - y +…
  20. Find the vector equation of the plane passing through the intersection of the…
  21. Find the equation of the plane through the line of intersection of the planes…
  22. Find the angle between the planes whose vector equations are
  23. 7x + 5y + 6z + 30 = 0 and 3x - y - 10z + 4 = 0 In the following cases,…
  24. 2x + y + 3z - 2 = 0 and x - 2y + 5 = 0 In the following cases, determine…
  25. 2x - 2y + 4z + 5 = 0 and 3x - 3y + 6z - 1 = 0 In the following cases,…
  26. 2x - 2y + 4z + 5 = 0 and 3x - 3y + 6z - 1 = 0 In the following cases,…
  27. 4x + 8y + z - 8 = 0 and y + z - 4 = 0 In the following cases, determine…
  28. Point Plane (0, 0, 0) 3x - 4y + 12 z = 3 In the following cases, find the…
  29. Point Plane (3, - 2, 1) 2x - y + 2z + 3 = 0 In the following cases, find the…
  30. Point Plane (2, 3, - 5) x + 2y - 2z = 9 In the following cases, find the…
  31. Point Plane (-6, 0, 0) 2x - 3y + 6z - 2 = 0 In the following cases, find the…
Miscellaneous Exercise
  1. Show that the line joining the origin to the point (2, 1, 1) is perpendicular…
  2. If l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually…
  3. Find the angle between the lines whose direction ratios are a, b, c and b - c,…
  4. Find the equation of a line parallel to x - axis and passing through the…
  5. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (-4, 3,…
  6. If the lines x-1/3k = y-2/1 = z-3/-5 and x-1/3k = y-2/1 = z-3/-5 are…
  7. Find the vector equation of the line passing through (1, 2, 3) and…
  8. Find the equation of the plane passing through (a, b, c) and parallel to the…
  9. Find the shortest distance between lines vector r = (6 i+2 j+2 k) + lambda (1…
  10. Find the coordinates of the point where the line through (5, 1, 6) and (3,…
  11. Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,…
  12. Find the coordinates of the point where the line through (3, -4, -5) and (2,…
  13. Find the equation of the plane passing through the point (-1, 3, 2) and…
  14. If the points (1, 1, p) and (-3, 0, 1) be equidistant from the plane vector r…
  15. Find the equation of the plane passing through the line of intersection of the…
  16. If O be the origin and the coordinates of P be (1, 2, -3), then find the…
  17. Find the equation of the plane which contains the line of intersection of the…
  18. Find the distance of the point (-1, -5, -10) from the point of intersection of…
  19. Find the vector equation of the line passing through (1, 2, 3) and parallel to…
  20. Find the vector equation of the line passing through the point (1, 2, - 4) and…
  21. Prove that if a plane has the intercepts a, b, c and is at a distance of p…
  22. Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 isA. 2…
  23. The planes: 2x - y + 4z = 5 and 5x - 2.5y + 10z = 6 areA. Perpendicular B.…

Exercise 11.1
Question 1.

If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines.


Answer:

Let the direction cosines of the line making ∠ α with x-axis, β – with y axis and γ- with z axis are l, m and n

⇒ l = cos α, m = cos β and n = cos γ


Here α = 90°, β = 135° and γ = 45°


So direction cosines are


l = cos 90° = 0


m = cos 135°= cos (180° - 45°) = -cos 45° =


n = cos 45° =


⇒ Direction cosines of the line



Question 2.

Find the direction cosines of a line which makes equal angles with the coordinate axes.


Answer:

Let the direction cosines of the line making ∠ α with x-axis, β – with y axis and γ- with z axis are l, m and n

⇒ l = cos α, m = cos β and n = cos γ


Here given α = β = γ (line makes equal angles with the coordinate axes) ……….1


Direction Cosines are


⇒ l = cos α, m = cos β and n = cos γ


We have


l2 + m 2 + n2 = 1


cos2 α + cos2β + cos2γ = 1


From 1 we have


cos2 α + cos2 α + cos2 α = 1


3 cos2 α = 1



The direction cosines are




Question 3.

If a line has the direction ratios –18, 12, –4, then what are its direction cosines?


Answer:

If the direction ratios of the line are a, b and c

Then the direction cosines are


Given direction ratios are – 18, 12 and – 4


⇒ a= -18, b = 12 and c= -4



=


=


Direction cosines are





Question 4.

Show that the points (2, 3, 4), (–1, –2, 1), (5, 8, 7) are collinear.


Answer:

If the direction ratios of two lines segments are proportional, then the lines are collinear.

Given A(2, 3, 4), B(−1, −2, 1), C(5, 8, 7)


Direction ratio of line joining A (2,3,4) and B (−1, −2, 1), are


(−1−2), (−2−3), (1−4)


= (−3, −5, −3)


So a1 = -3, b1 = -5, c1 = -3


Direction ratio of line joining B (−1, −2, 1) and C (5, 8, 7) are


(5− (−1)), (8−(−2)), (7−1)


= (6, 10, 6)


So a2 = 6, b2 = 10 and c2 =6


It is clear that the direction ratios of AB and BC are of same proportions


As




and



Therefore A, B, C are collinear.



Question 5.

Find the direction cosines of the sides of the triangle whose vertices are (3, 5, –4), (-1, 1, 2) and (–5, –5, –2).


Answer:


The direction cosines of the two points passing through A(x1, y1, z1) and B(x2, y2, z2) is given by


(x2 – x1), (y2-y1), (z2-z1)


And the direction cosines of the line AB is


Where AB =





Exercise 11.2
Question 1.

Show that the three lines with direction cosines are mutually perpendicular.


Answer:

We know that

If l1, m1, n1 and l2, m2, n2 are the direction cosines of two lines; and θ is the acute angle between the two lines; then cos θ = |l1l2 + m1m2 + n1n2|


If two lines are perpendicular, then the angle between the two is θ = 90°


⇒ For perpendicular lines, | l1l2 + m1m2 + n1n2 | = cos 90° = 0, i.e.


| l1l2 + m1m2 + n1n2 | = 0


So, in order to check if the three lines are mutually perpendicular, we compute | l1l2 + m1m2 + n1n2 | for all the pairs of the three lines.


Now let the direction cosines of L1, L2 and L3 be l1, m1, n1; l2, m2, n2 and l3, m3, n3.


First, consider




⇒ L1⊥ L2 ……(i)


Next, consider




⇒ L2⊥ L3 …(ii)


Now, consider




⇒ L1⊥ L3 …(iii)


∴ By (i), (ii) and (iii), we have


L1, L2 and L3 are mutually perpendicular.



Question 2.

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).


Answer:

We know that

Two lines with direction ratios a1, b1, c1 and a2, b2, c2 are perpendicular if the angle between them is θ = 90°, i.e. a1a2 + b1b2 + c1c2 = 0


Also, we know that the direction ratios of the line segment joining (x1, y1, z1) and (x2, y2, z2) is taken as x2 – x1, y2 – y1, z2 – z1 (or x1 – x2, y1 – y2, z1 – z2).


⇒ The direction ratios of the line through the points (1, –1, 2) and (3, 4, –2) is:


a1 = 3 – 1 = 2, b1 = 4 – (-1) = 4 + 1 = 5, c1 = -2 –2 = -4


and the direction ratios of the line through the points (0, 3, 2) and (3, 5, 6) is:
a2 = 3 – 0 = 3, b2 = 5 – 3 = 2, c2 = 6 – 2 = 4


Now, consider


a1a2 + b1b2 + c1c2 = 2 × 3 + 5 × 2 + (-4) × 4 = 6 + 10 + (-16) = 16 + (-16) = 0


⇒ 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).



Question 3.

Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (–1, –2, 1), (1, 2, 5).


Answer:

We know that

Two lines with direction ratios a1, b1, c1 and a2, b2, c2 are parallel if the angle between them is θ = 0°, i. e.



Also, we know that the direction ratios of the line segment joining (x1, y1, z1) and (x2, y2, z2) is taken as x2 – x1, y2 – y1, z2 – z1 (or x1 – x2, y1 – y2, z1 – z2).


⇒ The direction ratios of the line through the points (4, 7, 8) and (2, 3, 4) is:


a1 = 2 – 4 = -2, b1 = 3 – 7 = -4, c1 = 4 – 8 = -4


And the direction ratios of the line through the points (– 1, – 2, 1) and (1, 2, 5) is:


a2 = 1 – (-1) = 1 + 1 = 2, b2 = 2 – (-2) = 2 + 2 = 4, c2 = 5 – 1 = 4


Consider



∴ The line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (–1, –2, 1), (1, 2, 5).



Question 4.

Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector


Answer:

We know that

Vector equation of a line that passes through a given point whose position vector is and parallel to a given vector is .


So, here the position vector of the point (1, 2, 3) is given by and the parallel vector is .


∴ The vector equation of the required line is:


, where is a real number.



Question 5.

Find the equation of the line in vector and in cartesian form that passes through the point with position vector and is in the direction


Answer:

We know that

Vector equation of a line that passes through a given point whose position vector is and parallel to a given vector is .


Here, and


⇒ The vector equation of the required line is:



Also, we know that


The Cartesian equation of a line through a point (x1, y1, z1) and having direction cosines l, m, n is .


Also, we know that if the direction ratios of the line are a, b, c, then



⇒ The Cartesian equation of a line through a point (x1, y1, z1) and having direction ratios a, b, c is .


Here, x1 = 2, y1 = -1, z1 = 4 and a = 1, b = 2, c = -1


⇒ The Cartesian equation of the required line is:




Question 6.

Find the cartesian equation of the line which passes through the point (–2, 4, –5) and parallel to the line given by


Answer:

We know that

The Cartesian equation of a line through a point (x1, y1, z1) and having direction ratios a, b, c is .


Here, The point (x1, y1, z1) is (-2, 4, -5) and the direction ratios are:


a = 3, b = 5, c = 6


⇒ The Cartesian equation of the required line is:




Question 7.

The Cartesian equation of a line is . Write its vector form.


Answer:

We know that

The Cartesian equation of a line through a point (x1, y1, z1) and having direction cosines l, m, n is .


Comparing this standard form with the given equation, we get


x1 = 5, y1 = -4, z1 = 6 and l = 3, m = 7, n = 2


⇒ The point through which the line passes has the position vector and the vector parallel to the line is given by .


Now, ∵ Vector equation of a line that passes through a given point whose position vector is and parallel to a given vector is .


∴ The vector equation of the required line is:




Question 8.

Find the vector and the cartesian equations of the lines that passes through the origin and (5, –2, 3).


Answer:

We know that

The vector equation of as line which passes through two points whose position vectors are and is .


Here, the position vectors of the two points (0, 0, 0) and (5, -2, 3) are and , respectively.


So, The vector equation of the required line is:




Now, we know that


Cartesian equation of a line that passes through two points (x1, y1, z1) and (x2, y2, z2) is


So, the Cartesian equation of the line that passes through the origin (0, 0, 0) and (5, – 2, 3) is



Question 9.

. Find the vector and the cartesian equations of the line that passes through the points (3, –2, –5), (3, –2, 6).


Answer:

We know that

The vector equation of as line which passes through two points whose position vectors are and is .


Here, the position vectors of the two points (3, –2, –5) and (3, –2, 6) are and , respectively.


So, the vector equation of the required line is:





Now, we also know that


Cartesian equation of a line that passes through two points (x1, y1, z1) and (x2, y2, z2) is


So, the Cartesian equation of the line that passes through the origin (3, -2, -5) and (3, -2, 6) is




Question 10.

Find the angle between the following pairs of lines:



Answer:

We know that

If θ is the acute angle between and , then


……(i)


(i) and


Here and


So, from (i), we have


……(ii)




And


Now,



⇒ By (ii), we have




(ii) and


Here, and


So, from (i), we have


…(iii)




And


Now,



⇒ By (iii), we have





Question 11.

. Find the angle between the following pair of lines:



Answer:

We know that

If and are the equations of two lines, then the acute angle between the two lines is given by


cos θ = | l1l2 + m1m2 + n1n2 | ……(i)


(i) and


Here, a1 = 2, b1 = 5, c1 = -3 and a2 = -1, b2 = 8, c2 = 4


Now, ……(ii)


Here,


And


So, from (ii), we have



And


∴ From (i), we have




(ii) and


Here, a1 = 2, b1 = 2, c1 = 1 and a2 = 4, b2 = 1, c2 = 8


Here,


And


So, from (ii), we have



And


∴ From (i), we have





Question 12.

Find the values of p so that the lines are at right angles.


Answer:

For any two lines to be at right angles, the angle between them should be θ = 90°.

⇒ a1a2 + b1b2 + c1c2 = 0, where a1, b1, c1 and a2, b2, c2 are the direction ratios of two lines.


The standard form of a pair of Cartesian lines is:


and …(i)


Now, first we rewrite the given equations according to the standard form, i.e.


and , i.e.


and …(ii)


Now, comparing (i) and (ii), we get


and


Now, as both the lines are at right angles,


so a1a2 + b1b2 + c1c2 = 0






⇒ 11p = 70




Question 13.

Show that the lines are perpendicular to each other.


Answer:

We know that

Two lines with direction ratios a1, b1, c1 and a2, b2, c2 are perpendicular if the angle between them is θ = 90°, i.e. a1a2 + b1b2 + c1c2 = 0


Also, here the direction ratios are:


a1 = 7, b1 = -5, c1 = 1 and a2 = 1, b2 = 2, c2 = 3


Now, Consider


a1a2 + b1b2 + c1c2 = 7 × 1 + (-5) × 2 + 1 × 3 = 7 -10 + 3 = - 3 + 3 = 0


∴ The two lines are perpendicular to each other.



Question 14.

Find the shortest distance between the lines



Answer:

We know that

Shortest distance between two lines and is


…(i)


Here, , and


,


Now,


…(ii)


Now,



……….(iii)


……….(iv)


Now,


……….(v)


Now, using (i), we have


The shortest distance between the two lines, d [From (iv) and (v)]



Rationalizing the fraction by multiplying the numerator and denominator by √2,




Question 15.

Find the shortest distance between the lines



Answer:

We know that

Shortest distance between the lines:


and is


…(i)


The standard form of a pair of Cartesian lines is:


and


And the given equations are: and


Comparing the given equations with the standard form, we get


x1 = -1, y1 = -1, z1 = -1; x2 = 3, y2 = 5, z2 = 7


a1 = 7, b1 = -6, c1 = 1; a2 = 1, b2 = -2, c2 = 1


Now, consider








Next, consider






From (i), we have




Question 16.

Find the shortest distance between the lines whose vector equations are



Answer:

We know that

Shortest distance between two lines and is


……….(i)


Here, , and


,


Now,


……….(ii)


Now,



……….(iii)


……….(iv)


Now,


……….(v)


Now, using (i), we have


The shortest distance between the two lines,



Question 17.

Find the shortest distance between the lines whose vector equations are



Answer:

Firstly, consider







So, we need to find the shortest distance between and .


Now, We know that


Shortest distance between two lines and is


…(i)


Here, and



Now,


…(ii)


Now,



……….(iii)


……….(iv)


Now,


……….(v)


Now, using (i), we have


The shortest distance between the two lines,




Exercise 11.3
Question 1.

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

z = 2


Answer:

The eq. of the plane

z = 2


Direction ratio of the normal (0,0,1)





This is the form of


lx + my + nz = d (∴ d = Distance of the normal from the origin.)


Direction cosines = 0,0,1


Distance(d) = 2



Question 2.

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

x + y + z = 1


Answer:

The eq. of the plane

x + y + z = 1


Direction ratio of the normal (1,1,1)





This is the form of


lx + my + nz = d (∴ d = Distance of the normal from the origin.)





Question 3.

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

2x + 3y – z = 5


Answer:

The eq. of the plane

2x + 3y-z = 5


Direction ratio of the normal (2, 3, -1)





This is the form of


lx + my + nz = d (∴ d = Distance of the normal from the origin.)





Question 4.

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

5y + 8 = 0


Answer:

The eq. of the plane

0x-5y + 0z = 8


Direction ratio of the normal (0, -5, 0)




= 5



This is the form of


lx + my + nz = d (∴ d = Distance of the normal from the origin.)


Direction cosine = 0, -1, 0




Question 5.

Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector


Answer:

Vector eq. of the plane with position vector is







Question 6.

Find the Cartesian equation of the following planes:



Answer:

(Letbe the position vector of P(x,y,z)

Hence,




So, Cartesian eq. is


x + y - z = 2



Question 7.

Find the Cartesian equation of the following planes:


Answer:

(Letbe the position vector of P(x,y,z)

.



.


So, Cartesian eq. is


2x + 3y - 4z = 1


Question 8.

Find the Cartesian equation of the following planes:



Answer:

Letbe the position vector of P(x,y,z)

Hence,




So, Cartesian eq. is


(s-2t)x + (3-t)y + (2s + t)z=15



Question 9.

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

2x + 3y + 4z – 12 = 0


Answer:

Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x,y,z).

2x + 3y + 4z = 12


Direction ratio (2,3,4)






This is the form of


lx + my + nz = d (∴ d = Distance of the normal from the origin.)





Question 10.

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

2x + 3y + 4z – 12 = 0


Answer:

Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x,y,z).

0x + 3y + 4z = 6


Direction ratio (0,3,4)



.



= 5



his is the form of


lx + my + nz = d (∴ d = Distance of the normal from the origin.)



Coordinate of the foot (ld,md,nd)



Question 11.

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

3y + 4z – 6 = 0


Answer:

Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x,y,z).

x + y + z = 1


Direction ratio (1,1,1)






This is the form of


lx + my + nz = d (∴ d = Distance of the normal from the origin.)





Question 12.

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

x + y + z = 1


Answer:

Let the coordinate of the foot of ⊥ P from the origin to the given plane be P(x,y,z).

0x – 5y + 0z = 8


Direction ratio (0,-5,0)




= 5



This is the form of


lx + my + nz = d (∴ d = Distance of the normal from the origin.)



Coordinate of the foot (ld,md,nd)



Question 13.

Find the vector and cartesian equations of the planes

that passes through the point (1, 0, –2) and the normal to the plane is


Answer:

Let the position vector of the point


Normal⊥to the plane



Vector eq. of the plane,







x – 1 + y – z – 2 = 0


x + y – z – 3 = 0


Required Cartesian eq. of the plane


x + y – z = 3



Question 14.

Find the vector and cartesian equations of the planes

that passes through the point (1,4, 6) and the normal vector to the plane is


Answer:

Let the position vector of the point




Vector eq. of the plane,







x – 1 – 2y + 8 + z – 6 = 0


x – 2y + z + 1 = 0


Required Cartesian eq. of the plane


x – 2y + z = - 1



Question 15.

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

(1, 1, –1), (6, 4, –5), (–4, –2, 3)


Answer:

The given points are (1, 1, -1), (6, 4, -5), (-4, -2, 3).

Let,



= 1(12 - 10) – 1(18 - 20) -1 (-12 + 16)


= 2 + 2 – 4


= 0


Since, the value of determinant is 0.


Therefore, these points are collinear as there will be infinite planes passing through the given 3 points.



Question 16.

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

(1, 1, 0), (1, 2, 1), (–2, 2, –1)


Answer:

The given points are (1, 1, 0), (1, 2, 1), (-2, 2, -1).

Let,



= 1(-2 - 2) – 1(-1 + 2)


= -4 – 1


= -5 ≠ 0


There passes a unique plane from the given 3 points.


Equation of the plane passes through the points, (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3), i.e.,






⇒ (x - 1)(-2) – (y - 1)(3) + 3z = 0


⇒ -2x + 2 – 3y + 3 + 3z = 0


⇒ 2x + 3y – 3z = 5


This is the required eq. of the plane.



Question 17.

Find the intercepts cut off by the plane 2x + y – z = 5.


Answer:

We know that, the eq. of the plane in intercept form


where a, b, c are the intercepts cut-off by the plane at x, y and z axes respectively.


⇒ 2x + y – z = 5 (i)


Dividing both side of (i)eq. by 5, we get






Thus, the intercepts cut-off by the plane are 5/2, 5 and -5.



Question 18.

Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.


Answer:

We know that the eq. of the plane ZOX is

y = 0


Eq. of plane parallel to it is of the form, y = a


Hence, the required eq. of the plane is


y = 3



Question 19.

Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).


Answer:

Eq. of the plane passes through the intersection of the plane is given by

(3x – y + 2z – 4) + λ(x + y + z – 2) = 0


∵ Plane passes through the points (2,2,1)


(3 × 2 – 2 + 2 × 1 – 4) + λ(2 + 2 + 1 – 2) = 0


2 + 3λ = 0


3λ = -2


(i)


Hence, the required eq. of the plane




7x – 5y + 4z – 8 = 0


This is the required eq. of the plane.



Question 20.

Find the vector equation of the plane passing through the intersection of the planes and through the point (2, 1, 3).


Answer:

Let the vector eq. of the plane passing through the intersection of the planes


Here,






∵ Plane passes through points (2,1,3)



4 + 4λ + 2 + 5λ - 9 + 9λ - 7 - 9λ = 0


9 λ = 10








This is the required vector eq. of the plane.



Question 21.

Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.


Answer:

Let the eq. of the plane that passes through the two-given plane x + y + z = 1 and 2x + 3y + 4z = 5 is

(x + y + z – 1) + λ(2x + 3y + 4z – 5) = 0


(2λ + 1)x + (3λ + 1)y + (4λ + 1)z -1 - 5λ = 0 (i)


Direction ratio of the plane (2λ + 1, 3λ + 1, 4λ + 1)


and Direction ratio of another plane (1, -1, 1)


∵ Both are ⊥ hence


(2λ + 1 × 1) + (3λ + 1 × (-1)) + (4λ + 1 × 1) = 0


2λ + 1 - 3λ - 1 + 4λ + 1 = 0



Put the value of λ in (i)eq., we get




x – z + 2 = 0


This is the required eq. of the plane.



Question 22.

Find the angle between the planes whose vector equations are


Answer:

The eq. of the given planes are


If n1 and n2 are normal to the planes,


Angle between two planes








Question 23.

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0


Answer:

The eq. of the given planes are

7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0


Two planes are ⊥ if the direction ratio of the normal to the plane is


a1a2 + b1b2 + c1c2 = 0


21 – 5 – 60


-44 ≠ 0


∴ Both the planes are not ⊥ to each other.


Two planes are || to each other if the direction ratio of the normal to the plane is




∴ Both the planes are not || to each other.


The angle between them is given by








Question 24.

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

2x + y + 3z – 2 = 0 and x – 2y + 5 = 0


Answer:

The eq. of the given planes are

2x + y + 3z – 2 = 0 and x – 2y + 5 = 0


Two planes are ⊥ if the direction ratio of the normal to the plane is


a1a2 + b1b2 + c1c2 = 0


2 × 1 + 1 × (-2) + 3 × 0


= 0


Thus, the given planes are ⊥ to each other.



Question 25.

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0


Answer:

The eq. of the given planes are

2x – 2y + 4z + 5 =0 and x – 2y + 5 = 0


Two planes are ⊥ if the direction ratio of the normal to the plane is


a1a2 + b1b2 + c1c2 = 0


6 + 6 + 24


36 ≠ 0


∴ Both the planes are not ⊥ to each other.


Two planes are || to each other if the direction ratio of the normal to the plane is





Thus, the given planes are || to each other.



Question 26.

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0


Answer:

The eq. of the given planes are

2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0


Two planes are ⊥ if the direction ratio of the normal to the plane is


a1a2 + b1b2 + c1c2 = 0


2 × 2 + (-1) × (-1) + 3 × 3


14 ≠ 0


∴ Both the planes are not ⊥ to each other.


Two planes are || to each other if the direction ratio of the normal to the plane is





Thus, the given planes are || to each other.



Question 27.

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

4x + 8y + z – 8 = 0 and y + z – 4 = 0


Answer:

The eq. of the given planes are

4x + 8y + z – 8 = 0 and y + z – 4 = 0


Two planes are ⊥ if the direction ratio of the normal to the plane is


a1a2 + b1b2 + c1c2 = 0


0 + 8 + 1


9 ≠ 0


∴ Both the planes are not ⊥ to each other.


Two planes are || to each other if the direction ratio of the normal to the plane is




∴ Both the planes are not || to each other.


The angle between them is given by







= 450



Question 28.

In the following cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(0, 0, 0) 3x – 4y + 12 z = 3


Answer:

Distance of point P(x1,y1,z1) from the plane Ax + By + Cz – D = 0 is


Given point is (0,0,0) and the plane is 3x – 4y + 12z = 3






Question 29.

In the following cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(3, – 2, 1) 2x – y + 2z + 3 = 0


Answer:

Given point is (3,-2,1) and the plane is 2x – y + 2z + 3 = 0





Question 30.

In the following cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(2, 3, – 5) x + 2y – 2z = 9


Answer:

Given point is (2,3,-5) and the plane is x + 2y – 2z = 9




= 3



Question 31.

In the following cases, find the distance of each of the given points from the corresponding given plane.

Point Plane

(–6, 0, 0) 2x – 3y + 6z – 2 = 0


Answer:

Given point is (-6,0,0) and the plane is 2x – 3y + 6z – 2 = 0




= 2




Miscellaneous Exercise
Question 1.

Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, –1), (4, 3, –1).


Answer:

Let OA be the line joining the origin (0,0,0) and the point A(2,1,1).

Let BC be the line joining the points B(3,5,−1) and C(4,3,−1)


Direction ratios of OA = (a1, b1, c1) ≡ [(2 - 0), (1 - 0), (1 - 0)] ≡ (2,1,1)


Direction ratios of BC = (a2, b2, c2) ≡ [(4 - 3), (3 - 5), (-1 + 1)]


≡ (1, -2, 0)


Given-


OA is ⊥ to BC


∴ we have to prove that -


a1a2 + b1b2 + c1c2 = 0


L.H.S = a1a2 + b1b2 + c1c2 = 2 × 1 + 1 × (−2) + 1 × 0 = 2 - 2 = 0


R.H.S = 0


Thus, L.H.S = R.H.S ….PROVED


Hence OA is ⊥ to BC.



Question 2.

If l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are (m1n2 - m2n1), (n1l2 - n2l1), (l1m2 - l2m1)


Answer:

Let l, m, n be the direction cosines of the line perpendicular to each of the given lines. Then,

ll1 + mm1 + nn1 = 0 …(1)


and, ll2 + mm2 + nn2 = 0 …(2)


On solving (1) and (2) by cross - multiplication, we get -



Thus, the direction cosines of the given line are proportional to


(m1n2 - m2n1), (n1l2 - n2l1), (l1m2 - l2m1)


So, its direction cosines are



where .


we know that -


(l12 + m12 + n12) (l22 + m22 + n22) - (l1l2 + m1m2 + n1n2)2


= (m1n2 - m2n1)2 + (n1l2 - n2l1)2 + (l1m2 - l2m1)2 …(3)


It is given that the given lines are perpendicular to each other. Therefore,


l1l2 + m1m2 + n1n2 = 0


Also, we have


l12 + m12 + n12 = 1


and, l22 + m22 + n22 = 1


Putting these values in (3), we get -


(m1n2 - m2n1)2 + (n1l2 - n2l1)2 + (l1m2 - l2m1)2 = 1


⇒ λ = 1


Hence, the direction cosines of the given line are (m1n2 - m2n1), (n1l2 - n2l1), (l1m2 - l2m1)



Question 3.

Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.


Answer:

Angle between the lines with direction ratios a1, b1, c1 and a2, b2, c2 is given by

Given -


a1 = a, b1 = b, c1 = c


a2 = b - c, b2 = c - a, c2 = a - b


So,


= 0


∴ cosθ = 0


So, θ = 90°


Hence, Angle between the given pair of Lines is 90°.



Question 4.

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


Answer:

Equation of a line passing through (x1, y1, z1) and parallel to a line with direction ratios a,b,c is


Since the line passes through origin i.e. (0,0,0)


x1 = 0, y1 = 0, z1 = 0


Since line is parallel to x - axis,


a = 1, b = 0, c = 0


Equation of Line is given by –





Question 5.

If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (–4, 3, –6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.


Answer:

Angle between the lines with direction ratios a1, b1, c1 and a2, b2, c2 is given by


A line passing through A(x1, y1, z1) and B(x2, y2, z2) has direction ratios (x1 - x2), (y1 - y2), (z1 - z2)


Direction ratios of line joining the points A(1,2,3) and B(4,5,7)


= (4 - 1), (5 - 2), (7 - 3)


= (3,3,4)


∴ a1 = 3, b1 = 3, c1 = 4


Direction ratios of line joining the points C(-4, 3, -6) and B(2,9,2)


= (2 - (-4)), (9 - 3), (2-(-6))


= (6,6,8)


∴ a2 = 6, b2 = 6, c2 = 8


Now,







∴ cosθ = 1


So, θ = 0°


Hence, Angle between the lines AB and CD is 0°.



Question 6.

If the lines and are perpendicular, find the value of k.


Answer:

Two lines and

are perpendicular to each other if


a1a2 + b1b2 + c1c2 = 0


Given -



comparing with



we get -


x1 =1, y1 = 2, z1 = 3


& a1 = - 3, b1 = 2k, c1 = 2


Similarly,



comparing with



we get -


x2 = 1, y2 = 2, z2 = 3


& a2 = 3k, b2 = 1, c2 = -5


Since the two lines are perpendicular,


a1a2 + b1b2 + c1c2 = 0


⇒ (-3) × 3k + 2k × 1 + 2 × (-5) = 0


⇒ -9k + 2k - 10 = 0


⇒ -7k = 10


∴ k = -10/7


Hence, the value of k is -10/7.



Question 7.

Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane .


Answer:

The vector equation of a line passing through a point with position vector and parallel to vector is


Given, the line passes through (1,2,3)


So,


Finding normal of plane






Comparing with ,



Since line is perpendicular to plane, the line will be parallel of the plane



Hence,




This is the required vector equation of line.



Question 8.

Find the equation of the plane passing through (a, b, c) and parallel to the plane .


Answer:

The equation of a plane passing through (x1,y1,z1) and perpendicular to a line with direction ratios A, B, C is

A(x - x1) + B(y - y1) + C(z - z1) = 0


The plane passes through (a,b,c)


So, x1 = a, y1 = b, z1 = c


Since both planes are parallel to each other, their normal will be parallel


∴ Direction ratios of normal


= Direction ratios of normal of


Direction ratios of normal = (1,1,1)


∴ A = 1, B =1, C = 1


Thus,


Equation of plane in cartesian form is


A(x - x1) + B(y - y1) + C(z - z1) = 0


⇒ 1(x - a) + 1(y - b) + 1(z - c) = 0


⇒ x + y + z - (a + b + c) = 0


Thus, x + y + z = a + b + c is the required equation of plane.



Question 9.

Find the shortest distance between lines

and .


Answer:

Shortest distance between lines with vector equations

and is



Given -



Comparing with , we get -


&


Similarly,



Comparing with , we get -


&


Now,









Magnitude of =


= √144


= 12


Also,



= -80 + (-16) + (-12)


= -108


Shortest Distance = 9


Hence, the shortest distance between the given two lines is 9.



Question 10.

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


Answer:

The vector equation of a line passing through two points with position vectors & is


The position vector of point A(5,1,6) is given as -



The position vector of point B(3,4,1) is given as -






...(1)


Let the coordinates of the point where the line crosses the YZ plane be (0,y,z)


So, …(2)


Since point lies in line, it will satisfy its equation,


Putting (2) in (1)




Two vectors are equal if their corresponding components are equal


So,


0 = 5 - 2λ…(3)


y = 1 + 3λ…(4)


and, z = 6 - 5λ…(5)


From equation (3), we get -


λ = 5/2


Substitute the value of λ in equation (4) and (5), we get -


y = 1 + 3λ = 1 + 3 × (5/2) = 1 + (15/2) = 17/2


and


z = 6 - 5λ = 6 - 5 × (5/2) = 6 - (25/2) = - 13/2


Therefore, the coordinates of the required point is


(0, 17/2, -13/2).



Question 11.

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


Answer:

The vector equation of a line passing through two points with position vectors & is


The position vector of point A(5,1,6) is given as -


.


The position vector of point B(3,4,1) is given as -






...(1)


Let the coordinates of the point where the line crosses the ZX plane be (0,y)


So, …(2)


Since point lies in line, it will satisfy its equation,


Putting (2) in (1)




Two vectors are equal if their corresponding components are equal


So,


x = 5 - 2λ …(3)


0 = 1 + 3λ …(4)


and, z = 6 - 5λ …(5)


From equation (4), we get -


λ = -1/3


Substitute the value of λ in equation (3) and (5), we get -


x = 5 - 2λ = 5 - 2 × (-1/3) = 5 + (2/3) = 17/3


and


z = 6 - 5λ = 6 - 5 × (-1/3) = 6 + (5/3) = 23/3


Therefore, the coordinates of the required point is


(17/3, 0, 23/3).



Question 12.

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


Answer:

The equation of a line passing through two points A(x1,y1,z1) and B(x2,y2,z2) is


Given the line passes through the points A(3, –4, –5) and


B(2, –3, 1)


∴ x1 = 3, y1 = -4, z1 = -5


and, x2 = 2, y2 = -3, z2 = 1


So, the equation of line is




So,


x = -k + 3 | y = k - 4 | z = 6k - 5 …(1)


Let (x, y, z) be the coordinates of the point where the line crosses the plane 2x + y + z + 7 = 0


Putting the value of x,y,z from (1) in the equation of plane,


2x + y + z + 7 = 0


⇒ 2(-k + 3) + (k - 4) + (6k - 5) = 7


⇒ 5k - 3 = 7


⇒ 5k = 10


∴ k = 2


Putting the value of k in x, y, z


x = - k + 3 = - 2 + 3 = 1


y = k - 4 = 2 - 4 = - 2


z = 6k - 5 = 12 - 5 = 7


Hence, the coordinates of the required point are (1, -2,7).



Question 13.

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.


Answer:

The equation of a plane passing through (x1,y1,z1) is given by

A(x - x1) + B(y - y1) + C(z - z1) = 0


where, A, B, C are the direction ratios of normal to the plane.


Now the plane passes through (-1,3,2)


So, equation of plane is


A(x + 1) + B(y - 3) + C(z - 2) = 0 …(1)


Since this plane is perpendicular to the given two planes.


So, their normal to the plane would be perpendicular to normals of both planes.


we know that -


is perpendicular to both &


So, required normal is cross product of normals of planes


x + 2y + 3z = 5 and 3x + 3y + z = 0


Required Normal





Hence, direction ratios = -7, 8, -3


∴ A = -7, B = 8, C = -3


Putting above values in (1), we get -


A(x + 1) + B(y - 3) + C(z - 2) = 0


⇒ -7(x + 1) + 8(y - 3) + (-3)(z - 2) = 0


⇒ -7x - 7 + 8y - 24 - 3z + 6 = 0


⇒ -7x + 8y - 3z - 25 = 0


∴ 7x - 8y + 3z + 25 = 0


Therefore, equation of the required plane is 7x - 8y + 3z + 25 = 0.



Question 14.

If the points (1, 1, p) and (–3, 0, 1) be equidistant from the plane , then find the value of p.


Answer:

The distance of a point with position vector from the plane is .

The position vector of point (1,1,p) is given as -



The position vector of point (-3,0,1) is given as -



It is given that the points (1,1,p) and (-3,0,1) are equidistant from the plane






⇒ 20 - 12p = 8


⇒ 20 - 12p = 8 or, 20 - 12p = -8


⇒ 12p = 12 or, 12p = 28


∴ p = 1 or, p = 7/3


us, the possible values of p are 1 and 7/3.



Question 15.

Find the equation of the plane passing through the line of intersection of the planes and and parallel to x-axis.


Answer:

The equation of any plane through the line of intersection of the planes and is given by -

.


So, the equation of any plane through the line of intersection of the given planes is


.


.


. …(1)


Since this plane is parallel to x-axis.


So, the normal vector of the plane (1) will be perpendicular to x-axis.


Direction ratios of Normal (a1, b1, c1)≡ [(1 - 2λ), (1 - 3λ), (1 +)]


Direction ratios of x–axis (a2, b2, c2)≡ (1,0,0)


Since the two lines are perpendicular,


a1a2 + b1b2 + c1c2 = 0


(1 - 2λ) × 1 + (1 - 3λ) × 0 + (1 + λ) × 0 = 0


⇒ (1 - 2λ) = 0


∴ λ = 1/2


Putting the value of λ in (1), we get -





Hence, the equation of the required plane is



Question 16.

If O be the origin and the coordinates of P be (1, 2, –3), then find the equation of the plane passing through P and perpendicular to OP.


Answer:

The equation of a plane passing through (x1,y1,z1) and perpendicular to a line with direction ratios A, B, C is

A(x - x1) + B(y - y1) + C(z - z1) = 0


The plane passes through P(1,2,3)


So, x1 = 1, y1 = 2, z1 = - 3


Normal vector to plane =


where O(0,0,0), P(1,2, - 3)


Direction ratios of = (1 - 0), (2 - 0), (-3 - 0)


= (1,2, - 3)


∴ A = 1, B = 2, C = -3


Equation of plane in cartesian form is


1(x - 1) + 2(y - 2) - 3(z - (-3)) = 0


⇒ x - 1 + 2y - 4 - 3z - 9 = 0


⇒ x + 2y - 3z - 14 = 0



Question 17.

Find the equation of the plane which contains the line of intersection of the planes and . And which is perpendicular to the plane .


Answer:

The equation of any plane through the line of intersection of the planes and is given by -


So, the equation of any plane through the line of intersection of the given planes is




…(1)


Since this plane is perpendicular to the plane





…(2)


So, the normal vector of the plane (1) will be perpendicular to the normal vector of plane (2).


Direction ratios of Normal of plane (1) = (a1, b1, c1)


≡ [(1 - 2λ), (2 - λ), (3 + λ)]


Direction ratios of Normal of plane (2) = (a2, b2, c2)


≡ (-5, -3,6)


Since the two lines are perpendicular,


a1a2 + b1b2 + c1c2 = 0


⇒ (1 - 2λ) × (-5) + (2 - λ) × (-3) + (3 + λ) × 6 = 0


⇒ -5 + 10λ - 6 + 3λ + 18 + 6λ = 0


⇒ 19λ + 7 = 0


∴ λ = -7/19


Putting the value of λ in (1), we get -






Hence, the equation of the required plane is



Question 18.

Find the distance of the point (–1, –5, –10) from the point of intersection of the line and the plane .


Answer:

Given -

The equation of line is



and the equation of the plane is



To find the intersection of line and plane, putting value of from equation of line into equation of plane, we get -




⇒ (2 + 3λ) × 1 + (-1 + 4λ) × (-1) + (2 + 2λ) × 1 = 5


⇒ 2 + 3λ + 1 - 4λ + 2 + 2λ = 5


⇒ λ = 0


So, the equation of line is



Let the point of intersection be (x,y,z)


So,



Hence, x = 2, y = -1, z = 2


Therefore, the point of intersection is (2, -1, 2).


Now, the distance between points (x1, y1, z1) and (x2, y2, z2) is given by -


units


Distance between the points A(2, -1, 2) and B(-1, -5, -10) is given by -






= 13 units



Question 19.

Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes s and .


Answer:

The vector equation of a line passing through a point with position vector and parallel to a vector is


Given, the line passes through (1,2,3)


So,


Given, line is parallel to both planes


∴ Line is perpendicular to normal of both planes.


i.e is perpendicular to normal of both planes.


we know that -


is perpendicular to both &


So, is cross product of normals of planes


and


Required Normal





Thus,


Now, putting the value of & in formula




Therefore, the equation of the line is




Question 20.

Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:

and .


Answer:

The vector equation of a line passing through a point with position vector and parallel to a vector is


Given, the line passes through (1, 2, -4)


So,


Given, line is parallel to both planes


∴ Line is perpendicular to normal of both planes.


i.e is perpendicular to normal of both planes.


we know that -


is perpendicular to both &


So, is cross product of normals of planes


and


Required Normal





Thus,


Now, putting the value of & in formula






Therefore, the equation of the line is




Question 21.

Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then


Answer:

Distance of the point (x1,y1,z1) from the plane Ax + By + Cz = D is


The equation of a plane having intercepts a, b, c on the x-, y-, z- axis respectively is



Comparing with Ax + By + Cz = D, we get -


A = 1/a, B = 1/b, C = 1/c, D = 1


Given, the plane is at a distance of 'p' units from the origin.


So, The point is O(0,0,0)


∴ x1 = 0, y1 = 0, z1 = 0


Now,


Distance


Substituting all values, we get -







squaring both sides, we get -



Hence Proved.



Question 22.

Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is
A. 2 units

B. 4 units

C. 8 units

D. units


Answer:

Distance between two parallel planes Ax + By + Cz = d1 and Ax + By + Cz = d2 is


Given -


First Plane is


2x + 3y + 4z = 4


Comparing with Ax + By + Cz = d1, we get -


A = 2, B = 3, C = 4, d1 = 4


Second Plane is


4x + 6y + 8z = 12


After Dividing by 2,


2x + 3y + 4z = 6


Comparing with Ax + By + Cz = d1, we get -


A = 2, B = 3, C = 4, d2 = 6


So,


Distance between two planes




= 2/√29


Hence, (D) is the correct option.


Question 23.

The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are
A. Perpendicular

B. Parallel

C. intersect y–axis

D. passes through


Answer:

Given -

First Plane is


2x – y + 4z = 5


Multiply both sides by 2.5, we get -


5x - 2.5y + 10z = 12.5 …(1)


Second Plane is


5x – 2.5y + 10z = 6 …(2)


Clearly, the direction ratios of normals of both the plane (1) and (2) are same.


Hence, Both the given planes are parallel.