A plane is a surface such that if any two points are taken on it, the line segment joining them lies completely on the surface. OR
Every point on the line segment joining any two points on a plane lies on the plane.
Type-1
EQUATION OF A PLANE PASSING THROUGH A GIVEN POINT
1) The general equation of a plane passing through a point (x₁ , y₁ ,z₁) is
a(x - x₁)+ b(y - y₁)+ c(z - z₁)= 0, where a, b, and c are constants.
STEP-1.
Write the equation of a plane passing through (x₁ , y₁ ,z₁) as:
a(x - x₁)+ b(y - y₁)+ c(z - z₁)= 0. (i)
STEP -2
If the plane (i) passes through (x₂ , y₂ , z₂) and (x₃ , y₃ , z₃), then
a(x₂ - x₁)+ b(y₂ - y₁)+ c(z₂ - z₁)= 0. (ii) And
a(x₃ - x₁)+ b(y₃ - y₁) + c(z₃ - z₁)= 0. (iii)
STEP-3
Solve equation (ii) and (iii), obtained in STEP 2, by cross multiplication.
STEP -4
Substituting the value of a, b, c , obtained in STEP- 3, in equation (i) in STEP -1 to get the required plane.
REMARK: On eliminating a, b, c from equation (i),(ii),(iii), we get
x - x₁ y- y₁ z - z₁
x₂ - x₁ y₂- y₁ z₂ - z₁ = 0
x₃- x₁ y₃ - y₁ z₃ - z₁ in determinant form
As the equation of the plane passing through three given points (x₁ , y₁, z₁), (x₂, y₂, z₂), (x₃, y₃, z₃)
EXERCISE -A
1) Find the equation of the plane through the points
A) A(2,2, -1), B(3,4,2) and C(7,0,6). 5x + 2y- 3z= 17
B) M(1,1, 0), N(-2,2,-1) and P(1,2,2). 2x + 3y- 3z= 5
C) (2,1, 0), (3,-2, -2) and C(3,2,7). 7x + 3y- z= 17
D) A(1,1, 1), B(1,-1,2) and C(-2,-2,2). x - 3y- 6z+ 8= 0
E) A(0,-1, 0), B(3,3,0) and C(1,1,1). 4x - 3y + 2z= 3
2) Show that the following points are coplanar:
A) (0,-1,-1), (-4,4,4),(4,5,1).
B) (0,4,3),(-1,-5,-3),(-2,-2,1) and (1,1,-1).
C) (0,-1,0), (2,1,-1),(1,1,1) and (3,3,0)
3) Show that the four points (0,-1,-1),(4,5,1),(3,9,4) and (-4,4,4) are coplanar and find the equation of the common plane. 5x - 7y + 11z +4= 0
INTERCEPT FORM OF A PLANE
The equation of a plane intercepting lengths a, b, and c with x-axis, y-axis and z-axis respectively is
x/a + y/b + z/c = 1.
NOTE:
1) The above equation is known as the intercept form of the plane, because the plane intercepts lengths a, b and c with x-axis, y-axis and z-axis.
STEP
To determine the intercepts made by a plane with the coordinate axes
1) For x-intercept:
Put y= 0 , z= 0 in the equation of the plane and obtain the value of x. The value of x is the intercept on x-axis.
For y-intercept:
Put x= 0 , z= 0 in the equation of the plane and obtain the value of y. The value of y is the intercept on y-axis.
For z-intercept:
Put x= 0 , y= 0 in the equation of the plane and obtain the value of y. The value of y is the intercept on x-axis.
EXERCISE -B
1) Write the equation of the plane whose intercept on the coordinates axes are
A) -4, 2 and 3. -3x+6y+4z= 12
B) 2, -3 and 4. 6x-4y+3z= 12
2) Reduce the equation of the following planes in intercept form and find its intercept on the coordinate axes:
A) 2x+ 3y - 4z= 12. x/6 + y/4 + z/-3 = 1. 6, 4 , -3
B) 4x+ 3y - 6z= 12. x/3 + y/4 + z/-2 = 1. 3, 4 , -2
C) 2x+ 3y - z= 6. x/3 + y/2 + z/-6 = 1. 3, 2 , -6
D) 2x - y +z= 5. x/5/2 + y/-5 + z/5 = 1. 5/2, -5, 5
3) Find the equation of the plane passing through the point (2,4,6) and making equal intercept on the coordinate axes. x+y+ z= 12
4) A plane meets the coordinates axes are A, B and C respectively such that the centroid of triangle ABC is (1,-2,3). Find the equation of the plane. 6x- 3y +2z = 18
5) A plane meets the coordinates axes in A, B and C respectively such that the centroid of triangle ABC is (p,q,r). Show that the equation of the plane is x/p +y/q +z/r = 3.
Vector Equation Of A Plane Passing Through A Given Point And Normal To A Given Vector
* The vector equation of a plane passing through a point having position vector a and normal to vector n is (r - a) . n = 0
OR
r. n = a. n
Note 1: It is to note here that vector equation of a plane means a relation involving the position vector r of an arbitrary point on the plane.
Note 2: The above equation can also be written as r . n = d, where d= a. n. This is known as the scalar product form of a plane.
REDUCTION TO CARTESIAN FORM
If r= xi + yj + zk, a= a₁i+ a₂j + a₃k and n= n₁i+ n₂j + n₃k
Then, r - a= (x - a₁)i +(y - a₂)j +(z - a₃)k.
Substituting the value of (r - a) and n in equation (r - a). n = 0, we get
[(x -a₁)i +(y- a₂)j+(z- a₃)]. (n₁i+ n₂j + n₃k)= 0.
=> (x -a₁)n₁ +(y- a₂)n₂+(z- a₃)n₃= 0.
This is the Cartesian equation of a plane passing through (a₁ , a₂, a₃)
NOTE that the coefficient of x,y and z in equation are n₁, n₂ , n₃ respectively which are proportional to the direction ratios of vector n normal to the plane.
Thus, the coefficient of x, y and z in the Cartesian equation of a plane are proportional to the direction ratios of normal to the plane.
For Example:
The direction ratios of a vector normal to the plane 2x+ y - 2z - 5 = 0 are proportional to 2, 1, -2 and hence a vector normal to the plane is 2i + j - 2k.
EXERCISE - C
1) Find the vector equation of a plane passing through a point having
A) position vector 2i + 3j - 4k and Perpendicular to the vector 2i- j + 2k. Also, reduce it to Cartesian form. r. (2i - j + 2k)= -7. 2x- y +2z = -7
B) position vector 2i - j +k and Perpendicular to the vector 4i+ 2 j 3 k. Also, reduce it to Cartesian form. r. (4i + 2j - 3k)= 3.
2) Find the cartesian form of equation of a plane whose vector equation is
A) r. (12i -3j +4k)+5= 0. 12x - 3y + 4z +5= 0
B) r. (- i + j +2k)= 9. -x + y +2z = 9
3) Find the vector equation of each one of the following:
A) 2x - y +2z = 8. r. (2i - j +2k)=8
B) x +y -z = 5. r. (i +j -k)= 5
C) x +y = 3. r. (i + j)= 3
4) Find the vector and Cartesian equations of a plane passing through the
A) point (1,-1,1) and normal to the line joining the points (1,2,5) and (-1,3,1). r. (2i - j +4k)=7, 2x - y +4z = 7
B) point (3,-3,1) and normal to the line joining the points (3,4,-1) and (2,-1,5). r. (-i - 5j +6k)=18, x + 5y -6z = -18
5) a) The foot of the perpendicular drawn from the origin to the plane is (4,-2,-5). Find the equation of the plane. r. (4i - 2j -5k)=45, 4x -2y -5z = 45.
B) The coordinates of the foot of the perpendicular drawn from the origin to a plane is (12,-4,3). Find the equation of the plane. 12x -4y +3z = 169.
6) A) If the line drawn from the point (-2, -1,-3) meets a plane at right angle at the point (1,-3,3), find the equation of the plane. r. (3i - 2j +6k)=27
B) If the line drawn from the point (4, -1,2) meets a plane at right angle at the point (-10,5,4), find the equation of the plane. 7x - 3y - z+ 89= 0.
7)A) Find the equation of the plane which bisects the line segment joining the points A(2,3,4) and B(4, 5, 8) at right angles. r. (i +j +2k)=19, x+ y+ 2z= 19.
B) Find the equation of the plane which bisects the line segment joining the points A(12,3) and B(3,4, 5) at right angles. x+ y+ z= 9.
C) Find the equation of the plane passing through the point (1,-1,2) having 2,3,2 as direction ratios of normal to the plane. r. (2i +3j +2k)=3, 2x+ 3y+ 2z= 3.
D) Find the equation of the plane passing through the point (2,3,1) given that the direction ratios of normal to the plane are proportional to 5,3,2. 5x+ 3y+ 2z= 21
8) Show that the normals to the following of planes are perpendicular to each other:
A) x - y+ z= 2 and 3x+ 2y- z= -4
B) r. (2i -j +3k)=5, r . (2i - 2j -2k)=5.
9)
Incomplete
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* The vector equation of a plane normal to unit vector n and at a distance d from the origin= r. n = d.
* If l, m, n are direction of the normal to a given plane which is at a distance p from the origin, then the equation of the plane is lx + my + nz= p.
Note-1
The equation r.n= d is normal form if n is a unit vector and in such a case d on the right hand side denotes the distance of the plane from the origin. If n is not a vector, then to reduce the equation r. n = d to normal form divide both sides by |n| to obtain:
r. n/|n|= d/|n|
=> r. n= d/|n|
In order to reduce the Cartesian equation ax+ by + cz+ d= 0 of a plane to normal form,
Use following steps
Step-1: Keep terms containing x,y and z on LHS and shift the constant term d on the RHS.
Step-2 : If the constant term on RHS is not positive make it positive by multiplying both sides by -1.
Step-3: Divide each term on two sides by
√(a²+ b²+ c²)= √(coef. of x)²+ (Coeff. of y)²+ (Coeff. z)².
The coefficient of x,y and z in the equation so obtained will be the direction cosines of the normal to the plane and the RHS will be the distance of the plane from the origin.
REMARK: The Cartesian equations of the normal to the plane lx + my+ nz= p drawn from the origin are x/l = y/m = z/n and the coordinates of the foot N of the perpendicular drawn from the origin O are given x/l = y/m = z/n= p i.e., (lp, mp, np)
EXERCISE -D
1)a) Find the vector equation of a plane at a distance of 5 units from the origin and has i as the unit vector normal to it. r. i= 5
b) Find the vector equation of a plane at a distance of 3 units from the origin and has k as the unit vector normal to it. r. k = 5
2)a) find the vector equation of a plane which is at a distance of 8 units from the origin and which is normal to the vector to 2i + j +2k. r.(2i+ j+2k)= 24
b) find the vector equation of a plane which is at a distance of 5 units from the origin and which is normal to the vector to i -2 j -2k. r.(i/3- 2j/3-2k/3)= 5
3) a) Reduce the equation r.(3i- 4j+ 12k)= 5 to the Normal form and hence, find the length of the perpendicular from the origin to the plane. r.(3i/13- 4j/13+ 12k/13)= 5/13, 5/13
b) Reduce the equation r.(i- 2j+ 2k)= -6 to the Normal form and hence, find the length of the perpendicular from the origin to the plane. r.(-i/3+4j/3- 2k/3)= 2, 2
4) a) Reduce the equation of the plane x - 2y - 2z =12 to the normal form and hence find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane. x/2 - 2y/3 - 2z/3= 4, 4, 1/3,-2/3,-2/3
b) Reduce the equation of the plane 2x - 3y - 6z =14 to the normal form and hence find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane. 3, 2/7, -3/7, -6/7
5) a) Find the vector equation are plane which is at a distance of 6 units from the origin and has 2,-1,2 as the direction ratios of a normal to it. Also, find the coordinates of the foot of the normal drawn from the origin. r.(2i/3 - j/3+ 2k/3)= 6, (4,-2,4)
b) Find the vector equation are plane which is at a distance of 5 units from the origin and has 12,-3,4 as the direction ratios of a normal to it. Also, find the coordinates of the foot of the normal drawn from the origin. 12x/13- 3y/13 + 4z/13= 5
6) Find the co-ordinate of the foot of the perpendicular drawn from the origin to the plane 2x - 3y + 4z - 6= 0. (12/29, -18/29, 24/29)
7) Find the direction cosines of the perpendicular from the origin to the plane. r.(2i - 3j- 6k)+5= 0, -2/7, 3/7, 6/7
8) Find the equation of the plane passing through the point (-1,2,1) and the perpendicular to the line joining the points (-3,1,2) and (2,3,4). Find also the perpendicular distance of the origin from this plane. r.(5i/√33+ 2j/√33 + 2k/√33)= 1/√33, 1/√33
9) Write the normal form of the equation of the plane 2x - 3y +6z +14= 0. -2x/7 + 3y/7 -6z/7 = 2
10) Write a unit normal vector to the plane x + 2y +3z -6= 0. i/√14 + 2j/√14 + 3k/√14 = 2
11) a) Find the equation of the plane which is at a distance of 3√3 units from the origin and the normal to which is equally inclined with the coordinates axes. x +y +z = 9
b) Find the vector equation of the plane which is at a distance of 6/√29 units from the origin and the normal from the origin is 2i -3j + 4k. Also, find its Cartesian form. r.(2i/√29- 3j/√29 + 4k/√29)= 6/√29, 2x -3y + 4z = 6
12) 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. x -y +3z -2= 0, -2/√11
13) Find the distance of the plane 2x -3x +4z -6= 0 from the origin. 6/√29
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Vector Equation Of a Plane Passing Through 3 Given Points
n= AB x AC = (b-a) x(c -a)
ax b + bxc + cxa
Exercise -E
1) Find the vector equation of the plane passing through the points
A) A (2,2,-1), B( 3,4,2) and C(7,0,6). also, find the cartesian equation of the plane. r.(5i + 2j- 3k)= 17, 5x+2y-3z= 17
B) A (1,1,1), B(1,-1,1) and C(-7,-3,-5). r.(3i -4j- 4k)+1= 0
C) A (2,5-3), B(-2,-3,5) and C(5,3,-3). r.(2i +3j+4k)= 7
D) A (1,1,-1), B(6,4,-5) and C(-4,-2,3). 5a -3b-4c= 0
E) 3i +4j+2k, 2i -2j-k, and 7i + 6k. r. (9i +2j-7k)= 21
F) i +j-2k, 2i -j+k, and i +2j+ k. r. (9i + 3j- k)=14
2) Find the vector equation of the plane passing through the points A(a,0,0), B(0,b,0) and C(0,0,c). Reduce it to normal from. If plane ABC is at a distance p from the origin, prove that 1/p²= 1/a²+ 1/b²+ 1/c².
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Angle Between Two Planes
Definition: The angle between two planes is defined as the angle between their normals.
VECTOR FORM
The angle ¢ between the planes r. n₁ = d₁ and r. n₂= d₂ is
Cos¢= (n₁. n₂)/(|n₁| |n₂|)
CARTESIAN FORM
The angle ¢ between the planes a₁x+ b₁y + c₁z + d₁=0 and a₂x+ b₂y+ c₂z+ d₂= 0 is
Cos¢ =(a₁a₂ + b₁b₂+ c₁c₂)/{√(a₁²+ b₁²+ c₁²) √(a₂²+ b₂²+ c₂²)
Condition of perpendicularity:
* n₁. n₂= 0
* a₁a₂+ b₁b₂+ c₁c₂= 0
Condition of parallelism:
* a₁/a₂ = b₁/b₂= c₁/c₂
ON FINDING A PLANE PASSING THROUGH A GIVEN POINT AND PERPENDICULAR TO TWO GIVEN PLANES
The normal to the plane passing through a point having position vector a and perpendicular to the planes r. n₁= d₁ and r. n₂ = d₂ is perpendicular to the vector n₁ and n₂. So it is parallel to n₁ x n₂. We may use the following steps to find the plane passing through a given point and perpendicular to two planes.
Step 1: Obtain the position vector of the given point say, a
Step 2: Obtain the normal vector to two planes. Let the normal vector be n₁ and n₂.
Step 3: Compute n=n₁ x n₂. Clearly, n is normal to the required plane.
Step 4: write the equation of the desired plane as
(r-a).(n₁ x n₂)= 0 OR
r (n₁ xn₂)= a. (n₁ xn₂) OR
[r n₁n₂ ]=[a n₁n₂]
ON FINDING A PLANE PASSING THROUGH TWO GIVEN POINTS PERPENDICULAR TO A GIVEN PLANE
The normal to the plane passing through the two points P and Q having their position vectors a and b respectively and perpendicular to the plane r. n₁ = d₁ is perpendicular to the vectors PQ and n₁. So, the normal vector n to the plane is parallel to the vector PQ x n₁. Thus, use the following step to find the plane passing through two given points and perpendicular to a given plane.
Step 1: Obtain the position vector of the given points. Let the positions vectors of the given points P and Q be a and b respectively.
Step 2: Obtain the equation of the plane perpendicular to the required to the required plane. Let its equation be r. n₁= d.
Step-3: Let n be the normal vector to the required plane. Then, n is perpendicular to both n₁ and PQ.
So, Compute n= n₁ x PQ.
Step-4: Write the equation of the required plane as (r - a). n = 0 OR
r. n = a. n
EXERCISE -F
1) Find the angle between the planes
A) r. (2i- j+ k)= 6 and r. (i + j+ 2k)=5. π/3
B) r. (2i- 3j+ 4k)= 1 and r. (-i + j)=4. cos⁻¹(5/√58)
C) r. (2i- j+ 2k)= 6 and r. (3i + 6j- 2k)=9. cos⁻¹(-4/21)
D) r. (2i+ 3j- 6k)= 5 and r. (i -2 j+ 2k)=9 cos⁻¹(16/21)
2) Find the angle between the planes
A) x+ y+ 2z= 9 and 2 x- y+ z= 15. π/3
B) 2x- y+ z= 4 and x+ y+ 2z= 3. π/3
C) x+ y- 2z= 3 and 2 x- 2y+ z= 5. cos⁻¹(2/3√6)
D) x- y+ z= 5 and x+ 2y+ z= 9. π/2
E) 2x- 3y+ 4z= 1 and -x+ y= 4. cos⁻¹(5/√58)
F) 2x+ y- 2z= 5 and 3 x- 6y- 2z= 7. cos⁻¹(4/21)
3) Show that the planes
A) 2x+ 6y+ 6z= 7 and 3x+ 4y- 5z= 8
B) x-2y+ 4z= 10 and 18x+ 17y+ 4z= 49
C) r. (2i- j+k)= 5 and r. (-i - j+ k)=3 are at right angles.
4) Determine the value of λ for which the following planes are perpendicular to each other.
a) r. (2i- j+ λk)= 5 and r. (3i + 2j+ 2k)=4. -2
b) r. (i + 2j+ 3k)= 7 and r. (λi + 2j - 7k)= 26. 17
c) 2x - 4y + 3z =5 and x + 2y + λz = 5. 2
d) 3x - 6y - 2z =7 and 2x + y - λz = 5. 0
5) Find the equation of the plane
a) passing through the point (1,1,-1) and Perpendicular to the plane x + 2y + 3z -7=0 and 2x - 3y + 4z =0. 17x + 2y - 7z =26
b) passing through the point (-1,-1,2) and Perpendicular to the plane 3x + 2y - 3z -1=0 and 5x - 4y + z =5. 5x + 9y + 11z =8
c) passing through the point (1,-3,-2) and Perpendicular to the plane x + 2y + 2z -5=0 and 3x + 3y + 2z =8. 2x - 4y + 3z =8
d) passing through the origin and Perpendicular to each of the plane x + 2y - z = 1 and 3x - 4y + z =5. x + 2y - 3z +3 =0
6) Find the equation of the plane passing through the point (1,-1,2) and (2,-2,2) and which is Perpendicular to the plane 6x - 2y + 2z -9=0. x + y - 2z +4=0
7) Find the equation of the plane passing through the point (2,1,-1) and (-1,3,4) and Perpendicular to the plane x - 2y + 4z = 10, also that the plane thus obtained contains the line r= - i+ 3j + 4k + λ(3i - 2j - 5k). 18x +17y + 4z =49.
8) Find the equation of the plane passing through the point (2,2,1) and (9,3,6) and which is Perpendicular to the plane 2x + 6y + 6z = 1. 3x + 4y - 5z =9
9) Find the equation of the plane passing through the points (-1,1,1) and (1, -1,1) and Perpendicular to the plane x + 2y + 2z =5. 2x + 2y - 3z + 3=0
10) Find the equation of the plane passing through the point (3,4,2) and (7,0,6) and is Perpendicular to the plane 2x - 5y = 15. 5x + 2y - 3z -17 =0
11) Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane. y=3
12) Find the equation of the plane that contains the point (passing through the point (1,-1,2) and is perpendicular to each of the planes 2x +3y - 2z - 5=0 and x + 2y - 3z =8. 5x - 4y - z= 7
13) Find the equation of the plane passing through (a,b,c) and parallel to the plane r.(i + j + k)=2. x + y + z = a+ b+ c
14) 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 3 x + 3y +z=0. 7x - 8y + 3z +25=0
15) Find the vector equation of the plane passing through the point (2,1,-1) and (-1,3,4) and which is Perpendicular to the plane x - 2y + 4z = 10. 18x + 17y +14z =49
EXERCISE - G
1) Find the vector equation of the following plane in scalar product form:
a) r= (i - j)+ λ(i + j + k)+ μ(i - 2j+ 3k). r.(5i - 2j - 3k)= 7
b) r= (2i - k)+ λi + μ(i - 2j - k). r.(j - 2k)= 2
c) r= (1+ s - t)i + (2- s) j + (3- 2s + 2t)k. r.(2i + k)= 5
d) r= (i + j)+ λ(i + 2j - k)+ μ(-i + j - 2k). r.(-i +j +k)= 0
e) r= (i - j)+ λ(i + j + k)+ μ(4i - 2j+ 3k). r.(5i + j - 6k)= 4
2) Find the Cartesian form of the equation of the plane
a) r= (s - 2t) i+ (3- t) j + (2s + t)k. 2x - 5y - z = -15
b) r= (i - j)+ s(-i + j + 2k)+ t(i + 2j+ k). x - y + z= 2
c) r= (1+ s + t)i + (2- s + t)+ (3 - 2s + 2t)k. 2y - z = 1
3) Find the vector equation of the plane in non-parametric form:
a) r= (1+ s - t) i+ (2- s) j + (3- 2s + 2t)k. r.(2i + 0j + k)= 5
b) r= (λ - 2μ)i + (3- μ) j + (2λ + μ)k. r.(2i -5j - k)+15=0
c) r= (2i + 2j - k)+ λ(i + 2j + 3k)+ μ(5i - 2j+ 7k). r.(5i + 2j - 3k)= 17
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EXERCISE - H
1) Find the equation of a plane through the intersection of the planes r.(i + 3j - k)= 5 and r.(2i - j + k)=3 passing through the point (2, 1, -2). r.(3i + 2j + 0k)=8
2) Find the equation of the plane containing the line of intersection of the plane x + y + z -6=0 and 2x + 3y + 4z +5=0 and passing through the point (1,1,1). 20x + 23y + 26z -69=0
3) Find the direction ratios of the normal to the plane passing through the point (2,1,3) and the line of intersection of the plane x + 2y + z=3 and 2x - y - z =5. 13,6,1
4) Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z +8=0 and which contains the line of intersection of the plane x+ 2y + 3z -4=0 and 2x+ y - z +5 =0. 51x+ 15y - 50z +173=0
5) Find the equation of the plane through the intersection of r.(2i - 3j + 4k)=1 and r.(i -j) + 4 = 0 and perpendicular to r.(2i - j + k) + 8= 0. r.(-5i + 2j + 12k)=47
6) Find the equation of the plane passing through the intersection of the planes 2x - 3y + z - 4=0 and x - y + z +1= 0 and perpendicular to the plane x + 2y - 3z +6=0. x - 5y - 3z - 23= 0
7) Find the cartesian as well as vector equations of the planes through the intersection of the planes r.(2i + 6j) + 12= 0 and r.(3i - j + 4k)=0 which are at a unit distance from the origin. r.(2i + j + 2k) + 3=0 and r.(-i + 2j -2k)+ 3=0
8) The plane lx + my =0 is rotated through an angle α about its line of intersection with the plane z= 0. prove that the equation of the plane in its new position is lx + my ± √(l²+ m² tanα)z = 0.
9) The plane x - 2y + 3z =0 is rotated through a right angle about the line of intersection with the plane 2x + 3y - 4z - 5 =0, find the equation of the plane in its new position. 22x + 5y - 4z -35=0
10) Find the equation of the plane which is parallel to 2x - 3y + z=0 and which passes through (1,-1,2). 2x - 3y + z=7
11) Find the equation of the plane through (3,4,-1) which is parallel to the plane r.(2i - 3j + 5k)+ 2= 0. r.(2i - 3j + 5k)+ 11= 0.
12) Find the question of the plane passing through the line of intersection of the planes 2x -7y + 4z -3=0, 3x - 5y + 4z +11=0 and the point (-2,1,3). 15x -47y + 28z=7
13) Find the equation of plane through the point 2i + j - k. and passing through the line of intersection of the plane r.(i + 3j - k)= 0 and r.(j + 2k)= 0. r.(i + 9j + 11k)= 0
14) Find the equation of the plane passing through the line of intersection of the planes 2x - y =0 and 3z - y=0 and perpendicular to the plane 4x + 5y - 3z =8. 28x -17y +9z = 0
15) Find the equation of the plane which contains the line of intersection of the planes x+ 2y + 3z -4=0 and 2x + y - z+5=0 and which is perpendicular to the plane 5x + 3y - 6z +8=0. 33x + 45y + 50z - 41=0
16) Find the equation of the plane through the line of intersection of the planes x+ 2y + 3z +4=0 and x - y + z +3=0 and passing through the origin. x - 10y - 5z =0
17) Find the vector equation (in scalar product form) of the plane containing the line of intersection of the planes x - 3y + 2z -5=0 and 2x - y + 3z -1=0 and passing through (1,-2,3). r.(i + 7j)+ 13= 0.
18) Find the equation of the the plane which is perpendicular to the plane 5x + 3y + 6z +8=0 and which contains the line of intersection of the planes x + 2y + 3z -4=0, 2x + y - z +5=0. 51x + 15y - 50z + 173=0
19) Find the equation of the plane through the line of intersection of the planes r.(i + 3j)+ 6= 0 and r.(3i - j - 4k)= 0, which is at a distance from the origin. r.(-2i +4j + 4k)+ 6= 0 or r.(4i + 2j - 4k)+ 6= 0
20) Find the equation of the plane passing through the intersection of the planes 2x + 3y - z +1=0 and x + y - 2z +3=0 and perpendicular to the plane 3x - y - 2z -4=0. 7x + 13y + 4z -9=0
21) Find the equation the plane passing through the intersection of the planes 2x +3y - z +1=0, and x +y - 2z +3 =0 and perpendicular to the plane 3x - y - 2z -4=0. 7x + 13y +4z -9=0
22) Find the equation of the plane that contains the line of intersection of the planes r.(i + 2j + 3k) -4=0 and r.(2i + j - k)+5=0 and which is perpendicular to the plane r.(5i + 3j - 6k)+8=0. 33x + 45y + 50z -41=0
23) Find the equation of the plane passing through the intersection of the planes r.(i + j + k)- 6 =0 and r.(2i + 3j + 4k)+5=0 and the point (1,1,1). r.(20i + 22 j +26 k)= 69
24) Find equation of the plane passing through the intersection of the planes r.(2i + j + 3k) =7, r.(2i + 5j +3 k)=9 and the point (2,1,3). r.(2i -13 j +3 k)=0
25) Find the equation of the plane through the intersection of the planes 3x - y + 2z =4 and x + y + z =2 and the point (2,2,1). 7x - 5y + 4z =8
26) Find the vector 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. r.(i - k)+2=0
27) Find the equation of the plane passing through (a, b, c) and parallel to the plane r.(i + j + k)=2. x+ y+ z = a+ b+ c
EXERCISE - I
1) Find the distance of the point 2i + j - k from the plane r. (i - 2j + 4k)= 9. 13/√21
2) Find the distance of the point (2,1,0) from the plane 2x + y + 2z +5=0. 10/3
3) Prove that if a plane has an interceptive a,b,c and is at a distance of p units from the origin, then 1/a²+ 1/b²+ 1/c²= 1/p²
4) Show that the points i - j + 3k and 3(i + j + k) are equidistant from the plane r. (5i + 2j + k)+ 9 =0 and lie on opposite sides of it.
5) Find the equations of the Planes parallel to the plane x - 2y + 2z -3=0 which is at a unit distance from the point (1,2,3). x - 2y + 2z =0 and x - 2y + 2z - 6=0
6) if the points (1,1, M) and (-3.0.1) be equidistant from the plane r.(3i + 4j -12k)+ 13= 0, find the value of M. 1 or 7/3
7) Find the distance between the points P(6,5,9) and the plane determined the points A(3,- 1, 2), B(5,2,4) and C(-1, -1, 6). 6/√34
8) Find the equation of a plane through the point P(6,5, 9) and parallel to the plane determined by the point A(3, - 1,2) , B(5, 2, 4) and C(-1,-1,6). Also , find the distance of this plane from the point A. 3x - 4y + 3z = 25, 6/√34
7) Two systems of rectangular axes have the same origin. If a plane cuts them at distance of a,b,c and a', b', c' respectively prove that 1/a²+ 1/b²+ 1/c²= 1/a'² + 1/b'² + 1/c'².
8) A variable plane which remains at a constant distance 3p from the origin cut co-ordinate axes at A, B, C. Show that the locus of the centroid of triangle ABC is 1/x²+ 1/y²+ 1/z²= 1/p².
9) A variable plane is at the constant distance p from the origin and meets the coordinate axes in A, B, C. Show that the locus of the centroid of the tetrahedron OABC is 1/x²+ 1/y²+ 1/z²= 16/p².
10) If a variable plane at a constant distance p from the origin meets the coordinate axis in pointr A, B and C respectively. Through these points , planes are drawn parallel to the coordinate planes. Show that the locus of the point of intersection is 1/x²+ 1/y²+ 1/z²= 1/p².
11) Find the distance of the point 2i - j - 4k from the plane r.(3i - 4j +12k)- 9=0. 47/13
12) Show that the points i - j +3k and 3i + 3j +3k are equidistant from the plane r.(5i + 2j - 7k)+9=0.
13) Find the distance of the point (2,3,-5) from the plane x + 2y - 2z -9=0. 3
14) Find the equation of the planes parallel to the plane x + 2y - 2z + 8=0 which are at distance of 2 units from the point (2,1,1). x+ 2y - 2z +4=0 or x + 2y - 2z -8=0
15) Show that the points (1,1,1) and (-3,0,1) are equidistant from the plane 3x + 4y - 12z +13=0.
16) Find the equations of the planes parallel to the plane x - 2y + 2z -3=0 and which are at a unit distance from the point (1,1,1). x - 2y + 2z +2=0, x - 2y + 2z -4=0
17) Find the distance of the point (2,3,5) from the xy plane. 5
18) Find the distance of the point (3,3,3) from the plane r.(5i + 2j - 7k)+o=0. 9/√78
19) if the product of the distance of the point (1, 1,1) from the origin and the plane x - y + z + λ= 0 be 5, find the value of λ. 4
20) Find an equation for the set of all points that are equidistant from the Planes 3x - 4y + 12z = 6 and 4x + 3z =7. 36x + 20y - 21z =61, 67x -20y +99z =121
21) Find the distance between the point (7,2,4) and the plane determined by the points A(2,5,-3), B(-2,-3,5) and (5,3,-3). √29
22) A plane makes intersects -6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it. 12/√29
EXERCISE - J
1) Find the distance between the parallel planes x + y - z +4=0 and x + y - z +5=0. 1/√3
2) Find the distance between the parallel planes 2x - y +2z +3 =0 and 4x -2y +4 z +5=0. 1/6
3) Find the distance between the parallel planes r.(2i - 3j + 6k)=5 and r.(6i - 9j + 18k) + 20 =0. 5/3
4) Find the distance between the parallel planes 2x - y + 3z - 4=0 and 6x - 3y + 9z + 13 =0. 25/3√14
5) Find the equation of the plane passing through the point (3,4,-1) and is parallel to the plane 2x -3y +5z +7 =0. Also, find the distance between the two planes. 2x -3y +5 z +11=0, 4/√38
6) Find the equation of the plane mid-parallel to the plane 2x -2 y +z +3 =0 and 2x - 2y + z +9 =0. 2x - 2y + z +6=0
7) Find the distance between the planes r.(i + 2j + 3k)+ 7=0 and r.(2i + 4j + 6k) + 7 =0. 7/√56
Exercise - K
1) Reduce in symmetrical form, the equation of the line x - y + 2z =5, 3x + y + z =6. (4x-11)/-3= (4y+9)/5= (z -0)/1
2) Reduce in symmetrical form ,the equation of the line x = ay + b, z= cy + d. (x-b)/a = (y-0)/1= (z -d)/c
3) Find the angle between the lines x - 2y + z = 0= x + 2y - 2z and x+ 2y + z =0 = 3x + 9y + 5z. Cos⁻¹(8/√406)
4) Find the angles between the line r= (i+ 2j - k)+ λ(i - j + k) and the plane r.(2i - j + k)= 4. sin⁻¹(2√2/3)
5) Find the angle between the line (x+ 1)/3= (y - 1)/2 = (z -2)/4 and the plane 2x + y - 3z +4=0. sin⁻¹(-4/√406)
6) If the line r= (i- 2j + k)+ λ(2i + j + 2k) is parallel to the plane r. (3i- 2j + mk) = 14, find the value of m. -2
7) Show that the line whose vector equation is r= (2i- 2j +3k)+ λ(i - j + 4k) is parallel to the plane whose vector equations is r. (i+ 5j + k)= 5. Also, find the distance between them. 10/√27
8) Find the vector equation of the line passing through the point with position vector (2i- 3j - 5k) and perpendicular to the plane r. (6i - 3j + 5k) +2=0. r= (2i - 3j - 5k)+ λ(6i - 3j + 5k)
9) Find the equation of the line passing through the point (3, 0, 1) and parallel to the planes x + 2y=0 and 3y - z= 0. (x -3)/-2 = (y - 0)/1 = (z -1)/3
10) Find the equation of the plane passing through the line of intersection of the planes 2x + y - z =3, 5x - 3y + 4z +9=0 and parallel to the line (x -1)/2 = (y - 3)/4 = (z -5)/5. 7x + 9y - 10z -27 =0
11) Find the equation of the plane passing through the intersection of the Planes r. (i + j + k) -1 =0 and r. (2i + 3j -k) +4=0 and parallel to x-axis. r. (- j/2 + 3k/2) = 3 or r. (- j + 3k) =6
12) Find the equation of the plane through the point (1, 0, -1), (3, 2, 2) and parallel to the line (x -1)/1 = (y - 1)/-2 = (z -2)/3. 4x - y - 2z -6=0
13) State when the line r= a + λb is parallel to the plane r.n = d. Show that the line r= (i + j + λ(2i + j + 4k) is parallel to the plane r. (- 2i + k) = 5. Also, find the distance between the line and the plane. 7/√5
14) Find the equation of the plane passing through the intersection of the Planes 4x - y + z =10 and x + y - z = 4 and parallel to the line with direction ratios proportional to 2, 1,1. Find also the perpendicular distance of point (1,1,1) from this plane. 5y - 5z -6=0, 3√2/5
15) Find the equation of the plane passing through the point A(1, 2, 1) and perpendicular to the line joining the points P(1, 4, 2) and Q(2,3,5). Also, find the distance of this plane from the line (x +3)/2 = (y - 5)/-1 = (z -7)/-1. x - y +3z -2 =0, √11
16) Find an equation for the line that passes through the point (2, 3, 1) and is parallel to the of intersection of the planes x + 2y - 3z = 4 and x - 2y +z = 0. (x -2)/1 = (y - 3)/1 = (z -1)/1.
17) Find the plane passing through the point (4,-1,2) and parallel to the lines (x +2)/3 = (y - 2)/-1 = (z +1)/2 and (x -2)/1 = (y - 3)/2 = (z -4)/3. x + y - z -1= 0
18) Find the angle between the line r= (2i + 3j + 9k)+ λ(2i + 3j + 4k) and the plane r. (i + j + k) = 5. sin⁻¹(3√3/29)
19) Find the angle between and the line (x -1)/1 = (y - 2)/-1 = (z +1)/1 and the plane 2x + y - z = 4. 0
20) Find the angle between the line joining the points (3,-4,-2) and (12,2,0) and the plane 3x - y + z =1. sin⁻¹(23/11√11)
21) The line r= i + λ(2i - mj - 3k) is parallel to the plane r. (mi + 3j + k) = 4. find m. -3
22) Show that the line whose vector equation is r= (2i + 5j + 7k)+ λ(i + 3j + 4k) is parallel to the plane whose vector equation is r. (i + j - k) = 7. Also, find the distance between them. 7/√3
23) Find the vector equation of the line through the origin which is perpendicular to the plane r. (i + 2j + 3k) = 3. r= λ(i - 2j + 3k)
24) Find the equation the plane through (2, 3, -4) and (1,-1,3) and parallel to x-axis. 7y + 4z -5=0
25) Find the equation of a plane passing through the points (0, 0, 0) and (3, -1,2) and parallel to the line (x -4)/1 = (y + 3)/-4 = (z +1)/7. x - 19y - 11z = 0
26) Find the vector and cartesian equations of the line passing through (1, 2, 3) and parallel to the Planes r. (i - j + 2k) = 5 and r. (3i + j + 2k) = 6. r= (i + 2j + 3k)+ λ(-3i + 5j + 4k) ; (x -1)/-3 = (y -2)/5 = (z -3)/4
27) Prove that the line of section of the plane 5x + 2y - 4z +2=0 and 2x + 8y + 2z -1=0 is parallel to the plane 4x - 2y - 5z - 2=0.
28) Find the vector equation of the line passing through the point (1, -1, 2) and perpendicular to the plane 2x - y + 3z - 5 =0. r= (i - j + 2k)+ λ(2i - j + 3k)
29) Find the equation the plane through the points (2, 2, -1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6. 12x +15y - 14z - 68=0
30) Find the angle between the line (x -2)/3 = (y + 1)/- 1 = (z -3)/2 and the plane 3x + 4y + z + 5= 0. sin⁻¹√(7/52)
31) Find the equation 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. Find also the perpendicular distance of (1, 1, 1) from the plane. 9x - 8y + 7z = 21, 13/√194
32) State when the line r= a + λb is parallel to the plane r.n = d. Show that the line r= (i +j )+ λ(3i - j + 2k) is parallel to the plane r. (2i +k) = 3. Also , find the distance between the line and the plane. b.n =0, 1/√5
33) Show that the plane whose vector equation r. (i +2 j -k) =1 and the line whose vector equation is r= (-i +j + k)+ λ(2i + j + 4k) are parallel. Also , find the distance between them. 1/√6
34) Find the equation of the plane through the intersection of the plane 3x - 4y + 5z =10 and 2x + 2y - 3z = 4 and parallel to the line x = 2y = 3z. x - 20y + 27z = 14
35) Find the vector and cartesian forms of the equation of the plane passing through the point (1,2,-4) and parallel to the lines r= (i +2j -4k)+ λ(2i + 3j + 6k) and r= (i -3j + 5k)+ μ(i + j - k). Also, find the distance of the point (9,-8,-10) from the plane thus obtained. r. (-9i + 8j - k)= 11, 9x + 8y - z= 11, √146
36) Find the equation of the plane passing through the point (3,4,1) and (0,1,0) and parallel to the line (x +3)/2= (y -3)/7 = (z -2)/5. 8x - 13y + 15z +13=0
37) Find the coordinates of the point where the line (x -2)/3 = (y +1)/4 = (z -2)/2 interesects the plane x - y + z -5=0. Also, find the angle between the line and the plane. (2,-1,2), sin⁻¹(1/√87)
38) Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane r. (i + 2j - 5k) -9 = 0. r= (i +2j + 3k)+ λ(i + 2j - 5k)
39) Find the angle between the line (x +1)/2 = y/3 = (z -3)/6 and the plane 10x + 2y -11z =3. sin⁻¹(8/21)
40) Find the vector equation of the line passing through (1,,2,3) and parallel to the planes r (i - j + 2k) = 5 and r(3i + j + k) = 6. r= (i +2j + 3k)+ λ(-3i + 5j +4k)
41) Find the value of λ such that the line (x -2)/6 = (y-1)/λ = (z +5)/-4 is perpendicular to the plane 3x - y - 2z = 7. -2
42) Find the equation of the plane passing through the points (-1,2,0), (2,2,-1) and parallel to the line (x -1)/1 = (2y +1)/2 = (z +1)/-1. r= (i +2j + 3k)+ λ(-3i + 5j +4k)
EXERCISE - L
1) Find the coordinates of the point where the line through the points
A) A(3,4,1) and B (5,1,6) crosses the XY-plane. (13/5,23/5,0)
B) A(5,1,6) and B (3,4,1) crosses the yz-plane. (0,17/2,-13/2)
C) A(5,1,6) and B (3,4,1) crosses the zx plane. (17/3,0,23/3)
2) Find the coordinates of the point where the line through (3,-4,-5) and (2,-3,1) crosses the plane 2x+y+z= 7. (1,-2,7)
3) Find the distance between the point with position vector -i - 5j - 10k and the point of intersection of the line (x-2)/3= (y+1)/4 = (z-2)/12 with the plane x - y +z = 5. 13
4) Find the distance between the point with position vector (-1, - 5 - 10) from the point of intersection of the line r= (2i - j+ 2k)+ K(3i+ 4j+ 2k) and the plane r.(i - j+ k)= 5. 13
5) Find the distance between the point with position vector (2,12,5) from the point of intersection of the line r= (2i - 4j+ 2k)+ K(3i+ 4j+ 2k) and the plane r. (i - 2j+ k)= 0. 13
6) Find the distance between the point (1,-2,3) from the plane x- y +z= 5 measured parallel to the line whose direction cosines are proportional to 2,3,-6. 7/6
7) Find the distance between the point (-1,-5,-10) from the point of intersection of the joining the points A(2,-1,2) and B(5,3,4) with the plane x- y +z= 5. 13
8) Find the distance between the point (3,4,4) from the point, where the line joining the points A(3,-4,-5) and B(2,-3,1) intersect the plane 2x +y +z= 7. 7
CONDITION FOR A LINE TO LIE IN A PLANE:
Vector form:
If the line r= a + ¥b lies in the plane r. n= d, then
A) b.n= 0
B) a.n = d.
Cartesian form:
If the line (x-x₁)/l= (y -y₁)/m= (z -z₁)/n lies in the plane ax+ by+ cz + d= 0, then
A) ax₁ + by₁ + cz₁+ d= 0
B) al + bm+ cn = 0.
CONDITION OF PERPENDICULARITY OF TWO LINES AND EQUATION OF THE PLANE CONTAINING THEM
Vector form:
If the lines r= a₁ +¥b₁ and r= a₂ +€b₂ are coplanar, then
r₁. (b₁ x b₂= a₂. (b₁ x b₂)
OR
[r b₁ b₂ ]= [a₂ b₁ b₂]
And the equation of the plane containing them is
r. (b₁ x b₂) = a₁. (b₁ x b₂)
OR
r. (b₁ x b₂)= a₂. (b₁x b₂)
Cartesian form:
If the line (x- x₁)/l₁ = (y- y₁)/m₁ = (z- z₁)/n₁ and (x- x₂)/l₂ = (y- y₂)/m₂ = (z- z₂)/n₂ are coplanar, then
Mode of
(x₂- x₁) (y₂- y₁) (z₂- z₁)
l₁ m₁ n₁ = 0
l₂ m₂ n₂
And the equation of the plane containing them is
(x- x₁) (y- y₁) (z- z₁)
l₁ m₁ n₁ = 0
l₂ m₂ n₂
OR
(x-x₂) (y- y₂) (z - z₂)
l₁ m₁ n₁ = 0
l₂ m₂ n₂
EXERCISE-M
1)a) Show that the lines r= (i+ j- k)+ ¢(3i - j) and r= (4i- k)+ ¥(2i +3k) are coplanar. Also, find the plane containing these two lines. . r. (3i + 9j - 2k)= 14
b) Show that the lines r= (2i- 3k)+ ¢(i +2j+3k) and r= (2i+ 6j + 3k)+ ¥(2i+ 3j+ 4k) are coplanar . Also , find the equation of the plane containing them. r. (i -2j +k)= -7
c) show that the lines (x+1)/-3 = (y- 3)/2 = (z+2)/1 and x/1 = (y- 7)/-3 = (z+7)/2 are coplanar. Also, find the plane containing these two lines. x+ y+ z= 0
d) Show that the lines (x+1)/3 = (y+ 3)/5 = (z+5)/7 and (x-2)/1 = (y- 4)/4 = (z-6)/7 are coplanar. Also, find the plane containing these two lines. x- 2 y+ z= 0
2) Find the equation of the plane containing the line (x+1)/-3 = (y- 3)/2 = (z+2)/1and the point (0,7,-7) and show that the line x/1 = (y- 7)/-3 = (z+7)/2 also lies in the same plane. x+ y+ z= 0
3) Find the equation of the plane which contains two parallel lines (x-4)/1 = (y- 3)/-4 = (z-2)/5 and (x-3)/1 = (y+2)/-4 = z/5. 11x - y - 3z = 35
4) Show that the lines (x+4)/3 = (y+6)/5 = (z-2)/-2 and 3x-2y+ z= 0= 2x+ 3y+ 4z - 4 intersect. Find the equation of the plane in which they lie and also their point of intersection. (2,4,-3), 45x- 17y+ 25z+53= 0
5) a) Show that the plane whose vector equation is r. (i+2j- k)= 3 contains the line whose vector equation is r= (i +j)+ ¥(2i+ j + 4k).
6) Find the equation of the plane determined by the intersection of the lines (x+3)/3 = y/-2 = (z-7)/6 and (x+6)/1 = (y+5)/-3 = (z-1)/2. 2x - z +13=0
7) Show that the lines (x- a +d)/(a - δ)= (y - a)/α = (z - a - d)/(α+ δ) and (x - b + c)/(β- γ) = (y - b)/β = (z - b - c)/(β+ γ).
8) Find the vector equation of the plane that contains the lines r= (i+ j)+ ¥(i +2j - k) and r= (i+ j)+ ¢(-i + j -2k). Also , find the length of the perpendicular drawn from the point (2,1,4) to the plane thus obtained. r.(-i + j +k)= 0, √3
9) Find the equation of the plane passing through the point (0,7,-7) and containing the line (x+1)/-3 = (y- 3)/2 = (z+2)/1. x+ y + z=0
10) Show that the plane whose vector equation is r.(i + 2j - k)= 3 contains the line whose vector equation is r= (i+ j)+ λ(2i+ j + 4k).
11) Find the vector and cartesian equations of the plane containing the two lines r= (2i + j - 3k + λ(i+ 2j + 5k) and r= (3i + 3j + 2k) + μ(3i- 2j + 5k). r. (10i + 5j - 4k)= 37, 10x + 5y - 4z = 37
12) If 4x+ 4y + λz= 0 is the equation of the plane through the origin that contains the line (x-1)/2= (y+1)/3 = z/4, find the value of λ. 5
13) If the lines (x-1)/2 = (y+1)/3 = (z-1)/4 and (x-3)/1 = (y- k)/2 = z/1 intersect, then find the value of k, and hence find the equation of the plane containing these lines. 5x - 2y - z =6
14) Find the equation of the plane determined by the intersection of the lines (x+3)/3 = y/-2 = (z-7)/6 and (x+6)/1 = (y+5)/-3 = (z-1)/2. 2x - z +13=0
15) Find the vector equation of the plane passing through the points (3,4,2) and (7,0,6) and perpendicular to the plane 2x - 5y -15=0. Also, show that the plane thus obtained contains the line r= i + 3j - 2k + λ(i - j + k). r(5i + 2j - 3k)=17
16) If the lines (x-1)/-3 = (y-2)/-2k = (z-3)/2 and (x-1)/k = (y-2)/1 = (z-3)/5 are perpendicular, find the values of k and hence find the equation of the plane containing these lines. 2, -22x + 19y + 5z =31
17) Find the coordinates of the point where the line (x-2)/3 = (y+1)/4 = (z-2)/2 intersect the plane x - y + z -5=0. Also, find the angle between the line and the plane. (2,-1,2), sin⁻¹(1/√87)
18) Find the vector equation of the plane passing through three points with position vectors i+ j - 2k, 2i - j + k and I+ 2j + k. Also, find the coordinates of the point of intersection of this plane and the line r= 3i - j - k + λ(2i - 2j + k). r(9i + 3j - k)= 14, (1,1,-2)
19) Show that the lines (5-x)/-4 = (y-7)/4 = (z+3)/-5 and (x-8)/7 = (2y-8)/2 = (z-5)/3 are coplanar .
20) Find the equation of a plane which passes through the point (3,2,0) and contains the line (x- 3)/1 = (y - 6)/5 = (z- 4)/4. x - y + z -1=0
EXERCISE - N
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