Vector Equation of Plane in Normal Form Class 12 OP Malhotra Exe- 24B ISC Maths Solutions Ch-24 Plane. In this article you would learn to solve questions / Problems on vector equation of a plane in normal form. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Vector Equation of Plane in Normal Form Class 12 OP Malhotra Exe- 24B ISC Maths Solutions Ch-24 Plane

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-24	The Plane
Writer	OP Malhotra
Exe-24(b)	Vector Equation of Plane in Normal Form

Vector Equation of Plane in Normal Form

The Plane Class 12 OP Malhotra Exe-24B Solutions

Que-1: Find the vector equation of a plane at a distance of 6 units from the origin and has i as the unit vector normal to it.

Sol: Given p = length of ⊥ origin from plane
= 6 units
n̂ = î = unit vector normal to plane
∴ eqn. of required plane be
r→  n→ = p ⇒ r→  î = 6

Que-2: Find the vector equation of a plane which is at a distance of
(i) 8 units from the origin and which is normal to the vector î + 2ĵ – 2k̂.
(ii) 5 units from the origin and which is normal to the vector 2î + 6ĵ – 3k̂.

Sol: (i) Here p= length of ⊥ from origin to plane =8 units
Here n→   = î + 2ĵ – 2k̂ and
|n→ | = |î+2ĵ−2k̂|
= √1²+2²+(−2)² = 3
i.e. |n→ | < 1
⇒ n→  is not a unit vector Thus required eqn. of plane be
r→  n→  = p
i.e. r→  (î+2ĵ−2k̂)/3 = 8
⇒ r→  (î + 2ĵ – 2k̂) = 24

c(ii) Here p = length of ⊥ from origin to reqd. plane = 5 units
and n→  = 2î + 6ĵ – 3k̂
and |n→  | = |2î + ĵ – 3k̂|
= √2²+6²+(−3)²
= √49 = 7 < 1
∴ n→  is not a unit vector.
Thus required eqn. of plane be r→ n̂ = p
i.e. r→ (2î+6ĵ−3k̂)/7 = 5
⇒ r→ (2î + 6ĵ – 3k̂) = 35

Que-3: Find the cartesian equation of the following planes
(i) r→ (î + 3ĵ – 4k̂) = 1
(ii) r→ (3î – 7ĵ + k̂) + 3 = 0
(iii) r→ = (λ + 3µ)î + (2 – µ)ĵ + (µ + 2λ)k̂

Sol: (i) Given vector eqn. of plane be
r→ (2î + 3ĵ – 4k̂) = 1
Since r→ = x î + yĵ + zk̂ be the P.V. of any point (x, y, z) on plane (1). Thus eqn. (1) becomes:
(xî + yĵ + zk̂) (2î + 3ĵ – 4k̂) = 1
2 x + 3 y – 4 z = 1 be the required cartesian eqn. a of plane.

(ii) Given vector eqn. of plane be
r→ (3î – 7ĵ + k̂) + 3 = 0
Since r⃗ = xî + yĵ + zk̂ be the P.V. of any point (x, y, z) on plane (1).
Then eqn. (1) becomes;
(xî + yĵ + zk̂) (3î – 7ĵ + k̂) + 3 = 0
i.e. 3 x – 7 y + z + 3 = 0
which is the required carterian eqn. of plane.

(iii) r→ = (λ + 3µ )î + (2 – µ)ĵ + (µ + 2λ)k̂
Let r→= xî + yĵ + z k^ be the P.V. of any point on plane (1).
Thus, (xî + yĵ + zk̂)
= (λ + 3µ )î+ (2 – µ)ĵ + (µ + 2λ)k̂
on comparing the coefficients of î, ĵ k̂ on both sides; we have
x = λ + 3 µ
y = 2 – µ
z = µ + 2λ
eqn.(2) + eqn. (3); we have
y + z = 2 + 2λ
⇒ λ = y+z−2/2
∴ from (1); we have
3 µ = x – y+z−2/2 = 2x−y−z+2/2
⇒ µ = 2x−y−z+2/6
∴ from (2);
y = 2 – 2x−y−z+2/6
⇒ 6 y = 12 – 2 x + y + z – 2
⇒ 2 x + 5 y – z = 10
which is the required cartersian eqn. of plane.

Que-4: Find the vector equation of the following planes
(i) 5 x – 7 y + 2 z = 3
(ii) x – 2 y + 3 z + 1 = 0

Sol: (i) Given eqn. of plane be
5 x – 7 y + 2 z = 3
Let xî + yĵ + zk̂ be the P.V. of any point on plane (1).
∴ eqn. (1) can be written are :
(xî + yĵ + zk̂) (5î – 7ĵ + 2k̂) = 3
⇒ r→ (5î – 7ĵ + 2k̂) = 3

(ii) Given eqn. of plane in cartesian form be
x – 2 y + 3 z + 1 = 0
Let r⃗ = xî + yĵ + zk̂ be the P.V. of any point (x, y, z) on plane (1). Then eqn. (1) can be written as:
(xî + yĵ + zk̂) (î – 2ĵ + 3k̂) + 1 = 0
⇒ r→ (î – 2ĵ + 3k̂) = -1
which is the required vector eqn. of plane.

Que-5: Find the direction cosines of the perpendicular from the origin to the plane
(i) r→ (2î – 3ĵ – 6k̂) + 5 = 0
(ii) r→ (2î – 2ĵ + k̂) + 2 = 0

Sol: (i) Given vector eqn. of plane be
r→ (2î – 3ĵ – 6k̂) + 5 = 0
⇒ r→ (2î – 3ĵ – 6k̂) = -5
The normal form of the plane be
r→ n̂ = p
where p = + ve.
So eqn. (1) can be written as :
r→ (-2î + 3ĵ + 6k̂) = 5
Here n→ = -2î + 3ĵ + 6k̂ and
|n→ | = √4+9+36 = 7 < 1
∴ n→ is not a unit vector. Thus eqn. (2) can be written as
r→ (−2î+3ĵ+6k̂)/|n→ | = 5/|n→ |
i.e. r→ (−2î+3ĵ+6k̂)/7 = 5/7
which is the req. normal form of plane. Then direction cosines of the ⊥ from origin to plane are < −2/7, 3/7, 6/7 >

(ii) Given vector eqn. of plane be
r→ (2î – 2ĵ + k̂) = -2
The normal form of plane be
r→ n̂ = p
where p be positive.
but here p = -2 < 0
So eqn. (1) can be written as :
r→ (-2î + 2ĵ – k̂) = 2
Further n→ = -2î + 23ĵ – k̂
|n→ | = √4+4+1 = 3 < 1
∴ n⃗ is not a unit vector.
So dividing eqn. (2) throughout by |n→ | i.e. 3 ; we get
r→ (−2î+2ĵ−k̂)/3 = 2/3,
Here n̂ = −2/3î + 2/3ĵ – k̂/3
Thus direction cosines of the ⊥ from (0,0,0) to given plane are
< −2/3, 2/3, – 1/3 >

Que-6: Reduce the equation to normal form and hence find length of perpendicular from the origin to the plane.
(i) r→ (3î – 4ĵ + 12k̂) = 5
(ii) r→ (4î – 2ĵ + 4k̂) = 18

Sol: (i) Given vector eqn. of plane be
r→ (3î – 4ĵ + 12k̂) = 5
on comparing with r → n→ = d
i.e. n→= 3î – 4ĵ + 12k̂
and |n→| = √3²+(−4)²+12²
= √169 = 13 < 1
∴n→ is not a unit vector and required unit vector
n̂ = n→/|n→| = (3î−4ĵ+12k̂)/13
Thus normal form of plane be
r→ (3î−4ĵ+12k̂)13 = 5/13
Thus length of ⊥ from (0,0,0) to given plane = 5/13 units

(ii) Given vector eqn. of plane be
r→ (4î – 2ĵ + 4k̂) = 18
On comparing with r→  n→= d
More n→= 4î – 2ĵ + 4k̂
and
|n→| = √4²+(−2)²+4²
= √36 = 6 < 1
∴ n→  is not a unit vector and reqd. unit vector
n→ = n→/|n→| = (4î−2ĵ+4k̂)/6
Thus required normal form of given plane be given by
r→ (4î−2ĵ+4k̂)/6 = 18/6 = 3
Thus, reqd. length of ⊥ from origin to given plane = 3

Que-7: The vector equation of a plane is
r→ (2î – ĵ + 2k̂) = 9 , where (2î –ĵ + 2k̂) is normal to the plane. Find the length of perpendicular from the origin to the plane.

Sol: Given vector eqn. of plane be
r→ (2î – ĵ+ 2k̂) = 9
Here n⃗ = normal vector to plane (1)
= 2î – ĵ + 2k̂
and |n→| = √2²+(−1)²+2² = √9 = 3 < 1
Thus n→ is not a unit vector.
∴ required unit vector
n→= n→ /|n→ | = (2î−ĵ+2k̂)/3
Thus eqn. (1) can be written as :
r→ (2î−ĵ+2k̂)/3 = 9/3 = 3
which is of the normal form r→   n→= p
Thus, p = length of ⊥ from origin to given plane =3 units.

Que-8: Find the vector equation of a plane which is at a distance of 5 units from the origin and has -1,2,2 as the direction ratios of a normal to it.

Sol: Given p = length of ⊥ from (0,0,0) to required plane = 5 units.
Given direction ratios of normal to plane are < -1, 2, 2 >
∴ direction cosines of normal to plane are
< −1/√1+4+4, 2/√1+4+4, 2/√1+4+4 >
< −1/3, 2/3, 2/3 >
Thus n̂ = −3î + 23ĵ + 23k̂
Hence the required normal form of plane be
r→ n̂ = p r→ (−1/3î + 2/3ĵ + 2/3k̂) = 5

Que-9: Find a unit normal vector to the plane,
x + 2 y + 3 z – 5 = 0

Sol: Given cartesian eqn. of plane be
x + 2 y + 3 z – 5 = 0
Let r→ = xî + yĵ + zk̂ be the P.V. of any point (x, y, z) on plane (1).
∴ eqn. (1) can be written as :
(xî + yĵ + z k̂) (î + 2ĵ + 3k̂) – 5 = 0
i.e. r→ (î + 2ĵ + 3k̂) = 5
which is of the form r→  →n = 1→→
→ →Here n→ = î + 2ĵ + 3k̂
and n→= √1²+2²+3² = √14 < 1
∴ n⃗ is not a unit vector.
∴ required unit vector
= n̂ = n→/ |n→ | = î+2ĵ +3k̂/√1²+2²+3²
∴ n̂ = 1/√14 î + 2/√14 ĵ + 3/√14k̂

Que-10: Find the vector equation of a plane passing through a point having position vector 2 i + 3 j – 4 k and perpendicular to the vector 2 i – j + 2 k. Also reduce it to Cartesian form.

Sol: We know that eqn. of plane passing through the point whose P.V. is a→ and ⊥ to n→ is
(r→ – a→→→) n→ = 0
Here given a→ = 2î + 3ĵ – 4k̂ and n⃗ = 2î –ĵ + 2k̂
∴ eqn. (1) becomes,
r→  n→ = a→→ n→→
⇒ r→ (2î – ĵ + 2k̂)
= (2î + 3ĵ – 4k̂) (2î – ĵ + 2k̂)
⇒ r→ (2î – ĵ + 2k̂)
= 2(2) + 3(-1) – 4(2)
= -7
which is the required vector eqn. of plane.
Let r→ = xî + yĵ + zk̂ be the P.V. of any point (x, y, z) on required plane (2).
∴(xî + yĵ + zk̂) (2î – ĵ + 2k̂) = -7
⇒ 2 x – y + 2 z + 7 = 0
which is the required cartesian eqn. of plane.

Que-11: Find the equation of the plane through the points 2î + 3ĵ – k̂ and perpendicular to the 3î – 2ĵ – 2k̂. Determine the perpendicular distance of this plane from the origin.

Sol: We know that eqn. of plane passing through to the point whose P.V. a⃗ and normal to the vector n⃗ is given by
r→   n→= a→  n→
Here a→ = 2î + 3ĵ – k̂ and
n→ = 3î – 2ĵ – 2k̂
∴ eqn. (1) becomes ;
r⃗ (3î – 2ĵ – 2k̂)
= (2î + 3ĵ – k̂)(3î – 2ĵ – 2 k̂)
= 2(3) + 3(-2) – 1(-2) = 2
Which is the required vector eqn. of plane p = length of ⊥ from (0, 0, 0) to plane (2)
= |a→ n→ ||n→ | = 2/√3²+(−2)²+(−2)²
= 2/√17 units

Que-12: Find the equation of the plane through the points 2î + 3ĵ + 4k̂ and perpendicular to the vector 6î + 4ĵ + 3k̂. Put the above equation in normal form.

Sol: We know that eqn. of plane passing through the point whose P.V. is a⃗ and normal to n⃗ is given by
(r→ – a→ ) n→ = 0
⇒ r→ n→ = a→ n→
Here a→ = 2î + 3ĵ + 4k̂
and n→ = 6î + 4ĵ + 3k̂
∴ eqn. (1) becomes;
r→ (6î+ 4ĵ + 3k̂)
= (2î + 3ĵ + 4k̂) (6î + 4ĵ + 3k̂)
= 2(6) + 3(4) + 4(3) = 36
Which is of the form r→  n→ = d
Here n→ = 6î + 4ĵ + 3k̂
i.e. |n→ | = √36+16+9 = √61 < 1
∴ n⃗ is not a unit vector.
Thus the required normal form of plane be
r→  n →/ |n→ | = d/|n→ |
i.e. r→ (6î+4ĵ+3k̂)/√61 = 36/√61

Que-13: Find the vector equations of the coordinate planes.

Sol: (i) eqn. of XOY plane be z = 0
i.e. r→  k→ = 0
Where r→ = xî + yĵ + zk̂ be the P.V. of any point (x, y, z) on XOY plane.
(ii) eqn. of YOZ plane be x = 0 i.e. r⃗ î = 0
Where r→ = x î + yĵ + zk̂
(iii) eqn. of ZOX plane be y = 0 i.e. r⃗ ĵ = 0
Where r→ = x î+ yĵ + zk̂

Que-14: Show that the normals to the following pairs of plants are perpendicular to each other.
(i) x – y – z – 2 = 0 and 3 x + 2 y + z + 4 = 0
(ii) r→ (2î – ĵ + 3 k̂) = 5 and
r→ (2î – 2 ĵ – 2 k̂) = 5

Sol: (i) Given eqns. of planes are
x – y – z – 2 = 0
and 3 x + 2 y + z + 4 = 0
Vector form of planes (1) and (2) are :
r→ (î – ĵ – k̂) = 2
and
r→ (3î + 2ĵ + k̂) + 4 = 0
Here
n1 → =î – ĵ – k̂
n2 → = 3î + 2ĵ + k̂
n1 → n2 → = (î –ĵ – k̂) (3î + 2ĵ + k̂)
= 1(3) – 1(2) – 1(1) = 0
Thus n1 → ⊥ n2 →

(ii) Given eqn. of planes are
r→ (2î – ĵ + 3k̂) = 5
and r→ (2î – 2ĵ – 2k̂) = 5
Let n1 → and n2 → be the vectors normal to given planes (1) and (2).
i.e. n1 → = 2î –ĵ + 3k̂
and n2 → = 2î – 2ĵ – 2k̂
Here,
n1 → n2 → = (2î – ĵ + 3k̂) (2î – 2ĵ – 2k̂)
= 2(2) – 1(-2) + 3(-2)
= 6 – 6 = 0
Thus n1 → ⊥ n2 →

Que-15: Show that the normal vector to the plane 2 x + 2 y + 2 z = 3 is equally inclined with the coordinate axes.

Sol: Given eqn. of plane be
2 x + 2 y + 2 z = 3
So its vector form be
r→ (2î + 2ĵ + 2k̂) = 3
Where r⃗ = xî + yĵ + zk̂ be the P.V. of any point on plane (1).
Eqn. (2) is the form
r→  n→ = d
i.e. n→ = vector normal to plane (2)
= 2î + 2ĵ + 2k̂
Let α, β, γ be the angle made by n⃗ with x axis, y-axis and z-axis respectively.
∴ cosα = n→î/|n→||î| = (2î+2ĵ+2k̂)î/√4+4+4.1
= 2/√12 = 2/2√3 = 1/√3
and
cos β = n→ĵ/|n→||ĵ| = (2î+2ĵ+2k̂)⋅ĵ/√4+4+4 .1
= 2/2√3 = 1/√3
Also, cosα = n→ ⋅k̂/|n→ ||k̂| = (2î+2ĵ+2k̂)⋅k̂/√4+4+4⋅1
= 2/√12 = 2/2√3= 1/√3
∴ cosα = cosβ = cosγ = 1/√3
⇒ α = β = γ
Hence, the normal vector to given plane is equally inclined to coordinate axes.

–: End of Vector Equation of Plane in Normal Form Class 12 OP Malhotra Exe- 24B ISC Maths Solutions :–

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