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Given: Equation of plane is 5y + 4 = 0 

Formula Used: Equation of a plane is lx + my + nz = p where (l, m, n) are the direction cosines of the normal to the plane and (x, y, z) is a point on the plane and p is the distance of plane from origin. 

Explanation: Given equation is 5y = -4 

Dividing by -5,

- y = 4/5

which is of the form lx + my + nz = p where l = 0, m = -1, n = 0 

Therefore, direction cosines of the normal to the plane is (0, -1, 0)

We know that the equation of ZOX plane is y = 0 so a plane parallel to plane ZOX will have the equation y = constant

Now, it is given that the plane makes an intercept of 3 on y - axis so the value of constant is equal to 3.

Therefore, the equation of the required plane is y = 3.

Given \(\bar{r}\)(2\(\hat{i}\) - 3\(\hat{j}\) - \(\hat{k}\)) = 0 and the vector has position vector (\(\hat{i}\) - 2\(\hat{j}\) + 5\(\hat{k}\))

To find – The vector equation of the line passing through (1, - 2, 5) and perpendicular to the given plane 

Tip – The equation of a plane can be given by \(\bar{r}\).\(\bar{n}\) = d where \(\bar{n}\) is the normal of the plane 

A line parallel to the given plane will be in the direction of the normal and will have the direction ratios same as that of the normal. Formula to be used – If a line passes through the point (a, b, c) and has the direction ratios as (a’, b’, c’), then its vector equation is given by \(\bar{r}\) = (a\(\hat{i}\) + b\(\hat{j}\) + c\(\hat{k}\)) + λ(a'\(\hat{i}\) + b'\(\hat{j}\) + c'\(\hat{k}\)) where λ is any scalar constant 

The required equation will be \(\bar{r}\) = (\(\hat{i}\) - 2\(\hat{j}\) + 5\(\hat{k}\)) + λ(2\(\hat{i}\) - 3\(\hat{j}\) - \(\hat{k}\)) for some scalar λ

Given Eq. of plane is 2x – 3y + z = 0 …… (1)

We know that equation of a plane parallel to given plane (1) is

2x – 3y + z + k = 0 …… (2)

As given that , plane (2) is passing through the point (1, – 1, 2) so it satisfy the plane (2),

2(1) – 3( – 1) + (2) + k = 0

2 + 3 + 2 + k = 0

7 + k = 

 k = – 7

put the value of k in equation (2),

2x – 3y + z – 7 = 0

So, equation of the required plane is , 2x – 3y + z = 7

Plane passes through (1,1,2) and (2,- 2,2), 

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

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

Subtracting (1) from (2), 

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

A - 3B = 0 (3) 

Now plane is perpendicular to 

6x - 2y + 2z = 9 6A - 2B + 2C = 0 (4) 

Using (3) in (4) 

18A - 2B + 2C = 0 

16B + 2C = 0 

C = - 8B 

Putting values in equation (1) 

3B(x - 1) + B(y + 2) - 8B(z - 2) = 0 

B(3x - 3 + y + 2 - 8z + 16) = 0 

3x + y - 8z + 15 = 0

Given planes, 2x – 4y + 3z = 5 and x + 2y + λz = 5

We know that planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 are at right angles

 if, a₁a₂ + b₁b₂ + c₁c₂ = 0 ……(a)

We have, a₁ = 2, b₁ = – 4, c₁ = 3 and a₂ = 1, b₂ = 2, c₂ = λ

Using (a) we have, a₁a₂ + b₁b₂ + c₁c₂ = (2)(1) + (– 4)(2) + (3)(λ) = 0

⇒ 2 – 8 + 3λ = 0

⇒ 6 = 3λ

⇒ 2 = λ

Hence, for λ = 2 the given planes are perpendicular.

Let the equation of plane be

A₁x + B₁y + C₁z + D₁ = 0

Direction ratios of parallel planes are related to each other as

\(\frac{A_1}{A_2}\) = \(\frac{B_1}{B_2}\) = \(\frac{C_1}{C_2}\) = k (constant)

Putting the values from the equation of a given parallel plane,

\(\frac{A_1}{5}\) = \(\frac{B_1}{-3}\) = \(\frac{C_1}{7}\) = k

A₁ = 5k, B₁ = - 3k, C₁ = 7k

Putting in equation plane

5kx - 3ky + 7kz + D₁ = 0

As the plane is passing through (0,0,0), it must satisfy the plane,

5k.0 - 3k.0 + 7k.0 + D₁ = 0

D₁ = 0

5kx - 3ky + 7kz = 0 

5x - 3y + 7z = 0 

So, required equation of plane is 5x - 3y + 7z = 0.

Formula : Plane = r . (n) = d 

Where r = any random point 

n = normal vector of plane 

d = distance of plane from origin 

If two planes are parallel , then their normal vectors are same. 

Therefore , 

Parallel Plane r . (i + j + k) = 2 

Normal vector = (i + j + k) 

∴ Normal vector of required plane = (i + j + k) 

Equation of required plane r . (i + j + k) = d 

In cartesian form x + y + z = d 

Plane passes through point (a,b,c) therefore it will satisfy it. 

(a) + (b) + (c) = d 

d = a + b + c 

Equation of required plane r . (i + j + k) = a + b + c

Formula : Plane = r . (n) = d 

Where r = any random point 

n = normal vector of plane 

d = distance of plane from origin 

If two planes are parallel , then their normal vectors are same. 

Therefore , 

Parallel Plane r . (2i - j + 2k) = 5 

Normal vector = (2i - j + 2k) 

∴ Normal vector of required plane = (2i - j + 2k) 

Equation of required plane r . (2i - j + 2k) = d 

In cartesian form 2x - y + 2z = d 

Plane passes through point (1,1,1) therefore it will satisfy it. 

2(1) - (1) + 2(1) = d 

d = 2 – 1 + 2 = 3 

Equation of required plane r . (2i - j + 2k) = 3

Formula : Plane = r . (n) = d 

Where r = any random point 

n = normal vector of plane 

d = distance of plane from origin 

If two planes are parallel , then their normal vectors are same. 

Therefore , 

Parallel Plane 2x – 3y + 7z + 13 = 0 

Normal vector = (2i - 3j + 7k) 

∴ Normal vector of required plane = (2i - 3j + 7k) 

Equation of required plane r . (2i - 3j + 7k) = d 

In cartesian form 2x - 3y + 7z = d 

Plane passes through point (0,0,0) therefore it will satisfy it. 

2(0) - (0) + 3(0) = d 

d = 0 

Equation of required plane 2x - 3y + 7z = 0

Formula : Plane = r . (n) = d 

Where r = any random point 

n = normal vector of plane 

d = distance of plane from origin 

If two planes are parallel , then their normal vectors are same. 

Therefore , 

Parallel Plane 2x – y + 3z + 7 = 0 

Normal vector = (2i - j + 3k) 

∴ Normal vector of required plane = (2i - j + 3k) 

Equation of required plane r . (2i - j + 3k) = d 

In cartesian form 2x - y + 3z = d 

Plane passes through point (1,4, - 2) therefore it will satisfy it. 

2(1) - (4) + 3( - 2) = d 

d = 2 – 4 – 6 = - 8 

Equation of required plane 2x - y + 3z = - 8 

2x - y + 3z + 8 = 0

The equation of the family of planes through the line of intersection of planes 

x + y + z = 1 and 2x + 3y + 4z = 5 is,

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

(2k + 1)x + (3k + 1)y + (4k + 1)z = 5k + 1

It is perpendicular to the plane x – y + z = 0

(2k + 1)(1) + (3k + 1)( – 1) + (4k + 1)(1) = 5k + 1

2k + 1 – 3k – 1 + 4k + 1 = 5k + 1

K = \(-\cfrac13\) 

Substituting k = in eq.(1) , We get, x – z + 2 = 0 as the equation of the required plane

And its vector equation is  \(\vec r.(\hat i-\hat k)+2=0\) 

The equation of the family of a plane parallel to  \(\vec r.(\hat i+\hat j+\hat k)=0\)  is  

\(\vec r.(\hat i+\hat j+\hat k)=0\)…… (1) 

If it passes through (a, b, c) then 

(\(a\hat i+b\hat j+c\hat k\) )(\(\hat i+\hat j+\hat k\) ) = d

a + b + c = d 

Substituting a + b + c = d in eq.(1), we get,

\(\vec r.(\hat i+\hat j+\hat k)\) = a + b + c

x + y + z = a + b + c as the equation of the required plane.

Given: Equation of plane is \(\bar{r}\)(2\(\hat{i}\) - 3\(\hat{j}\) + 4\(\hat{k}\)) = 12

To find: Intercepts made by the plane. 

Formula Used: Equation of plane is \(\frac{x}a\) + \(\frac{y}b\) + \(\frac{z}c\) = 1 where (x, y, z) is a point on the plane and a, b, c are intercepts on x-axis, y-axis and z-axis respectively. 

Explanation: 

The equation of the plane can be written as 

2x – 3y + 4z = 12 

Dividing by 12,

\(\frac{x}6\) + \(\frac{y}{-4}\) + \(\frac{z}3\) = 1 which is of the form \(\frac{x}a\) + \(\frac{y}b\) + \(\frac{z}c\) = 1

Therefore the intercepts made by the plane are 6, -4, 3

The equation of the family of planes through the line of intersection of planes

3x – y + 2z = 4 and x + y + z = 2 is,

(3x – y + 2z – 4) + k(x + y + z – 2) = 0 ……(1)

If it passes through (2, 2, 1) then,

(6 – 2 + 2 – 4) + k(2 + 2 + 1 – 2) = 0

k = –\(\cfrac23\) 

Substituting k = – \(\cfrac23\)in eq.(1) We get,

7x – 5y + 4z = 0 as the equation of the required plane.

Given: Plane passes through the point A(1, 0, -1). 

Plane is perpendicular to the line

\(\frac{x+1}2\) = \(\frac{y+3}4\) = \(\frac{z+7}{-3}\) 

To find: Equation of the plane. 

Formula Used: Equation of a plane is ax + by + cz = d where (a, b, c) are the direction ratios of the normal to the plane. 

Explanation: 

Let the equation of the plane be 

ax + by + cz = d … (1) 

Substituting point A, 

a – z = d 

Since the given line is perpendicular to the plane, it is the normal. 

Direction ratios of line is 2, 4, -3 

Therefore, 2 + 3 = d 

d = 5 

So the direction ratios of perpendicular to plane is 2, 4, -3 and d = 5 

Substituting in (1), 

2x + 4y – 3z = 5 

Therefore, equation of plane is 2x + 4y – 3z = 5

Given : line \(\frac{x-1}2\) = \(\frac{y-2}4\) = \(\frac{z-3}{-3}\) meets plane 2x + 3y – z = 14

 To find: Point of intersection of line and plane. 

Explanation: 

Let the equation of the line be

\(\frac{x-1}2\) = \(\frac{y-2}4\) = \(\frac{z-3}{-3}\) = 2

Therefore, any point on the line is (2λ + 1, 4λ +2, -3λ +3) 

Since this point also lies on the plane, 

2(2λ + 1) + 3(4λ +2) – (-3λ +3) =14 

4λ + 2 + 12λ + 6 + 3λ – 3 = 14 

19λ + 5 = 14

λ = \(\frac{19}{19}\) = 1

Therefore the required point is (3, 5, 7).

Given: Plane passes through the intersection of planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0. Point A(2, 2, 1) lies on the plane. 

To find: Equation of the plane. 

Formula Used: Equation of plane passing through the intersection of 2 planes P₁ and P₂ is given by P₁ + λP₂ = 0 

Explanation: 

Equation of plane is 

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

Since A(2, 2, 1) lies on the plane, 

6 – 2 + 2 – 4 + λ (2 + 2 + 1 – 2) = 0 

2 + 3λ = 0

λ  = \(\frac{-2}3\)

Substituting in (1) and multiplying by 3, 

9x – 3y + 6z – 12 – 2 (x + y + z - 2) = 0 

9x – 3y + 6z – 12 – 2x – 2y – 2z + 4 = 0 

7x – 5y + 4z – 8 = 0 

Therefore the equation of the plane is 7x – 5y + 4z – 8 = 0

(i) Here if θ = 90° we get cos 90° = 0

So we get

A₁A₂ + B₁B₂ + C₁C₂ = 0

By comparing with the standard equation of a plane we get

A₁ = 3, B₁ = 4, C₁ = 5

A₂ = 2, B₂ = 6, C₂ = 6

Consider LHS = A₁A₂ + B₁B₂ + C₁C₂

Substituting the values

= (3 × 2) + (4 × 6) + (-5 × 6)

= 6 + 24 - 30

= 0

= RHS

Therefore, it is proved that the angle between the planes is 90°.

(ii) Here if θ = 90° we get cos 90° = 0

So we get

A₁A₂ + B₁B₂ + C₁C₂ = 0

By comparing with the standard equation of a plane we get

A₁ = 1, B₁ = -2, C₁ = 4

A₂ = 18, B₂ = 17, C₂ = 4

Consider LHS = A₁A₂ + B₁B₂ + C₁C₂

Substituting the values

= (1 × 18) + (-2 × 17) + (4 × 4)

= 18 + (-34) + 16

= 0

= RHS

Therefore, it is proved that the angle between the planes is 90°.

To show that planes are perpendicular

A₁A₂ + B₁B₂ + C₁C₂ = 0

Where A₁, B₁, C₁ are direction ratios of plane and A₂, B₂, C₂ are of other plane. 

2.1 + 2.2 + 4.2 = 2 + 4 + 8 =14 ≠ 0 

Hence, planes are not perpendicular. 

Similarly for the other plane 

2.5 + 2.6 + 2.7 =10 + 12 + 14 = 36 ≠ 0 

Hence, planes are not perpendicular.

We know that the lines \(\cfrac{\text x-\text x_1}{l_1}=\cfrac{y-y_1}{m_1}=\cfrac{z-z_1}{n_1}\) lies in plane ax + by + cz + d = 0, then

ax₁ + by₁ + cz₁ + d = 0 and al + bm + cn = 0

Here,

x₁ = 3, y₁ = –2, z₁ = –4 and l = 2, m = –1, n = 3

a = l, b = m, c = –1, d = –9

i.e, 3l + (– 2)m + (– 4)(– 1) – 9 = 0 and 2l – m – 3 = 0

3l – 2m = 5 and 2l – m = 3

3l – 2m = 5 …… (1)

2l – m = 3 ……(2)

Multiply eq.(1) by 2 and eq.(2) by 3 and then subtract we get

m = –1

l = 1

l² + m² = 2

Correct option is D. none of these

The given equation plane is, 2x–(1 + λ)y + 3λz = 0

We can rewrite the equation of the given plane as,

2x – (1 + λ)y + 3λz = 0

2x - y - λ(y - 3z) = 0

So, the given plane passes through the intersection of

the planes 2x - y = 0 and y - 3z = 0.

Plane passes through (3,4,2) and (7,0,6), 

A(x - 3) + B(y - 4) + C(z - 2) = 0 (1) 

A(x - 7) + B(y - 0) + C(z - 6) = 0 (2) 

Subtracting (1) from (2), 

A(x - 7 - x + 3) + B(y - y + 4) + C(z - 6 - z + 2) = 0 

-4A + 4B - 4C = 0 

A - B + C = 0 

B = A + C (3) 

Now plane is perpendicular to 2x - 5y =15 

2A - 5B = 0 (4) 

Using (3) in (4) 

2A-5(A + C) = 0 

2A - 5A - 5C = 0

-3A - 5C = 0

C = \(\frac{-3}5\)A

B = A + \(\frac{-3}5\)A

⇒ \(\frac{2}5\)A

Putting values in equation (1)

A(x - 3) + \(\frac{2}5\)A (y - 4) +   \(\frac{-3}5\)A(z - 2) = 0

A(5(x - 3) + 2(y - 4) - 3(z -2) = 0 

5x + 2y - 3z -15-8 + 6 = 0 

5x + 2y - 3z -17 = 0 

So, required equation of plane is 5x + 2y - 3z -17 = 0.

Given: P(1, 2, -3) is a point on the plane. OP is perpendicular to the plane. 

Explanation: 

Let equation of plane be ax + by + cz = d … (1) 

Substituting point P, 

⇒ a + 2b -3c = d … (2)

\(\overset\rightarrow{OP}\) = \(\hat{i}\) + 2\(\hat{j}\) - 3\(\hat{k}\)

Since OP is perpendicular to the plane, direction ratio of the normal is (1, 2, -3) 

Substituting in (2) 

1 + 4 + 9 = d 

d = 14 

Substituting the direction ratios and value of ‘d’ in (1), we get 

x + 2y – 3z = 14 

Therefore equation of plane is x + 2y – 3z = 14

Let direction ratios of normal of required plane are A, B and C

Since, required plane is perpendicular to the plane 

x - 2y + 4z = 10

∴ Their normal are perpendicular to each other

∴ A -2B + 4C = 0 ...(i)

Since required plane is passing through the points 

A(2, 1, -1) & B (-1, 3, 4)

∴ Normal of required plane is perpendicular to line passing through points A (2, 1, -1) & B (-1, 3, 4). 

And direction ratios of line AB are 2-(-1), 1-3, -1-4 or 3, -2, -5

∴ 3A-2B -5C = 0...(ii)

Now from (i) and (ii)

\(\frac {A}{-2\times-5 + 2 \times4} = \frac {B}{4\times3+5\times1}= \frac{C}{1\times-2-3\times-2}\)(By cross multiplication method)

= A/18 = B/17 = C/4 =  λ  Lets

∴ A = 18λ , B 17λ  & C = 4λ .

Hence, direction ratios of required plane are 18λ , 17λ  & C4λ and required plane is passing through point A(1, 1,-1).

∴ Equation of required plane is 

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

= 18λ (x-2) + 17λ (y-1) + 4λ (z+1) = 0

= 18(x-2) + 17(y-1) + 4(z+1) = 0

= 18x + 17y+4z - 36-17+4=0

= 18x + 17y + yz -49 = 0...(iii)

Given line \(\oversetλ r\) = (\((\hat-i +3\hat j + 4\hat k) +λ (3\hat i-2\hat j-5\hat k)\)

∴ direction vector of line \(\oversetλ r\) is \(\hat b = 3\hat i- 2\hat j-5\hat k\) and line \(\oversetλ r\) is passing through point whose position vector is \(\oversetλ a\) = \(-\hat i + 3\hat j + 4\hat k\)

(By comparing \(\oversetλ r\) = \(\oversetλ a\) + λ \(\oversetλ b\))

Thus, the direction ratios of line \(\oversetλ r\) = are 3, -2, -5 & line \(\oversetλ r\) is direction vector of line \(\oversetλ r\) is perpendicular to normal of plane (3) and point (-1, 3, 4) lies on the plane (3)

Since, 18 x 3 + 17 x -2 + 4 x -5 = 54 -34 -20

= 54 - 54 = 0

It implies that direction vector of line \(\oversetλ r\) is perpendicular to the normal of the plane.

Also, 18 x -1 + 17 x 3+ 4 x 4 -49 = -18 + 51 + 16 - 49 = 67 - 67 = 0

It implies that point  (-1, 3, 4) lies on the plane (3) we conclude that plane 18 x + 17 y + 4z - 49 = 0

contains the line \(\oversetλ r\) = \((-\hat i+ 3\hat j + 4\hat k) + λ (3\hat i-2 \hat j- 5\hat k)\)

Hence Proved.

Plane passes through (2,1,-1) and (-1,3,4), 

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

A(x + 1) + B(y - 3) + C(z - 4) = 0 (2) 

Subtracting (1) from (2), 

A(x + 1 - x + 2) + B(y - 3 - y + 1) + C(z - 4 - z -1) = 0 

3A - 2B - 5C = 0 (3) 

Now plane is perpendicular to x - 2y + 4z = 10 

A - 2B + 4C = 0 (4) 

Using (3) in (4) 

2A - 9C = 0

C = \(\frac{2}9\)A

2B = A + 4. \(\frac{2}9\)A

⇒ \((\frac{9+8}9)\)A = \(\frac{17}9\)A

B = \(\frac{17}{18}\)A

Putting values in equation (1)

A(x - 2) + \(\frac{17}{18}\)A (y - 1) + \(\frac{2}9\)A (z + 1) = 0

A(18(x-2) + 17(y-1) + 4(z + 1) = 0 

18x + 17y + 4z - 36 -17 + 4 = 0 

18x + 17y + 4z - 49 = 0 

So, the required equation of plane is 18x + 17y + 4z - 49 = 0

If plane contains  \(\bar{r}\) = -\(\hat{i}\) + 3\(\hat{j}\) + 4\(\hat{k}\) + (3\(\hat{i}\) - 2\(\hat{j}\) - 5\(\hat{k}\)) then (-1,3,4) satisfies plane and normal vector of plane is perpendicular to vector of line

LHS =18(-1) + 17.3 + 4.4 - 49 

= -18 + 51 + 16 - 49 

= -2 + 2 = 0 = RHS 

In vector form normal of plane 

\(\bar{n}\) = 18\(\hat{i}\) + 17\(\hat{j}\) + 4\(\hat{k}\)

LHS =18.3 + 17(-2) + 4.(- 5) = 54 - 34 - 20 = 0 = RHS 

Hence line is contained in plane.