Students must practice these Maths 1B Important Questions TS Inter 1st Year Maths 1B The Plane Important Questions to help strengthen their preparations for exams.

## TS Inter 1st Year Maths 1B The Plane Important Questions

Question 1.

Reduce the equation x + 2y – 3z – 6 = 0 of the plane to the normal form. [Mar. ’19 (TS); Mar. ’16 (AP & TS), ’14, ’01; May ’13]

Solution:

Given the equation of the plane is,

x + 2y – 3z – 6 = 0

x + 2y – 3z = 6

On dividing both sides by

∴ which is the normal form.

Question 2.

Find the equation of the plane whose intercepts on X, Y, Z axes are 1, 2, 4 respectively. [May ’14; Mar ’10]

Solution:

Given that,

x-intercept, a = 1

y-intercept, b = 2

z-intercept, c = 4

∴ The equation of the plane is

4x + 2y + z = 4

Question 3.

Find the directions of the normal to the plane x + 2y + 2z – 4 = 0. [May ’15 (TS); Mar. ’13; May ’12]

Solution:

Given, equation of the plane is x + 2y + 2z – 4 = 0

Directions ratio’s of the normal to the plane are (a, b, c) = (1, 2, 2).

Directions cosines of the normal

Question 4.

Write the equation of the plane 4x – 4y + 2z + 5 = 0 in the intercept form. [Mar. ’19 (AP); Mar. ’12]

Solution:

Given, equation of the plane is 4x – 4y + 2z + 5 = 0

4x – 4y + 2z = -5

\(\frac{4 x}{-5}-\frac{4 y}{-5}+\frac{2 z}{-5}=1\)

\(\frac{x}{\frac{-5}{4}}-\frac{y}{\frac{-5}{4}}+\frac{z}{\frac{-5}{2}}=1\)

which is the intercept form.

Question 5.

Find the angle between the planes x + 2y + 2z – 5 = 0 and 3x + 3y + 2z – 8 = 0. [Mar. ’17, ’15 (TS), ’09 ; May ’15 (AP)]

Solution:

Given equations of the planes are

x + 2y + 2z – 5 = 0 ……….(1)

3x + 3y + 2z – 8 = 0 ……..(2)

Comparing (1) with a_{1}x + b_{1}y + c_{1}z + d_{1} = 0, we get

a_{1} = 1, b_{1} = 2, c_{1} = 2, d_{1} = -5

Comparing (2) with a_{2}x + b_{2}y + c_{2}z + d_{2} = 0, we get

a_{2} = 3, b_{2} = 3, c_{2} = 2, d_{2} = -8

If ‘θ’ is the angle between the planes (1) & (2) then

Question 6.

Find the equation of the plane passing through (1, 1, 1) and parallel to the plane x + 2y + 3z – 7 = 0. [May ’13 (Old), ’11, ’10, ’09]

Solution:

Given, the equation of the plane is

x + 2y + 3z – 7 = 0 ……….(1)

Let the given point P = (1, 1, 1)

The equation of any plane parallel to x + 2y + 3z – 7 = 0 is x + 2y + 3z + k = 0

If equation (1) passes through the point p(1, 1, 1), then

1 + 2(1) + 3(1) + k = 0

1 + 2 + 3 + k = 0

6 + k = 0

k = -6

Substituting the value of ‘k’ in equation (1)

∴ The equation of the required plane is x + 2y + 3z – 6 = 0

Question 7.

Find the equation of the plane bisecting the line segment joining (2, 0, 6) and (-6, 2, 4) and perpendicular to it.

Solution:

Let A(2, 0, 6), B(-6, 2, 4) are the given points

Directions of \(\overline{\mathrm{AB}}\) are (x_{2} – x_{1}, y_{2} – y_{1}, z_{2} – z_{1})

= (-6 – 2, 2 – 0, 4 – 6)

= (-8, 2, -2)

= (a, b, c)

Since the plane is ⊥ to \(\overline{\mathrm{AB}}\) then the line \(\overline{\mathrm{AB}}\) is normal to the plane.

Also, ‘C'(-2, 1, 5) lies in the plane.

∴ The equation of the plane passing through C(-2, 1, 5) & having direction ratios of the normal to the plane is (-8, 2, -2) is a(x – x_{1}) + b(y – y_{1}) + c(z – z_{1}) = 0

⇒ -8(x + 2) + 2(y – 1) – 2(z – 5) = 0

⇒ -8x – 16 + 2y – 2 – 2z + 10 = 0

⇒ -8x + 2y – 2x – 8 = 0

⇒ 4x – y + z + 4 = 0

Question 8.

Find the equation of the plane through the points (2, 2, -1), (3, 4, 2), (7, 0, 6). [May ’00]

Solution:

Let the given points are

A(x_{1}, y_{1}, z_{1}) = (2, 2, -1)

B(x_{2}, y_{2}, z_{2}) = (3, 4, 2)

C(x_{3}, y_{3}, z_{3}) = (7, 0, 6)

The equation of the plane passing through the points A(2, 2, -1), B(3, 4, 2), C(7, 0, 6) is

(x – 2)(14 + 6) – (y – 2)(7 – 15) + (z + 1)(-2 – 10) = 0

(x – 2)(20) – (y – 2)(-8) + (z + 1)(-12) = 0

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

5x – 10 + 2y – 4 – 3z – 3 = 0

5x + 2y – 3z – 17 = 0

Question 9.

A plane meets the coordinate axes in A, B, C. If the centroid of ∆ABC is (a, b, c) show that the equation to the plane is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=3\). [Mar. ’01]

Solution:

Let the equation of the plane be

\(\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1\) ……..(1)

Plane (1) cuts the X-axis, Y-axis & Z-axis at A, B, C respectively.

Then, A = (p, 0, 0), B = (0, q, 0), C = (0, 0, r)

The centroid of triangle ABC is

Given that, the centroid of triangle ABC is G = (a, b, c)

∴ (a, b, c) = \(\left(\frac{\mathrm{p}}{3}, \frac{\mathrm{q}}{3}, \frac{\mathrm{r}}{3}\right)\)

a = \(\frac{p}{3}\), b = \(\frac{q}{3}\), c = \(\frac{r}{3}\)

p = 3a, q = 3b, r = 3c

∴ The equation to the required plane is from (1)

### Some More Maths 1B The Plane Important Questions

Question 10.

Find the angle between the planes, 2x – y + z = 6 and x + y + 2z = 7. [Mar. ’11; B.P.]

Solution:

Given equations of the plane are

2x – y + z = 6 ……..(1)

x + y + 2z = 7 ……..(2)

Comparing (1) with a_{1}x + b_{1}y + c_{1}z + d_{1} = 0, we get

a_{1} = 2, b_{1} = -1, c_{1} = 1, d_{1} = -6

Comparing (2) with a_{2}x + b_{2}y + c_{2}z + d_{2} = 0, we get

a_{2} = 1, b_{2} = 1, c_{2} = 2, d_{2} = -7

If ‘θ’ is the angle between the planes (1) & (2) then

Question 11.

Find the equation of the plane through the point (α, β, γ) and parallel to the plane ax + by + cz = 0.

Solution:

Given the equation of the plane is

ax + by + cz = 0 ………(1)

Let the given point p = (α, β, γ)

∴ The equation of any plane parallel to ax + by + cz = 0 is ax + by + cz + k = 0

If equation (1) passes through the point p(α, β, γ) then

a(α) + b(β) + c(γ) + k = 0

⇒ k = -aα – bβ – cγ

Substituting the value of ‘k’ in equation (1)

∴ The equation of the required plane is ax + by + cz – aα – bβ – cγ = 0

a(x – α) + b(y – β) + c(z – γ) = 0

Question 12.

Show that the plane through (1, 1, 1), (1, -1, 1), and (-7, -3, -5) is parallel to Y-axis.

Solution:

Let the given points are

A(x_{1}, y_{1}, z_{1}) = (1, 1, 1)

B(x_{2}, y_{2}, z_{2}) = (1, -1, 1)

C(x_{3}, y_{3}, z_{3}) = (-7, -3, -5)

The equation of the plane passing through the points A(1, 1, 1), B(1, -1, 1), C(-7, -3, -5) is

(x – 1)(12 – 0) – (y – 1)(0 – 0) + (z – 1)(0 – 16) = 0

(x – 1)(12) – 0 + (z – 1)(-16) = 0

12x – 12 – 16z + 16 = 0

12x – 16z + 4 = 0

3x – 4z + 1 = 0

Since the coefficient of y is zero, the plane is perpendicular to xz-plane and hence the plane is parallel to the y-axis.

Question 13.

Find the equation to the plane through the points (0, -1, -1), (4, 5, 1) and (3, 9, 4).

Solution:

Let the given points are

A(x_{1}, y_{1}, z_{1}) = (0, -1, -1)

B(x_{2}, y_{2}, z_{2}) = (4, 5, 1)

C(x_{3}, y_{3}, z_{3}) = (3, 9, 4)

The equation of the plane passing through the points A(0, -1, -1), B(4, 5, 1), C(3, 9, 4) is

x(30 – 20) – (y + 1)(20 – 6) + (z + 1)(40 – 18) = 0

x(10) – (y + 1)(14) + (z + 1)(22) = 0

10x – 14y – 14 + 22z + 22 = 0

10x – 14y + 22z + 8 = 0

5x – 7y + 11z + 4 = 0

Question 14.

Find the equation of the plane if the foot of the perpendicular from the origin to the plane is (1, 3, -5).

Solution:

Let O(0, 0, 0) be the origin.

P(1, 3, -5) is the foot of the perpendicular.

The plane passing through P and is perpendicular to the line segment \(\overline{\mathrm{OP}}\) is normal to the plane.

Directions of \(\overline{\mathrm{OP}}\) = (x_{2} – x_{1}, y_{2} – y_{1}, z_{2} – z_{1})

= (1 – 0, 3 – 0, -5 – 0)

= (1, 3, -5)

∴ The equation of the plane passing through P(1, 3, -5) having direction ratios of the normal to the plane are (1, 3, -5) is a(x – x_{1}) + b(y – y_{1}) + c(z – z_{1}) = 0

⇒ 1(x – 1) + 3(y – 3) – 5(z + 5) = 0

⇒ x – 1 + 3y – 9 – 5z – 25 = 0

⇒ x + 3y – 5z – 35 = 0

⇒ x + 3y – 5z = 35

Question 15.

Find the equation of the plane if the foot of the perpendicular from the origin to the plane is (2, 3, -5).

Solution:

Let O(0, 0, 0) be the origin.

P(2, 3, -5) is the foot of the perpendicular.

The plane passing through P is perpendicular to \(\overline{\mathrm{OP}}\) the line segment \(\overline{\mathrm{OP}}\) is normal to the plane.

Directions of \(\overline{\mathrm{OP}}\) are (x_{2} – x_{1}, y_{2} – y_{1}, z_{2} – z_{1})

= (2 – 0, 3 – 0, -5 – 0)

= (2, 3, -5)

The equation of the plane passing through P(2, 3, -5) having direction ratios of the normal to the plane are (2, 3, -5) is a(x – x_{1}) + b(y – y_{1}) + c(z – z_{1}) = 0

2(x – 2) + 3(y – 3) + (-5)(z + 5) = 0

2x – 4 + 3y – 9 – 10 – 5z = 0

2x + 3y – 5z = 0

Question 16.

Find the intercepts of the plane 4x + 3y – 2z + 2 = 0 on the coordinate axes. [Mar. ’18 (AP & TS)]

Solution:

Given, the equation of the plane is

4x + 3y – 2z + 2 = 0

4x + 3y – 2z = -2

x-intercept, a = \(\frac{-1}{2}\)

y-intercept, b = \(\frac{-2}{3}\)

z-intercept, c = 1

Question 17.

Find the equation of the plane passing through (-2, 1, 3) and having (3, -5, 4) as directions of its normal.

Solution:

Let the given point (x_{1}, y_{1}, z_{1}) = (-2, 1, 3)

Direction ratio’s of the normal to the plane are (a, b, c) = (3, -5, 4)

∴ The equation of the plane passing through the point (-2, 1, 3) and having directions of (3, -5, 4) is a(x – x_{1}) + b(y – y_{1}) + c(z – z_{1}) = 0

3(x + 2) + (-5)(y – 1) + 4(z – 3) = 0

3x + 6 – 5y + 5 + 4z – 12 = 0

3x – 5y + 4z – 1 = 0

Question 18.

Find the equation of the plane passing through (2, 3, 4) and perpendicular to the X-axis.

Solution:

If the plane is perpendicular to X-axis then the plane is parallel to yz-plane.

∴ The equation of a plane parallel to yz axis plane is x = k ……….(1)

If plane (1) passes through the point (2, 3, 4) then

2 = k

∴ k = 2

Substituting the value of k in equation (1)

∴ The equation of the required plane is x = 2.

Question 19.

Find the equation to the plane parallel to the zx-plane and passing through (0, 4, 4).

Solution:

The equation of the plane parallel to zx plane is y = k ……..(1)

If plane (1) passes through point (0, 4, 4) then

4 = k

∴ k = 4

Substituting ‘k’ in equation (1), y = 4

The equation of the required plane is y = 4.

Question 20.

Find the equation of the plane through (-1, 6, 2) and perpendicular to the join of (1, 2, 3) and (-2, 3, 4).

Solution:

Let A(1, 2, 3), B(-2, 3, 4) are the given points

Directions of \(\overline{\mathrm{AB}}\) are (x_{2} – x_{1}, y_{2} – y_{1}, z_{2} – z_{1})

= (-2 – 1, 3 – 2, 4 – 3)

= (-3, 1, 1)

Directions of normal to the plane are (a, b, c) = (-3, 1, 1)

The equation of the plane passing through (-1, 6, 2) and having directions of normal to the plane are (-3, 1, 1) is a(x – x_{1}) + b(y – y_{1}) + c(z – z_{1}) = 0

-3(x + 1) + 1(y – 6) + 1(z – 2) = 0

-3x – 3 + y – 6 + z – 2 = 0

-3x + y + z – 11 = 0

3x – y – z + 11 = 0