Quantitative Aptitude PYQs

Concept:

Angle of elevation: The term angle of elevation denotes the angle from the horizontal upward to an object. An observer's line of sight would be above the horizontal.

Calculation:

Given: height of the building = h

Let BC be the height of the hill = H

And distance between hill and building = x

In triangle ABC,

tan π/3 = BC/AB = H/x

√3 = H/x

⇒ x = H/√3       ….

(1) Now, In triangle DEC,

AB = DE = x

⇒ tan π/6 = CE/DE = (H – h)/x

⇒ 1/√3 = (H – h)/x

Put the value of x from equation 1^st

\(\frac{1}{{\sqrt 3 }} = {\rm{;}}\frac{{{\rm{H}} - {\rm{h}}}}{{\left( {\frac{{\rm{H}}}{{\sqrt 3 }}} \right)}}{\rm{;}}\)

\(\Rightarrow \frac{1}{{\sqrt 3 }}{\rm{}} \times {\rm{}}\left( {\frac{{\rm{H}}}{{\sqrt 3 }}} \right){\rm{}} = {\rm{;H}} - {\rm{h}}\)

⇒ H /3 = H – h

⇒ H = 3H – 3h

⇒ 3h = 2H

∴ H = 3h/2

Concept:

• cosec² x – cot² x = 1
• a² – b² = (a – b) (a + b)

Calculation:

Given:

cosec x + cot x = √3       ….

(1) We know that, cosec² x – cot² x = 1

⇒ (cosec x - cot x) (cosec x + cot x) = 1

⇒ (cosec x - cot x) × √3 = 1

⇒ cosec x - cot x = 1/√3       …. (2) Adding equation 1 and equation 2;

2 cosec x = √3 + 1/√3  

⇒ 2 cosec x = 4/√3

⇒ cosec x = 2/√3

⇒ sin x = √3/2

∴ x = π/3

Concept:

• \({\cos ^2}\frac{\pi }{8} = ;\frac{{2 + ;\sqrt 2 }}{4};\)
• \(\cos \frac{\pi }{4} = ;\frac{1}{{\sqrt 2 }}\)

Calculation:

We have to find the value of (2cos θ + 1)¹⁰ (2 cos 2θ – 1)¹⁰ (2 cos θ – 1)¹⁰ (2 cos 4θ – 1)¹⁰

(2cos θ + 1)¹⁰ (2 cos 2θ – 1)¹⁰ (2 cos θ – 1)¹⁰ (2 cos 4θ – 1)¹⁰

= [(2cos θ + 1) (2 cos θ – 1)]¹⁰ (2 cos 2θ – 1)¹⁰ (2 cos 4θ – 1)¹⁰

= (4 cos² θ – 1)¹⁰ (2 cos 2θ – 1)¹⁰ (2 cos 4θ – 1)¹⁰

Putting, θ = π /8

2 θ = π/4 and 4 θ = π /2

= (4 cos² π /8 – 1)¹⁰ (2 cos π/4 – 1)¹⁰ (2 cos π /2 – 1)¹⁰

\(= {\left( {4; \times ;\frac{{2 + ;\sqrt 2 }}{4} - 1} \right)^{10}}{\left( {2; \times ;\frac{1}{{\sqrt 2 }} - 1} \right)^{10}}{\left( {0; - ;1} \right)^{10}}\)

= (√2 + 1)¹⁰ (√2 – 1)¹⁰ × 1 

= [(√2 + 1) (√2 – 1)]¹⁰

= (2 – 1)¹⁰

= 1

Concept:

Let us consider the standard form of a quadratic equation, ax² + bx + c =0

Where a, b and c are constants with a ≠ 0

Let α and β be the two roots of the above quadratic equation. 

• The sum of the roots of a quadratic equation are:

\({\rm{\alpha }} + {\rm{\beta }} = - \frac{{\rm{b}}}{{\rm{a}}} = - \frac{{{\rm{coefficient;of;x}}}}{{{\rm{coefficient;of;}}{{\rm{x}}^2}}}\)

• The product of the roots is given by:

\({\rm{\alpha \beta }} = \frac{{\rm{c}}}{{\rm{a}}} = \frac{{{\rm{constant;term}}}}{{{\rm{coefficient;of;}}{{\rm{x}}^2}}}\)

Calculation:

Given: cos α and cos β (0 < α < β < π) are the roots of the quadratic equation.

4x² – 3 = 0

Product of the roots = cos α × cos β = - 3/4

\(\Rightarrow {\rm{;}}\frac{1}{{\sec {\rm{\alpha ;}}\sec {\rm{\beta }}}} = {\rm{;}}\frac{{ - 3}}{4}\)

∴ sec α sec β = -4/3

Concept:

• \({\tan ^{ - 1}}x + ;{\tan ^{ - 1}}y = ;{\tan ^{ - 1}}\frac{{x; + ;y}}{{1; - ;xy}}\)
• \({\tan ^{ - 1}}x - ;{\tan ^{ - 1}}y = ;{\tan ^{ - 1}}\frac{{x; - ;;y}}{{1;; + ;;xy}}\)

Calculation:

\({\tan ^{ - 1}}\left( {2x} \right) + {\tan ^{ - 1}}\left( {3x} \right) = \frac{\pi }{4}\)

\( \Rightarrow ;{\tan ^{ - 1}}\frac{{2x + 3x}}{{1 - \left( {2x} \right)\left( {3x} \right)}} = ;\frac{\pi }{4}\)

\(\Rightarrow ;{\tan ^{ - 1}}\frac{{5x}}{{1 - 6{x^2}}} = ;\frac{\pi }{4}\)        ….

(1) \(\Rightarrow ;\frac{{5x}}{{1 - 6{x^2}}} = ;\tan \frac{\pi }{4} = 1\)

⇒ 5x = 1 – 6x²

⇒ 6x² + 5x - 1 =0

⇒ 6x² + 6x – x – 1 = 0

⇒ 6x (x + 1) – 1(x + 1) = 0

⇒ (6x – 1) (x + 1) = 0

So, \(x=\frac 16\)and -1

x= - 1 (Rejected)

∴ \(x=\frac 16\)

Concept:

If a, ar, ar², ….is terms an infinite GP, then the sum of infinite geometric series is given by.

Sum of infinite GP = \({s_\infty } = ;\frac{a}{{1; - ;r}}\) ; |r| < 1

Calculation:

If a, ar, ar², ….is terms an infinite GP.

Given: Second term of GP = 2

⇒ ar = 2      ….

(1) Sum of infinite GP = \({s_\infty } = ;\frac{a}{{1; - ;r}} = 8\)

⇒ a = 8 × (1 – r)

Multiplying r both sides,

⇒ ar = 8r × (1- r)

From equation 1^st

⇒ 2 = 8 (r – r²)

⇒ 1 = 4r – 4r²

⇒ 4r² – 4r + 1 = 0

⇒ 4r² – 2r – 2r + 1 = 0

⇒ 2r (2r – 1) – 1(2r – 1) = 0

⇒ (2r – 1) (2r – 1) = 0

⇒ 2r – 1 =0

So, r = ½

Put in equation 1^st

a × ½ = 2

So, a = 4

Then the GP is a, ar ,ar²…….

So 4,2,1…..are in GP

Explanation:

Case: 1

If a, b, c are in AP,

Then b – a = c – b

⇒ a – b = b – c

∴ \(\frac{{a - b}}{{b - c}} = 1\)

Case: 2

If a, b, c are in GP,

Then \(\frac{{\rm{b}}}{{\rm{a}}} = {\rm{;}}\frac{{\rm{c}}}{{\rm{b}}}\)

Subtracting 1 from both sides,

\( \Rightarrow ;\frac{{\rm{b}}}{{\rm{a}}} - 1 = {\rm{;}}\frac{{\rm{c}}}{{\rm{b}}} - 1\)

\(\Rightarrow {\rm{;}}\frac{{{\rm{b;}} - {\rm{;a}}}}{{\rm{a}}} = {\rm{;}}\frac{{{\rm{c;}} - {\rm{;b}}}}{{\rm{b}}}\)

\(\Rightarrow ;\frac{{b - a}}{{c - b}} = ;\frac{a}{b}\)

∴ \(\frac{{a - b}}{{b - c}} = ;\frac{a}{b}\)

Case: 3

If a, b, c are in HP

\(\Rightarrow ;b = ;\frac{{2ac}}{{a; + ;c}}\)

⇒ ba + bc = 2ac

⇒ ba + bc = ac + ac

⇒ ba – ac = ac – bc

⇒ a (b – c) = c (a – b)

∴ \(\frac{{a - b}}{{b - c}} = ;\frac{a}{c}\)

Concept:

• If all the possible n-digit numbers using n distinct digits are formed

The sum of all the numbers so formed = (n-1)! × (sum of all the digits) × {111…….} n times.

Calculation:

We have to find sum of all three-digit numbers that can be formed using all the digits 3, 4 and 5

Here n = 3 and Sum of 3 digits = 12

The sum of all the numbers which can be formed by using the digits 3, 4 and 5

= (3 – 1)! × (3 + 4 + 5) × (111)

= 2 × 12 × 111

= 2664

Explanation:

Let α₁ and β₁ are roots of ax² + bx + c = 0 and

α₂ and β₂ are roots of  px² + qx + r = 0

⇒ ax² + bx + c = 0

α₁ + β₁ = -b/a

α₁β₁ = c/a

D₁ = b² -4ac

⇒ px² + qx + r = 0

α₂ + β₂ = -q/p

α₂β₂ = r/p

D₂ = q² -4pr

Given: \(\frac{{{\alpha _1}}}{{{\beta _1}}} = ;\frac{{{\alpha _2}}}{{{\beta _2}}}\)

Using componendo and dividendo we get,

\(\frac{{{\alpha _1}; + ;{\beta _1};}}{{{\alpha _1}; - ;;{\beta _1}}} = ;\frac{{{\alpha _2}; + ;{\beta _2};}}{{{\alpha _2}; - ;{\beta _2}}}\)

Squaring both sides,

\({\left( {\frac{{{\alpha _1}; + ;{\beta _1};}}{{{\alpha _1}; - ;;{\beta _1}}}} \right)^2} = ;{\left( {\frac{{{\alpha _2}; + ;;{\beta _2};}}{{{\alpha _2}; - ;{\beta _2}}}} \right)^2}\)

\( \Rightarrow ;\frac{{{{\left( {{\alpha _1}; + ;{\beta _1}} \right)}^2}}}{{{{\left( {{\alpha _1}; - ;;{\beta _1}} \right)}^2}}} = ;\frac{{{{\left( {{\alpha _2}; + ;;{\beta _2}} \right)}^2}}}{{{{\left( {{\alpha _2}; - ;{\beta _2}} \right)}^2}}}\)

\(\Rightarrow ;\frac{{{{\left( {{\alpha _1}; + ;{\beta _1}} \right)}^2}}}{{{{\left( {{\alpha _1}; + ;;{\beta _1}} \right)}^2}; - ;4{\alpha _1}{\beta _1}}} = ;\frac{{{{\left( {{\alpha _2}; + ;{\beta _2}} \right)}^2}}}{{{{\left( {{\alpha _2}; + ;;{\beta _2}} \right)}^2} - ;4{\alpha _2}{\beta _2}}};\)

Shortcut trick:

\(\frac{{{{\rm{D}}_1}}}{{{{\rm{D}}_2}}} = {\rm{;}}{\left( {{\rm{Ratio;of;coefficient;of;x}}} \right)^2} = {\rm{;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{q}}^2}}}\)

Concept:

• 0 ≤ sin² θ ≤ 1
• sin2 θ = 2 sin θ cos θ
• sin² θ + cos² θ = 1

Calculation:

Given:

A = sin² θ + cos⁴ θ

= 1 – cos² θ + cos⁴ θ

= 1 - cos² θ (1 - cos² θ)

\(= 1 - ;\frac{1}{4} \times 4;{\sin ^2}\theta ;{\cos ^2}\theta\)

\(= 1 - ;\frac{1}{4} \times ;{\left( {2\sin \theta \cos \theta } \right)^2}\)

\(= 1 - ;\frac{1}{4} \times {(\sin 2\theta )^2}\)

= \(1 - ;\frac{{{{\sin }^2}2\theta }}{4}\)

Now, we know that,

0 ≤ sin² 2θ ≤ 1

\(\Rightarrow 0; \le \frac{{{{\sin }^2}2\theta }}{4}; \le \frac{1}{4}\)

\(\Rightarrow ; - \frac{1}{4}; \le ; - ;\frac{{{{\sin }^2}2\theta }}{4}; \le 0\)

Adding one both sides,

\(\Rightarrow 1 + \left( { - \frac{1}{4};} \right) \le ;1 - ;\frac{{{{\sin }^2}2\theta }}{4}; \le 1; + ;0\)

\( \Rightarrow \frac{3}{4}; \le ;1 - ;\frac{{{{\sin }^2}2\theta }}{4}; \le 1\)

⇒ 3/4 ≤ A ≤ 1

Concept:

• Angles formed by the ends of the diameter of a circle to its circumference form a right angle.
• When two lines are perpendicular, the product of their slope is -1.

Calculation:

Let a point p(x, y) on circumference of circle and AB is diameter of circle.

We know that Angles formed by the ends of the diameter of a circle to its circumference form a right angle.

So, ∠ APB = 90°

Line AP is perpendicular on BP,

Slope of AP × Slope of BP = -1

\(\frac{{\left( {y - ;{y_1}} \right)}}{{\left( {x - ;{x_1}} \right)}}; \times \frac{{\left( {y - ;{y_2}} \right)}}{{\left( {x - ;{x_2}} \right)}} = ; - 1\)

⇒ (y – y₁) (y – y₂) = - (x – x₁) (x – x₂)

⇒ (y – y₁) (y – y₂) + (x – x₁) (x – x₂) =0

∴ Option 4 is correct.

Concept:

Let second-degree equation be,

ax² + by² + 2hxy + 2gx + 2fy + c = 0

Discriminant = Δ = \(\left| {\begin{array}{*{20}{c}} a&h&{\rm{g}}\ h&{\rm{b}}&{\rm{f}}\ {\rm{g}}&{\rm{f}}&{\rm{c}} \end{array}} \right| = abc + 2fgh - a{f^2} - b{g^2} - c{h^2}\)

Case: 1

If discriminant (Δ) of this equation doesn’t equal to zero (Δ ≠ 0)

• If h² – ab > 0, it represents a hyperbola and a rectangular hyperbola (a + b = 0).
• If h² – ab = 0, it represents a parabola.
• If h² – ab < 0, it represents an ellipse. (a ≠ b)
• If h² – ab < 0, it represents a circle. (a = b)

Case: 2

If discriminant (Δ) of this equation equal to zero (Δ = 0)

• If h² – ab > 0, it represents two distinct real lines.
• If h² – ab = 0, it represents a parallel lines.
• If h² – ab < 0, it represents non-real lines.

 

Calculation:

Given: Second degree equation, x² + 4y² – 2x – 4y + 2 = 0

Compare with second-degree equation ax²+by²+2hxy+2gx+2fy+c=0

So, a = 1, b = 4, h = 0, g = -1, f = -2 and c = 2

Now, Δ = abc + 2fgh - af² - bg² - ch²

Δ = 1 × 4 × 2 + 0 – 1 × (- 2)² – 4 × (-1)²

= 8 – 4– 4 = 0

So, Δ = 0

Now, h² – ab = 0 – 1 × 4 = - 4

So, h² – ab < 0

It represents non-real lines.

Concept:

• Angle between two lines: The angle θ between the lines having slope m₁ and m₂ is given by

\(\tan \theta = ;\left| {\frac{{{m_2} - {m_1}}}{{1 + ;{m_1};{m_2}}}} \right|\)

Calculation:

The lines lx + my + n = 0 and l’x + m’y + n’ = 0 having slope m₁ and m₂.

lx + my + n = 0

my = -lx – n

⇒ y = (-l/m) x – n

So, m₁ = -l/m

Again,

l’x + m’y + n’ = 0

m’y = - l’x - n’

⇒ y = (- l’/m’) x – n’

So, m₂ = -l’/m’

Angle between two lines = \(\tan \theta = ;\left| {\frac{{{m_2} - {m_1}}}{{1 + ;{m_1};{m_2}}}} \right|\)

⇒ \(\tan ({\tan ^{ - 1}}\theta ) = \left| {\frac{{\frac{{ - ;{l'}}}{{m'}} - ;\frac{{ - ;l}}{m};}}{{;1 + \frac{{ - ;{l'}}}{{m'}}; \times ;\frac{{ - ;l}}{m}}}} \right|\) 

\(\Rightarrow \theta = \left| {\frac{{ - l'm + lm'}}{{mm' + ll'}}} \right|\)

\(\Rightarrow \theta = \left| {\frac{{lm' - l'm}}{{ll' + ;mm'}}} \right|\)

∴ Option 3 is correct.

Concept:

Distance between parallel lines:

• The distance between the lines y = mx + c₁ and y = mx + c₂ is \(\frac{{\left| {{c_1} - {c_2}} \right|}}{{\sqrt {1{\rm{;}} + {\rm{;}}{m^2}} }}\)
• The distance between the lines ax + by + c₁ = 0 and ax + by + c₂ = 0 is \(\frac{{\left| {{c_1} - {c_2}} \right|}}{{\sqrt {{a^2} + {b^2}} }}\)
• The distance between the line x = c₁ and x = c₂ is |c₁ – c₂|

∴ Statement 2 and 3 are correct.

Concept:

• If the two lines are parallel, then their slopes will be equal.
• The "point-slope" form of the equation of a straight line is:

y – y₁ = m (x – x₁)

Here (x₁, y₁) is point on the line and m is the slope of line.

• Equation of a Line Parallel to a Line

Let, ax + by + c = 0 (b ≠ 0) be the equation of the given straight line.

So, equation of a line parallel to a given line ax + by + λ = 0, where λ is a constant.

Calculation:

Given: Lines are \(\frac{x}{2} + \frac{y}{3} = 1{\rm{;}}and\frac{x}{3} + \frac{y}{2} = 1\)

Or 3x + 2y = 6 and 2x + 3y = 6

3x + 2y = 6       ….

(1) 2x + 3y = 6       …. (2) Now, applying equation (1) × 2 - equation (2) × 3

6x + 4y - 6x – 9y = 12 – 18

⇒ - 5y = -6

So, y = 6/5

Put the value of y in equation 1^st.

3x + 2 × 6/5 = 6

⇒ 3x = 6 – 12/5

⇒ 3x = 18/5

So, x = 6/5

Point of intersection = (6/5, 6/5)

Given line is parallel to the line 4x + 5y – 6 = 0

So, equation of the line be 4x + 5y + k =0

Line passing through point of intersection (6/5, 6/5)

So, 4 × 6/5 + 5 × 6/5 + k =0

⇒ 24/5 + 6 + k = 0

⇒ 54/5 + k =0

∴ k = - 54/5

Hence, equation of line be 4x + 5y + (- 54/5) = 0

⇒ 20x + 25y -54 = 0

Concept:

Perpendicular Distance of a Point from a Plane

Let us consider a plane given by the Cartesian equation, Ax + By + Cz = d

And a point whose coordinate is, (x₁, y₁, z₁)

Now, distance = \(\left| {\frac{{A{x_1}; + ;B{y_1}; + ;C{z_1} - ;d}}{{\sqrt {{A^2}; + ;;{B^2}; + ;{C^2}} }}} \right|\)

Calculation:

We have to find the distance of the point (2, 3, 4) form the plane 3x – 6y + 2z + 11 = 0

Distance \(= \left| {\frac{{3; \times ;2; + \left( { - 6} \right); \times ;3; + ;2; \times ;4; + ;11}}{{\sqrt {{3^2}; + ;;{{\left( { - 6} \right)}^2}; + ;{{\left( 2 \right)}^2}} }}} \right|\)

\(= \left| {\frac{{6; - 18; + ;8; + ;11}}{{\sqrt {9; + ;;36; + ;4} }}} \right|\)

\(= ;\left| {\frac{7}{{\sqrt {49} }}} \right|\) = 1

Concept:

Let A(x₁, y₁, z₁) and B(x₂, y₂, z₂)

• \({\rm{Distance;between;A;and;B;}} = \sqrt {{\rm{;}}{{\left( {{{\rm{x}}_2} - {\rm{;}}{{\rm{x}}_1}} \right)}^2} + {\rm{;}}{{\left( {{{\rm{y}}_2} - {\rm{;}}{{\rm{y}}_1}} \right)}^2} + {{\left( {{{\rm{z}}_2} - {\rm{;}}{{\rm{z}}_1}} \right)}^2}} \)
• \({\rm{Midpoint;}} = {\rm{;}}\left( {\frac{{{x_1}; + ;;{x_2}}}{2},;\frac{{{y_1}; + ;;{y_2}}}{2},\frac{{{z_1}; + ;;{z_2}}}{2}} \right)\)

 

Calculation:

Given:  M, N and K points on the side OR, OP, PQ and QR respectively such that LMNK is a parallelogram.

And length LK = LM = √2 

L, M and K are midpoint of OR, OP and RQ

So coordinate of L = (1, 2, 3), M = (2, 3, m) and K = (2, 2, 3+n)

LM = √2

\(\Rightarrow \sqrt {;{{\left( {2 - 1} \right)}^2} + ;{{\left( {3 - ;2} \right)}^2} + {{\left( {m - ;3} \right)}^2}} = ;\sqrt 2 \)

Squaring both sides,

⇒ 1 + 1 + (m – 3)² = 2

⇒ (m – 3)² = 0

⇒ m – 3 = 0

∴ m = 3

Again,

Lk = √2

\(\Rightarrow \sqrt {;{{\left( {2 - 1} \right)}^2} + ;{{\left( {2 - ;2} \right)}^2} + {{\left( {3 + n - ;3} \right)}^2}} = ;\sqrt 2\)

Squaring both sides,

⇒ 1 + 0 + n² = 2

⇒ n² = 1

∴ n = 1

Concept:

• The equation of a line with direction ratio (a, b, c) that passes through the point (x₁, y₁, z₁) is given by the formula: \(\frac{{x - {{\rm{x}}_1}}}{{\rm{a}}} = \frac{{y - {{\rm{y}}_1}}}{{\rm{b}}} = \frac{{z - {{\rm{z}}_1}}}{{\rm{c}}}\)
• Equation of plane in 3-D: lx + my + nz = d

If line given by plane then following condition will be hold

1. Parallel vector of line and normal vector of plane perpendicular to each other

          So, al + bm + nz = 0

1. Point satisfy the equation of plane lx₁ + my₁ + nz₁ – d =0

Calculation:

Given: Equation of line \(\frac{{x - 1}}{2} = \frac{{y - 2}}{3} = \frac{{z - 3}}{4}\)

So, The direction ratio of line is 2, 3, and 4

And The line passing through point (1, 2, 3)

According to the concept used 

The point (1, 2, 3) will satisfy the planes

Option 1:

x + y + z = 6, x + 2y – 3y = -4

On putting (1, 2, 3) we get 

1 + 2 + 3 = 6 and 1 + 2 × 2 - 3 × 3 = - 4

So, option 1 is correct

Option 2:

x + 2y – 2z = -1, 4x + 4y – 5z – 3 = 0

On putting (1, 2, 3) we get 

1 + 2 × 2 – 2 × 3 = -1, 4 × 1 + 4 × 2 – 5 × 3 – 3 = -6 ≠ 0

So, option 2 is incorrect

Option 3:

3x + 2y – 3z = 0, 3x – 6y + 3z = -2

On putting (1, 2, 3) we get 

3 × 1 + 2 × 2 – 3 × 3 = -2 ≠ 0

So, option 3 is incorrect

Option 4:

3x + 2y – 4z = -2, 3x – 6y + 4z = 0

On putting (1, 2, 3) we get 

→ 3 × 1 + 2 × 2 – 4 × 3 = -2 = - 2 

But: On putting (1, 2, 3) in 3x – 6y + 4z = 0 we get 

→ 3 × 1 - 6 × 2 + 4 × 3 = 3 ≠ 0  

So, option 4 is incorrect.

Concept:

1. If a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 are a plane equations, then angle between planes can be found using the following formula

\(\cos \theta = ;\frac{{{a_1}{a_2} + ;{b_1}{b_2} + ;{c_1}{c_2};}}{{\sqrt {{{({a_1})}^2} + ;{{({b_1})}^2}; + {{({c_1})}^2};} ;\sqrt {{{({a_2})}^2} + ;{{({b_2})}^2}; + {{({c_2})}^2}} }}\)

1. Condition of parallel plane:

If plane a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 are a parallel

i.e. \(\frac{{{a_1}}}{{{a_2}}} = ;\frac{{{b_1}}}{{{b_2}}} = ;\frac{{{c_1}}}{{{c_2}}}\)

1. Distance between two parallel plane ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0 is

\(Distance; = ;\left| {\frac{{{d_1} - ;{d_2}}}{{\sqrt {{a^2} + ;{b^2} + ;{c^2}} ;}}} \right|\)

Calculation:

We have to find the angle between the planes 2x – y + z = 1 and x + y + 2z = 3

So, a₁ = 2, b₁ = -1, c₁ = 1 and a₂ = 1, b₂ = 1 and c₂ = 2

Let θ will be angle between these two plane

So, \(\cos \theta = ;\frac{{{a_1}{a_2} + ;{b_1}{b_2} + ;{c_1}{c_2};}}{{\sqrt {{{({a_1})}^2} + ;{{({b_1})}^2}; + {{({c_1})}^2};} ;\sqrt {{{({a_2})}^2} + ;{{({b_2})}^2}; + {{({c_2})}^2}} }} = ;\frac{{2 \times 1; + ;\left( { - 1} \right) \times 1; + ;1 \times 2}}{{\sqrt {{2^2} + ;{{\left( { - 1} \right)}^2}; + ;{1^2};} ;\sqrt {{1^2} + ;{1^2}; + ;{2^2};} }}\)

\(\Rightarrow ;\cos \theta = ;\frac{{2 - 1 + 2}}{{\sqrt 6 ;\sqrt 6 }} = ;\frac{3}{6} = ;\frac{1}{2} = ;\cos \frac{\pi }{3}\)

∴ θ = π/3

So, statement 1 is true.

Now, we have to the find distance between the planes

6x – 3y + 6z + 2 = 0 and

2x – y + 2z + 4 = 0

Rewrite the equation 6x – 3y + 6z + 2 = 0

⇒ 3 (2x – y + 2z + 2/3) = 0

⇒ 2x – y + 2z + 2/3 = 0

So, 2x – y + 2z + 2/3 = 0 and 2x – y + 2z + 4 = 0 are parallel

a = 2, b = -1, c = 2, d₁ = 2/3 and d₂ = 4

\(Distance; = ;\left| {\frac{{{d_1} - ;{d_2}}}{{\sqrt {{a^2} + ;{b^2} + ;{c^2}} ;}}} \right| = ;\left| {\frac{{\frac{2}{3} - ;4}}{{\sqrt {{2^2} + ; - {1^2} + ;{2^2}} ;}}} \right| = \left| {\frac{{\frac{{ - 10}}{3}}}{{\sqrt 9 ;}}} \right| = ;\frac{{10}}{{3; \times 3}} = ;\frac{{10}}{9}\)

So, statement 2 is true.

Concept:

Cosine formula or cosine rule

• a² = b² + c² – 2bc cos A

          Or, \(\cos A = ;\frac{{{b^2} + ;{c^2}; - ;{a^2}}}{{2bc}}\)

• b² = a² + c² – 2ac cos B
• c² = a² + b² – 2ab cos C

Calculation:

 

In triangle OPQ,

\(OP = \sqrt {{m^2} + ;{n^2}} \)   ⇒ OP² = m² + n²

\(OQ = \sqrt {{r^2} + ;{s^2}}\)    ⇒ OQ² = r² + s²

\(PQ = ;\sqrt {{{(m - ;r)}^2} + {{(n - s)}^2}}\)   ⇒ PQ² = (m – r)² + (n – s)² = m² + r² – 2mr + n² + s² – 2ns

From cosine formula.

\(\cos \alpha = \frac{{O{P^2} + ;O{Q^2}; - ;P{Q^2}}}{{2 \times OP \times OQ}}\)

\(\cos \alpha = \frac{{{m^2} + ;{n^2} + {r^2} + ;{s^2} - {m^2} - {r^2} + 2mr - ;{n^2} - {s^2} + 2ns}}{{2 \times \sqrt {{m^2} + ;{n^2}} \times \sqrt {{r^2} + ;{s^2}} }}\)

\(\cos \alpha = \frac{{2mr + ;2ns}}{{2\sqrt {\left( {{m^2} + {n^2}} \right)({r^2} + {s^2}} }} = \frac{{mr + ;ns}}{{\sqrt {\left( {{m^2} + {n^2}} \right)({r^2} + {s^2}} }};\)

So, statement 1 is wrong

In any triangle ABC, a² = b² + c² – 2bc cos A.

By cosine rule it is true so statement 2 is true.