Quantitative Aptitude PYQs

Concepts:

Suppose there is a given statement P (n) involving the natural number n such that

• The statement is true for n = 1, i.e., P (1) is true, and
• If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., truth of P (k) implies the truth of P (k + 1).

Then, P (n) is true for all natural numbers n

Calculation:

Given:

P (n) = 121ⁿ – 25ⁿ + 1900ⁿ – (-4) ⁿ

Now, P (1) = 121¹ – 25¹ + 1900¹ – (-4)¹

⇒ P (1) = 121 – 25 + 1900 + 4

⇒ P (1) = 2000

Therefore we can say that P (n) is divisible by 2000.

Calculation:

Numbers between 100 and 1000 are three digits number, so we have to fill three places.

Total numbers = 5

(They are 5, 6, 7, 8 and 9)

Number of choices for the 1^st digit = 5

Repetition of digits is not allowed therefore available numbers for 2^nd place is four

⇒ Number of choices for the 2^nd digit = 4

Repetition of digits is not allowed therefore available numbers for 1^st place is three

⇒ Number of choices for the 3^rd digit = 3

∴ Total possible combinations = 5 × 4 × 3 = 60

Concepts:

Let z = x + iy be a complex number.

• Modulus of z = \(\left| \text{z} \right|=\text{ }\!\!~\!\!\text{ }\sqrt{{{\text{x}}^{2}}+{{\text{y}}^{2}}}=\text{ }\!\!~\!\!\text{ }\sqrt{\text{Re}{{\left( \text{z} \right)}^{2}}+\text{ }\!\!~\!\!\text{ Im }\!\!~\!\!\text{ }{{\left( \text{z} \right)}^{2}}}\)
• arg (z) = arg (x + iy) = \({\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)

Calculation:

We have to find value of ′x′ for which should be a non-zero integral.

Given: ∣1 − 2i∣^x = 5^x

Let first find value of ∣1 − 2i∣

\( \Rightarrow {\rm{\;}}\mid 1{\rm{\;}} - {\rm{\;}}2{\rm{i}}\mid {\rm{\;}} = {\rm{\;}}\sqrt {{{\left( 1 \right)}^2} + {{\left( { - 2} \right)}^2}} {\rm{\;}} = {\rm{\;}}\sqrt 5\)​     ...(∵ z = x + iy ⇒ \(\left| {\rm{z}} \right| = {\rm{\;}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}}\))

Now, ∣1 − 2i∣^x = 5^x

⇒ (√5) ^x = 5^x

⇒ 5^x/2 = 5^x

Base is same so power is also equal

Therefore x/2 = x

⇒ x = 2x

∴ x = 0

Hence no non-zero integral solutions.

Concepts:

• Arithmetic mean of two positive numbers a and b is \(\frac{{a\; + \;b}}{2}\)
• Geometric mean of two positive numbers a and b is \(\sqrt {ab}\)

Calculation:

Given: Ratio of AM to GM of two positive numbers a and b is 5 : 3

\(\Rightarrow \;\frac{{\left( {\frac{{a\; + \;b}}{2}} \right)}}{{\sqrt {ab} }} = \;\frac{5}{3}\)

\(\Rightarrow \;\frac{{a\; + \;b}}{{2\sqrt {ab} }} = \;\frac{5}{3}\)

\(\Rightarrow \;\frac{{\frac{{a\; + \;b}}{b}}}{{\frac{{\sqrt {ab} }}{b}}} = \;\frac{{10}}{3}\)

\(\Rightarrow \;\frac{{\frac{a}{b} + 1}}{{\sqrt {\frac{a}{b}} }} = \;\frac{{10}}{3}\)

\(\Rightarrow \;\frac{a}{b} + 1 = \frac{{10}}{3}\sqrt {\frac{a}{b}}\)

Let a/b = x²     ...(Assume that a > b)

⇒ x² + 1 = (10/3) x

⇒ 3x2 +3 = 10x

⇒ 3x2 -10x + 3 = 0

⇒ 3x2 - 9x - x + 3 = 0

⇒ (3x – 1) (x – 3) = 0

⇒ x = 1/3 and 3

⇒ x² = 1/9 and 9

⇒ a/b = 9     ...(∵ a > b)

Therefore, a : b = 9 : 1.

Concepts:

1. Let z = x + iy be a complex number.

• Modulus of z = \(\left| {\rm{z}} \right| = {\rm{\;}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{\;}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{\;Im\;}}{{\left( {\rm{z}} \right)}^2}}\)
• arg (z) = arg (x + iy) = \({\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)
• Conjugate of z =

 = x – iy

• \(z\bar z = {\left| z \right|^2}\)
• \(\left| {\rm{z}} \right| = \left| {\bar z} \right|\)
• \(\overline {\left( {\bar z} \right)} = z\)

2. Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂ be any two complex numbers.

• |z₁z₂| = |z₁||z₂|
• \(\left| {\frac{{{z_1}}}{{{z_2}}}} \right| = \frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}},\;\left| {{z_2}} \right| \ne 0\)
• \(\overline {{z_1} \pm {z_2}} = \;\overline {{z_1}} \pm \overline {{z_2}}\)

 

Calculation:

Given: |α | = 1

We have to find the value of \(\left| {\frac{{\alpha - \beta }}{{1 - \alpha \bar \beta }}} \right|\)

Multiple with \(\bar \alpha\) in both Numerator and denominator

\(\Rightarrow \;\left| {\frac{{\alpha - \beta }}{{1 - \alpha \bar \beta }}} \right| = \left| {\frac{{\left( {\alpha - \beta } \right)\; \times \;\bar \alpha }}{{\left( {1 - \alpha \bar \beta } \right)\; \times \;\bar \alpha }}} \right|\)

\(= \left| {\frac{{\left( {\alpha \bar \alpha - \beta \bar \alpha } \right)\;}}{{\left( {1 - \alpha \bar \beta } \right) \times \;\bar \alpha \;}}} \right|\)

We know that \(z\bar z = {\left| z \right|^2}\)

So, \(\alpha \bar \alpha = {\left| \alpha \right|^2} = 1\)     ...(∵ |α| = 1)

\( = \left| {\frac{{\left( {1 - \beta \bar \alpha } \right)\;}}{{\left( {1 - \alpha \bar \beta } \right)\; \times \;\bar \alpha \;}}} \right|\)

We know that \(\overline {\left( {\bar z} \right)} = z\;and\;\overline {{z_1} - {z_2}\;} = \;\overline {{z_1}} - \;\overline {{z_2}} \)

So, \(\left( {1 - \beta \bar \alpha } \right) = \overline {\left( {1 - \alpha \bar \beta } \right)}\)

\( = \left| {\frac{{\overline {\left( {1 - \alpha \bar \beta } \right)} \;}}{{\left( {1 - \alpha \bar \beta } \right)\; \times \;\bar \alpha \;}}} \right|\)

\(=\frac{\left| \overline{\left( 1-\alpha \bar{\beta } \right)}~ \right|}{\left| \left( 1-\alpha \bar{\beta } \right) \right|\left| {\bar{\alpha }} \right|}\text{ }\!\!~\!\!\text{ }\left( \because \left| \frac{{{z}_{1}}}{{{z}_{2}}} \right|=~\frac{\left| {{z}_{1}} \right|}{\left| {{z}_{2}} \right|}~and~\left| z1z2 \right|=~\left| z1 \right|\left| z2 \right| \right)\)

We know that \(\left| {\rm{z}} \right| = \left| {\bar z} \right|\)

\(\therefore \left| {{\rm{\alpha }}} \right| = {\rm{}}\left| {\bar \alpha } \right| = 1\;and\;\left| {\left( {1 - \alpha \bar \beta } \right)} \right| = \left| {\overline {\left( {1 - \alpha \bar \beta } \right)} \;} \right|\)

\(\Rightarrow \;\left| {\frac{{\alpha - \beta }}{{1 - \alpha \bar \beta }}} \right| = \frac{{\left| {\overline {\left( {1 - \alpha \bar \beta } \right)} \;} \right|}}{{\left| {\left( {1 - \alpha \bar \beta } \right)} \right|\left| {\bar \alpha } \right|}} = 1\)

Concepts:

Convert decimal to binary

Conversion steps:

• Divide the number by 2.
• Get the integer quotient for the next iteration.
• Get the remainder for the binary digit.
• Repeat the steps until the quotient is equal to 0.

 

Calculation:

Convert 31₁₀ to binary

 

So, binary form of 31 is 11111.

Concepts:

1. Let z = x + iy be a complex number.

• Modulus of z = \(\left| {\rm{z}} \right| = {\rm{\;}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{\;}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{\;Im\;}}{{\left( {\rm{z}} \right)}^2}}\)
• arg (z) = arg (x + iy) = \({\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)

2. Modulus- amplitude form a complex number ⇔ z = r (cos θ + i sin θ)

Where r = |z| and θ = arg z, (-π  < θ < π)

Calculation:

Let z = √3 + i = r (cos θ + i sin θ)

Comparing real and imaginary part,

⇒ r cos θ = √3     ….1)

⇒ r sin θ = 1     ….2)

Squaring and adding above two equations,

⇒ r² (cos² θ + sin² θ) = 3 + 1 = 4

⇒ r² = 4     ...(∵ cos² θ + sin² θ = 1)

∴ r = 2

Now dividing equation 2 by equation 1

\(\frac{{{\rm{r\;}}\sin {\rm{\theta }}}}{{{\rm{r\;}}\cos {\rm{\theta }}}} = {\rm{}}\frac{1}{{\sqrt 3 }}\)

⇒ tan θ = tan (π/6)

∴ θ = π/6

Hence z = r (cos θ + i sin θ) = 2 (cos π/6 + i sin π/6)

Concepts:

(1 + x)ⁿ = 1 + ⁿc₁ x + ⁿc₂ x² + ⁿc₃ x³ + …. + ⁿc_n xⁿ

(1 - x)ⁿ = 1 - ⁿc₁ x + ⁿc₂ x² - ⁿc₃ x³ + …. + (-1)^{n n}c_n xⁿ

Total number of terms in the expansion of (1 ± x)ⁿ is n + 1

(1 + x) ⁿ + (1 - x) ⁿ = 2 [1 + ⁿc₂ x² + …]

∴ Total number of non-zero terms in the expansion of (1 + x)ⁿ + (1 - x)ⁿ is (n + 1)/2

Calculation:

We have to find the number of non-zero terms in the expansion of (1 + 2√3x)¹¹ + (1 – 2√3x)¹¹

We know that total number of non-zero terms in the expansion of (1 + x)ⁿ + (1 - x)ⁿ is (n +1)/2

Here n = 11

∴ Total number of non-zero terms in the expansion = (11 + 1)/2 = 6

Concepts:

1. Let Standard form of quadratic Functions: ⇒ f(x) = ax² + bx + c

Steps to Graphing Quadratic Equations

Step-1: Check the sign of a

• If a > 0 the graph will be a convex parabola.
• If a < 0 the graph will be a concave parabola

Step- 2: Vertex = \(\rm \left( {\frac{{ - b}}{{2a}},\frac{{ - D}}{{4a}}} \right)\), Where D is Discriminants (D = b² – 4ac)

Step- 3: Parabola is symmetric about the line that intersects x = \(\rm \frac{{ - b}}{{2a}}\)

Step- 4: Find the coordinates of the points of intersection with the x axis, we must solve the equation f(x) = 0

 

2. Greatest Integer Function: Greatest Integer Function [x] indicates an integral part of the real number x which is nearest and smaller integer to x. It is also known as floor of x

• In general, If \(\rm n \le x\; \le n + 1\) Then [x] = n (n ∈ Integer)
• Means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

 

Example:

Calculation:

We have to find number of solutions of the equation x² - 4x + [x] = 0 in the interval [0, 2]

x² - 4x + [x] = 0

⇒ [x] = -x² + 4x

Let g(x) = [x] and f(x) = -x² + 4x

Graphing of function f(x):

f(x) = -x² + 4x

Above equation is compare with ax² + bx + c,

a = -1, b = 4 and c = 0

Now, Discriminants = D = b² – 4ac = 4² – 0 = 16

Here a < 0 the graph will be a concave parabola

Vertex = \(\rm \left( {\frac{{ - b}}{{2a}},\frac{{ - D}}{{4a}}} \right) = \;\left( {\frac{{ - \;4}}{{2\; \times \; - 1}},\frac{{ - 16}}{{4\; \times \; - 1}}} \right) = \left( {2,\;4} \right)\)

Points of intersection with the x axis: f(x) = 0

⇒ -x² + 4x = 0

∴ x = 0 and 4

We can see that x = 0 is only one point is common in both curves in the interval [0, 2]

∴ Number of solutions of the equation is one.

Concepts:

• \({\log _e}y = x \Leftrightarrow y = {e^x}\)

 

Calculation:

Given: log₁₀ a is negative.

⇒ log₁₀ a < 0

Let log₁₀ a = - b (Here b is positive)

⇒ a = 10^-b

Given a lies between 0 to 1

⇒ 0 < a < 1

⇒ 0 < 10^-b < 1

Hence it is negative power of 10 is between 0 and 1

∴ Option 2 is correct.

Concepts:

Let us consider sequence a₁, a₂, a₃ …. a_n is an G.P.

• Common ratio = r = \(\frac{{{a_2}}}{{{a_1}}} = \frac{{{a_3}}}{{{a_2}}} = \ldots = \frac{{{a_n}}}{{{a_{n - 1}}}}\)
• n^th term of the G.P. is an = arⁿ⁻¹
• Sum of n terms = s = \(\frac{{a\;\left( {{r^n} - 1} \right)}}{{r - 1}}\); where r >1
• Sum of n terms = s = \(\frac{{a\;\left( {1 - {r^n}} \right)}}{{1 - r}}\); where r <1
• Sum of infinite GP = \({{\rm{s}}_\infty } = {\rm{}}\frac{{\rm{a}}}{{1{\rm{}} - {\rm{r}}}}{\rm{}}\); |r| < 1

Where a is 1^st term and r is common ratio.

Calculation:

Given: The third term of a GP is 3

Let 'a' be the first term and 'r' be the common ratio.

∴ T₃ = ar² = 3

We know that T_n = a r^n-1

So, T₁ = a, T₂ = ar, T₃ = ar², T₄ = ar³, T₅ = ar⁴

Now, Product of the first five terms = a × ar × ar² × ar³ × ar⁴ = a⁵r¹⁰ = (ar²)⁵ = 3⁵ = 243

Concepts:

• If a, b, c are in A.P ⇔ b = \(\frac{{a\; + \;c}}{2}\)
• If a, b, c are in G.P ⇔ b² = ac
• If a, b, c are in H.P ⇔ \(b = \frac{{2ac}}{{a\; + \;c}}\)

 

Calculation:

Given: x, 3/2, z are in AP

∴ 2 × (3/2) = x + z

⇒ x + z = 3     ….1)

Now, x, 3, z are in GP

∴ 3² = xz

⇒ xz = 9     ….2)

Let x, y, z are in HP

∴ \(y = \frac{{2xz}}{{x + z}} = \frac{{2 \times 9}}{3} = 6\)

Hence x, 6, z are in HP

Concepts:

• Sine law ⇔ \(\frac{{\rm{a}}}{{\sin {\rm{A}}}} = {\rm{\;}}\frac{{\rm{b}}}{{\sin {\rm{B}}}} = {\rm{\;}}\frac{{\rm{c}}}{{\sin {\rm{C}}}}\)

Where a, b and c are sides and A, B and C are angles.

Here; Side a faces angle A, side b faces angle B and side c faces angle C

Calculation:

Given: a = 2, b = 3 and sin A = 2/3

Apply sine rule,

\(\frac{{\rm{a}}}{{\sin {\rm{A}}}} = \frac{{\rm{b}}}{{\sin {\rm{B}}}}\)

\( \Rightarrow \sin {\rm{B}} = \frac{{b\sin {\rm{A}}}}{a}\)

\( \Rightarrow \sin {\rm{B}} = \frac{{3\; \times \left( {\frac{2}{3}} \right)}}{2} = 1\)

∴ B = π/2

Concepts:

• sin (x + y) = sin x cos y + cos x sin y
• sin (x – y) = sin x cos y - cos x sin y

 

Calculation:

Given:

\(\frac{{\sin \left( {x\; + \;y} \right)}}{{\sin \left( {x\; - \;y} \right)}} = \;\frac{{a\; + \;b}}{{a\; - \;b}}\)

\(\Rightarrow \;\frac{{\sin x\cos y\; + \;\cos x\sin y}}{{\sin x\cos y\; - \;\cos x\sin y}} = \;\frac{{a\; + \;b}}{{a\; - \;b}}\)

Using Componendo and dividendo Rule, we get

\(\Rightarrow \frac{{(\sin x\cos y + \cos x\sin y) + (\sin x\cos y - \cos x\sin y)}}{{(\sin x\cos y + \cos x\sin y) - (\sin x\cos y - \cos x\sin y)}} = \frac{{\left( {a + b} \right) + \left( {a - b} \right)}}{{\left( {a + b} \right) - \left( {a - b} \right)}}\)

\(\Rightarrow \;\frac{{2{\rm{\;sin}}x\cos y}}{{2\cos x\sin y}} = \frac{{2a}}{{2b}}\)

\(\therefore \frac{{\tan x}}{{\tan y}} = \frac{a}{b}\)

Concepts:

• 1 + cos A = 2 cos² (A/2)
• 1 – cos A = 2 sin² (A/2)

Calculation:

We know that 2 cos² (A/2) = 1 + cos A

\(\Rightarrow \cos \left( {\frac{A}{2}} \right) = \pm \sqrt {\frac{{1 + \cos A}}{2}}\)

Here only one value of cos A possible then cos A must be -1,

Therefore ‘A’ must be odd multiple 180°

Verification:

Let A = 180°

L.H.S = cos (A/2) = cos 90° = 0

R.H.S = \(\sqrt {\frac{{1 + \cos A}}{2}} = \sqrt {\frac{{1 + \cos 180}}{2}} = \sqrt {\frac{{1 - 1}}{2}} = 0\)

So, L.H.S = R.H.S

Concept:

Ellipse: The locus of a point which moves such that its distance from a fixed point and a fixed straight line is always less than one (Eccentricity = e < 1)

 

Calculation:

Given:

Vertices = (± 5, 0) and foci = (± 4, 0)

⇒ (± a, 0) = (± 5, 0) and (± ae, 0) = (± 4, 0)

∴ a = 5 and ae = 4

⇒ ae = 4

Squaring both sides,

⇒ a²e² = 4² (∵ a²e² = a² – b²)

⇒ a² – b² = 16

⇒ 5² – b² = 16

⇒ 25 – 16 = b²

⇒ b² = 9

Now, Equation of ellipse \(\frac{{{\rm{\;}}{x^2}}}{{{a^2}}} + \frac{{{{\bf{y}}^2}}}{{{b^2}}} = 1\) (a > b)

⇒ \(\frac{{{\rm{\;}}{x^2}}}{{25}} + \frac{{{{\bf{y}}^2}}}{9} = 1\)

∴ Option 1 is correct.

Concept:

Vector equation of a plane passing through three points with position vectors \(\vec a,{\rm{\;}}\vec b,{\rm{\;}}\vec c\) is

\(\left( {\vec r - \vec a} \right) \cdot [\left( {\vec b - \vec a} \right) \times \left( {\vec c - \vec a} \right) = 0\)

Cartesian equation of a plane passing through three points:

Let us now consider the coordinates of the three noncollinear points as A(x₁, y₁​, z₁​), B(x₂​, y₂​, z₂​), and C(x₃​, y₃​, z₃​).

Equation of plane is:

\(\left| {\begin{array}{*{20}{c}} {x - {\rm{\;}}{x_1}}&{y - {\rm{\;}}{y_1}}&{z - {\rm{\;}}{z_1}}\\ {{x_2} - {\rm{\;}}{x_1}}&{{y_2} - {\rm{\;}}{y_1}}&{{z_2} - {\rm{\;}}{z_1}}\\ {{x_3} - {\rm{\;}}{x_1}}&{{y_3} - {\rm{\;}}{y_1}}&{{z_3} - {\rm{\;}}{z_1}} \end{array}} \right| = 0\)

If \(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]\) then determinant of A is given by:

|A| = a₁₁ × {(a₂₂ × a₃₃) – (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) – (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) – (a₂₂ × a₃₁)}

Calculation:

We have to find the equation of the plane which is passing through the point (-2, 6, -6), (-3, 10, -9), and (-5, 0, -6)

So, (x₁, y₁, z₁) = (-2, 6, -6), (x₂, y₂, z₂) = (-3, 10, -9) and (x₃, y₃, z₃) = (-5, 0, -6)

We know that the Equation of plane is:

 \(\left| {\begin{array}{*{20}{c}} {x - {\rm{\;}}{x_1}}&{y - {\rm{\;}}{y_1}}&{z - {\rm{\;}}{z_1}}\\ {{x_2} - {\rm{\;}}{x_1}}&{{y_2} - {\rm{\;}}{y_1}}&{{z_2} - {\rm{\;}}{z_1}}\\ {{x_3} - {\rm{\;}}{x_1}}&{{y_3} - {\rm{\;}}{y_1}}&{{z_3} - {\rm{\;}}{z_1}} \end{array}} \right| = 0\)

So, the plane passing through points (-2, 6, -6), (-3, 10, -9) and (-5, 0, -6) is

\(\left| {\begin{array}{*{20}{c}} {x - \left( { - 2} \right)}&{y - {\rm{\;}}6}&{z - \left( { - 6} \right)}\\ { - 3 - {\rm{\;}}\left( { - 2} \right)}&{10 - {\rm{\;}}6}&{ - 9 - {\rm{\;}}\left( { - 6} \right)}\\ { - 5 - {\rm{\;}}\left( { - 2} \right)}&{0 - {\rm{\;}}6}&{ - 6 - {\rm{\;}}\left( { - 6} \right)} \end{array}} \right| = 0\)

\(\left| {\begin{array}{*{20}{c}} {x + 2}&{y - 6}&{z + 6}\\ { - 1}&4&{ - 3}\\ { - 3}&{ - 6}&0 \end{array}} \right| = 0\)

Now, Expanding along R₁

(x + 2) (0 – 18) – (y – 6) (0 - 9) + (z + 6) (6 + 12) = 0

⇒ -18 (x + 2) + 9 (y – 6) + 18 (z + 6) = 0

⇒ -9 [2 (x + 2) – (y – 6) – 2 (z + 6)] = 0

⇒ 2x + 4 – y + 6 - 2z – 12 = 0

⇒ 2x – y – 2z – 2 = 0

⇒ 2x – y – 2z = 2

Shortcut Method: Substitute the point (-2, 6, -6), (-3, 10, -9) and (-5, 0, -6) in equation of plane. If points satisfy the equation of the plane then it will be the correct option.

Concept:

1. The intercept form of the line ⇔ \(\frac{x}{a} + \frac{y}{b} = 1\)

Where a is the x- intercept and b is the y- intercept

1. Section Formula: Section formula is used to determine the coordinate of a point that divides a line into two parts such that ratio of their length is m : n
2. Let P and Q be the given two points (x₁, y₁) and (x₂, y₂) respectively and M(x, y) be the point dividing the line- segment PQ internally in the ratio m: n
3. Internal Section Formula: When the line segment is divided internally in the ration m: n, we use this formula. ⇔ \(\left( {x,{\rm{\;}}y} \right) = \left( {\frac{{m{x_2}{\rm{\;}} + {\rm{\;}}n{x_1}}}{{m\; + {\rm{\;}}n}},\frac{{m{y_2}{\rm{\;}} + {\rm{\;}}n{y_1}}}{{m\; + {\rm{\;}}n}}} \right)\)
4. External Section Formula: When the point M lies on the external part of the line segment. ⇔ \(\left( {x,{\rm{\;}}y} \right) = \left( {\frac{{m{x_2} - {\rm{\;}}n{x_1}}}{{m\; - {\rm{\;}}n}},\frac{{m{y_2} - {\rm{\;}}n{y_1}}}{{m\; - {\rm{\;}}n}}} \right)\)

 

Calculation:

Intercept form of the line is \(\frac{x}{a} + \frac{y}{b} = 1\)

At x axis point is A (a, 0)

At y axis point is B (0, b)

Case 1: Given point P (2, 3) divides A and B in ratio of m : n = 3 : 2

Applying Internal Section Formula,

We know that \(\left( {x,{\rm{\;}}y} \right) = \left( {\frac{{m{x_2}{\rm{\;}} + {\rm{\;}}n{x_1}}}{{m\; + {\rm{\;}}n}},\frac{{m{y_2}{\rm{\;}} + {\rm{\;}}n{y_1}}}{{m\; + {\rm{\;}}n}}} \right)\)

\(\Rightarrow \;\left( {2,\;3} \right) = \;\left( {\frac{{3\left( 0 \right) + \;2\left( a \right)}}{{3 + 2}},\frac{{3\left( b \right) + 2\left( 0 \right)}}{{3 + 2}}} \right)\)

\( \Rightarrow \;\left( {2,\;3} \right) = \;\left( {\frac{{2a}}{5},\frac{{3b}}{5}} \right)\)

Now,

\(\frac{{2a}}{5} = 2\;and\;\frac{{3b}}{5} = 3\)

⇒ a = 5 and b = 5

∴ Equation of line is \(\frac{x}{5} + \frac{y}{5} = 1\) ⇒ x + y = 5

Case 2: Given point P (2, 3) divides A and B in ratio of m : n = 2 : 3

\( \Rightarrow \;\left( {2,\;3} \right) = \;\left( {\frac{{2\left( 0 \right) + \;3\left( a \right)}}{{3 + 2}},\frac{{2\left( b \right) + 3\left( 0 \right)}}{{3 + 2}}} \right)\)

\( \Rightarrow \;\left( {2,\;3} \right) = \;\left( {\frac{{3a}}{5},\frac{{2b}}{5}} \right)\)

Now,

\(\frac{{3a}}{5} = 2\;and\;\frac{{2b}}{5} = 3\)

⇒ a = 10/3 and b = 15/2

∴ Equation of line is \(\frac{x}{\left( {}^{10}\!\!\diagup\!\!{}_{3}\; \right)}+\frac{y}{\left( {}^{15}\!\!\diagup\!\!{}_{2}\; \right)}=1\)

\(\Rightarrow {\rm{\;}}\frac{{3x}}{{10}} + \frac{2}{{15}} = 1\)

⇒ 9x + 4y = 30

Shortcut Method:

• Check through the options: Substitute the point (2, 3) in all options

⇒ Only option 4 satisfied the point (2, 3), so option 4 is correct.

Concept:

• General equation of a sphere cantered at the point (x₀, y₀, z₀) with radius R is given by
• (x – x₀)² + (y – y₀)² + (z – z₀)² = R²

 

Calculation:

Given: Centre is at (-2, 3, 4) and radius is 6

⇒ (x₀, y₀, z₀) = (-2, 3, 4) and R = 6

We know that general equation of sphere is (x – x₀)² + (y – y₀)² + (z – z₀)² = R²

⇒ (x – (-2)) ² + (y – 3)² + (z – 4)² = 6²

⇒ (x + 2) ² + (y – 3)² + (z – 4)² = 6²

⇒ x² + 4x + 4 + y² – 6y + 9 + z² – 8z + 16 = 36

⇒ x² + y² + z² + 4x – 6y – 8z + 29 = 36

⇒ x² + y² + z² + 4x – 6y – 8z = 7

∴ Option 1 is correct.

Concept:

• A cross or vector product is not commutative ⇔ \(\vec a \times \vec b = - {\rm{\;}}\vec b \times \vec a\)
• \(\vec a{\rm{\;and\;}}\vec b\) are two vectors parallel to each other ⇔ \(\vec a \times \vec b = 0\)
• Cross product of parallel vectors is zero ⇔ \(\vec a \times \vec a = 0,{\rm{\;}}\vec b \times \vec b = 0{\rm{\;and\;}}\vec c \times \vec c = 0\)

 

Calculation:

Given: \(\vec a + 2\vec b + 3\vec c = \vec 0\)

\(\Rightarrow \vec a = \; - \left( {2\vec b + 3\vec c} \right)\)

Now, \(\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a\)

\(= - \left( {2\vec b + 3\vec c} \right) \times \vec b + \vec b \times \vec c - \vec c \times \left( {2\vec b + 3\vec c} \right)\)

\(= - 2\vec b \times \vec b - 3\vec c \times \vec b + \vec b \times \vec c - \vec c \times 2\vec b -\vec c \times 3\vec c\)

We know that \(\vec a \times \vec b = - {\rm{\;}}\vec b \times \vec a\)

So, \(- k\vec c \times \vec b = \;k\vec b \times \vec c\)

\(= \;0 + \;3\left( {\vec b \times \vec c} \right) + \left( {\vec b \times \vec c} \right) + 2\left( {\vec b \times \vec c} \right) + 0{\rm{\;}}\left( {\vec b \times \vec b = 0{\rm{\;and\;}}\vec c \times \vec c = 0} \right)\)

\(= 6\left( {\vec b \times \vec c} \right)\)

∴ λ = 6