Quantitative Aptitude PYQs

Concept:

• The slope of a line passing through the distinct points (x₁, y₁) and (x₂, y₂) is \({\rm{m}} = \;\frac{{{y_2}{\rm{\;}} - {\rm{\;}}{y_1}}}{{{x_2}{\rm{\;}} - {\rm{\;}}{x_1}}}\)
• Equation of line is (y – y₁) = m (x – x₁)
• Collinear points are points that all lie in the same line.

 

Calculation:

Let, a line passing through the points (0, 0) and (a, b)

⇒ Slope of the line  = \({\rm{m}} = \;\frac{{{b}{\rm{\;}} - {\rm{\;}}{0}}}{{{a}{\rm{\;}} - {\rm{\;}}{0}}} = \frac{b}{a}\)

Now, Equation of line is (y – y1) = m (x – x1)

⇒ (y - 0) = (b/a) × (x - 0)

⇒ ay = bx

All given points lie on the line of the equation,

∴ All points are collinear.

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 = \(\left| {\frac{{{\rm{A}}{{\rm{x}}_1}{\rm{\;}} + {\rm{\;B}}{{\rm{y}}_1}{\rm{\;}} + {\rm{\;C}}{{\rm{z}}_1} - {\rm{\;d}}}}{{\sqrt {{{\rm{A}}^2}{\rm{\;}} + {\rm{\;\;}}{{\rm{B}}^2}{\rm{\;}} + {\rm{\;}}{{\rm{C}}^2}} }}} \right|\)

Calculation:

We know that normal is always perpendicular to the plane,

Given: Equation of the plane is x + 2y – 2z = 9

⇒ x + 2y – 2z - 9 = 0

Now we have to find the distance from the origin (0, 0, 0)

We know that distance = \(\left| {\frac{{{\rm{A}}{{\rm{x}}_1}{\rm{\;}} + {\rm{\;B}}{{\rm{y}}_1}{\rm{\;}} + {\rm{\;C}}{{\rm{z}}_1} - {\rm{\;d}}}}{{\sqrt {{{\rm{A}}^2}{\rm{\;}} + {\rm{\;\;}}{{\rm{B}}^2}{\rm{\;}} + {\rm{\;}}{{\rm{C}}^2}} }}} \right|\)

∴ Distance = \(\left| {\frac{0 + 0 - 0 - 9}{{\sqrt {{{\rm{1}}^2}{\rm{\;}} + {\rm{\;\;}}{{\rm{2}}^2}{\rm{\;}} + {\rm{\;}}{{\rm{(-2)}}^2}} }}} \right| = \frac{9}{3} = 3\)

Concept:

1. Direction angles: If α, β, and γ are the angles made by the line segment with the coordinate axis then these angles are termed to be the direction angles.
2. Direction cosines:  The cosines of direction angles are the direction cosines of the line. Hence, cos α, cos β and cos γ are called as the direction cosines, It is denoted by l, m and n. ⇔ l = cos α, m = cos β and n = cos γ

   

1. The sum of squares of the direction cosines of a line is equal to unity.
2. l² + m² + n² = 1 or cos² α + cos² β + cos² γ = 1
3. Direction ratios: Any numbers which are proportional to the direction cosines of a line are called as the direction ratios. It is denoted by ‘a’, ‘b’ and ‘c’.
4. a ∝ l, b ∝ m and c ∝ n ⇔ a = kl, b = km and c = kn Where k is a constant.

 

Calculation:

We know that sum of squares of the direction cosines of a line is equal to unity.

⇒ cos² α + cos² β + cos² γ = 1

⇒ cos2 α + cos2 β  = 1 - cos2 γ

∴ cos2 α + cos2 β = sin2 γ                               (∵ sin2 θ + cos2 θ = 1)

So, Statement 1 is correct.

Again, cos2 α + cos2 β + cos2 γ = 1

⇒ 1 - sin ² α + 1 - sin² β + 1 - sin² γ = 1            (∵ sin² θ + cos² θ = 1)

⇒ 3 – (sin² α + sin² β + sin² γ) = 1

⇒ 3 – 1 = sin² α + sin² β + sin² γ

∴ sin² α + sin² β + sin² γ =  2

So, Statement 3 is correct.

Concept:

• Angle between two lines: The angle θ between the lines having slope m₁ and m₂ is given by \(\tan {\rm{\theta }} = {\rm{\;}}\left| {\frac{{{{\rm{m}}_2} - {{\rm{m}}_1}}}{{1 + {\rm{\;}}{{\rm{m}}_1}{\rm{\;}}{{\rm{m}}_2}}}} \right|\)

 

Calculation:

Let's find the slope,

⇒ Let slope of line x + y – 3 = 0 is m₁ and slope of line x – y + 3 = 0 is m₂

∴ m₁ = -1 and m₂ = 1

Given angle between the lines x + y – 3 = 0 and x – y + 3 = 0 is α 

Hence tan α = \(\left| {\frac{{1 - \left( { - 1} \right)}}{{1 + ( - 1) \times {1_\;}}}} \right| = \infty \)

∴ α = 90° 

Again,

⇒ Let slope of line  x- √3y + 2√3 = 0 is m1 and √3x – y + 1 = 0 is m₂

∴ m1 = (1/√3) and m2 = √3 

Given angle between the lines  x- √3y + 2√3 = 0 and √3x – y + 1 = 0 is β

Hence tan β = \(\left| {\frac{{\sqrt 3 - {\rm{\;}}\frac{1}{{\sqrt 3 }}}}{{1 + \left( {\sqrt 3 \; \times \;\frac{1}{{\sqrt 3 }}} \right)}}} \right| = \frac{1}{{\sqrt 3 }}\)

∴ β = 30° 

Therefore α > β

Concept:

1. Cross multiplication method:

Let the system of equations be,

a₁x + b₁y + c₁z = 0

a₂x + b₂y + c₂z = 0

By cross multiplication, we get

\(\frac{{\rm{x}}}{{\left| {\begin{array}{*{20}{c}} {{{\rm{b}}_1}}&{{{\rm{c}}_1}}\\ {{{\rm{b}}_2}}&{{{\rm{c}}_2}} \end{array}} \right|}} = {\rm{\;}}\frac{{ - {\rm{y}}}}{{\left| {\begin{array}{*{20}{c}} {{{\rm{a}}_1}}&{{{\rm{c}}_1}}\\ {{{\rm{a}}_2}}&{{{\rm{c}}_2}} \end{array}} \right|}} = {\rm{\;}}\frac{{\rm{z}}}{{\left| {\begin{array}{*{20}{c}} {{{\rm{a}}_1}}&{{{\rm{b}}_1}}\\ {{{\rm{a}}_2}}&{{{\rm{b}}_2}} \end{array}} \right|}} = {\rm{Constant\;}}\)

 

2.  \({\rm{\vec a\;and\;\vec b}}\) are two vectors perpendicular to each other ⇔  \({\rm{\vec a}}{\rm{.\vec b}} = 0\)

 

Calculation:

Given:  \({\rm{\vec \alpha }}\) and \({\rm{\vec \beta }}\) are both perpendicular to the vector \({\rm{\vec δ }}\)

 ∴ \({\rm{\vec \alpha }} \cdot {\rm{\vec δ }} = 0\) and \({\rm{\vec \beta }} \cdot {\rm{\vec δ }} = 0\)

Let \({\rm{\vec δ }} = {\rm{a\hat i}} + {\rm{b\hat j}} +{\rm{c\hat k}}\)

\(⇒ {\rm{\vec \alpha }} \cdot {\rm{\vec δ }} = ({\rm{\hat i}} + 2{\rm{\hat j}} - {\rm{\hat k}}).({\rm{a\hat i}} + {\rm{b\hat j}} +{\rm{c\hat k}})\) = = a + 2b - c = 0

\(⇒ {\rm{\vec \alpha }} \cdot {\rm{\vec δ }} \)= 0

∴  a + 2b - c = 0        ....

(1) \(⇒ {\rm{\vec \beta }} \cdot {\rm{\vec δ }} = (2{\rm{\hat i}} - {\rm{\hat j}} + 3{\rm{\hat k}}).({\rm{a\hat i}} + {\rm{b\hat j}} +{\rm{c\hat k}})\) = 2a - b + 3c

⇒ \({\rm{\vec \beta }} \cdot {\rm{\vec δ }} = 0\)

∴ 2a - b + 3c = 0      .... (2) Solving equation 1 and 2 by cross multiplication, we get

\(\frac{{\rm{a}}}{{\left| {\begin{array}{*{20}{c}} 2&{ - 1}\\ { - 1}&3 \end{array}} \right|}} = {\rm{\;}}\frac{{ - {\rm{b}}}}{{\left| {\begin{array}{*{20}{c}} 1&{ - 1}\\ 2&3 \end{array}} \right|}} = {\rm{\;}}\frac{{\rm{c}}}{{\left| {\begin{array}{*{20}{c}} 1&2\\ 2&{ - 1} \end{array}} \right|}} = {\rm{Constant}}\)

\( ⇒ \frac{{\rm{a}}}{5} = {\rm{\;}}\frac{{ - {\rm{b}}}}{5} = {\rm{\;}}\frac{{\rm{c}}}{{ - 5}} = {\rm{p}}\)

∴ a = 5p, b = -5p and c = -5p or a = p, b = -p and c = -p     (∵ 5p = constant = p)

Now, \({\rm{\vec δ }} = {\rm{p\hat i}} - {\rm{p\hat j}} -{\rm{p\hat k}}\)

Given: \({\rm{\vec δ }} \cdot {\rm{\vec \gamma }} = 10\)

\(⇒ {\rm{\vec δ }} \cdot {\rm{\vec \gamma }} = ( {\rm{p\hat i}} - {\rm{p\hat j}} -{\rm{p\hat k}}).(2{\rm{\hat i}} + {\rm{\hat j}} + 6{\rm{\hat k}}) = 10\)

⇒ 2p - p - 6p = 10

⇒ -5p = 10

∴ p = -2

∴ \({\rm{\vec δ }} = {\rm{-2\hat i}} + {\rm{2\hat j}}+{\rm{2\hat k}}\)

Magnitude of \({\rm{\vec δ }}\)= |δ| = \(\sqrt {{{\left( { - 2} \right)}^2} + \;{2^2} + \;{2^2}\;} = 2\sqrt 3 \)

Concept:

Vector Triple Product: Vector Triple Product is a vector quantity.

Vector triple product of three vectors a, b, c is defined as the cross product of vector a with the cross product of vectors b and c, i.e. a × (b × c)

• a × (b × c) = (a . c) b – (a . b) c

Calculation:

Given: â and b̂ are two unit vectors

∴ |â| = |b̂| = 1

Now, (â + b̂) × (â × b̂) 

= â × (â × b̂) + b̂ × (â × b̂)

= (â . b̂) â - (â . â) b̂ + (b̂ . b̂) â - (b̂ . â) b̂

= (â . b̂) (â - b̂) - |â|²b̂ + |b̂|² â (∵ â . b̂ = b̂ . â)

= (â . b̂) (â - b̂) - b̂ + â

= (â . b̂) (â - b̂) + (â - b̂)

= (â . b̂ + 1) (â - b̂) (Let â . b̂ + 1 = λ)

= λ (â - b̂)

∴ (â + b̂) × (â × b̂) is parallel to (â - b̂)

Concept:

Work is done when a force is applied, at least partially, in the direction of the displacement of the object.

 If that force is constant then the work done by the force is the dot product of the force with the displacement:

 

Calculation:

Given: \({\rm{\vec F}} = {\rm{\hat i}} + 3{\rm{\hat j}} + 2{\rm{\hat k}}\)

Now, \({\rm{\vec r}} = \overrightarrow {{\rm{AB}}} = {\rm{\vec B}} - {\rm{\vec A}} = {\rm{\;}}\left( {3{\rm{\hat i}} - {\rm{\hat j}} + 5{\rm{\hat k}}} \right) - {\rm{\;}}\left( {{\rm{\hat i}} + 2{\rm{\hat j}} - 3{\rm{\hat k}}} \right) = {\rm{\;}}\left( {2{\rm{\hat i}} - 3{\rm{\hat j}} + 8{\rm{\hat k}}} \right)\)

Work done by the force = \({\rm{\vec F}}\;.\;{\rm{\vec r}} = \left( {{\rm{\hat i}} + 3{\rm{\hat j}} + 2{\rm{\hat k}}} \right).\; {\rm{\;}}\left( {2{\rm{\hat i}} - 3{\rm{\hat j}} + 8{\rm{\hat k}}} \right)\)

= 2 - 9 + 16 = 9 units

Concept: Cross product of two vectors: The cross product can be written as a determinant.

Let \(\vec a\) and \(\vec b\) be the two vectors. ⇒ \(\vec a = \;{a_1}\vec i + {b_1}\vec j + \;{c_1}\vec k\;\) and \(\vec b = \;{a_2}\vec i + {b_2}\vec j + \;{c_2}\vec k\)

\(\therefore \;\vec a\; \times \;\vec b = \;\left| {\begin{array}{*{20}{c}} {\vec i}&{\vec j}&{\vec k\;}\\ {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right|\)

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:

Let \(\vec a = x\;\vec i + y\;\vec j + z\;\vec k\) 

Magnitude of vector of a = \(\left| {\vec a} \right| = \;\sqrt {{x^2} + \;{y^2} + {z^2}} \)

 

\({\rm{\vec a\;}} \times {\rm{\vec i}} = {\rm{\;}}\left| {\begin{array}{*{20}{c}} {{\rm{\vec i}}}&{{\rm{\vec j}}}&{{\rm{\vec k\;}}}\\ {\rm{x}}&{\rm{y}}&{\rm{z}}\\ 1&0&0 \end{array}} \right| = {\rm{z\vec j}} - {\rm{y\vec k}}\)

\({\left| {{\rm{\vec a}} \times {\rm{\hat i}}} \right|^2} = \left( {{\rm{\vec a}} \times {\rm{\hat i}}} \right).\left( {{\rm{\vec a}} \times {\rm{\hat i}}} \right) = \left( {{\rm{z\vec j}} - {\rm{y\vec k}}} \right).\left( {{\rm{z\vec j}} - {\rm{y\vec k}}} \right) = {{\rm{z}}^2} + {{\rm{y}}^2}{\rm{\;}}\)

Similarly,

\({\left| {{\rm{\vec a}} \times {\rm{\hat j}}} \right|^2} = {{\rm{x}}^2} + {{\rm{z}}^2}\)

\({\left| {{\rm{\vec a}} \times {\rm{\hat k}}} \right|^2} = {{\rm{x}}^2} + {{\rm{y}}^2}\)

Now, \({\left| {{\rm{\vec a}} \times {\rm{\hat i}}} \right|^2} + {\left| {{\rm{\vec a}} \times {\rm{\hat j}}} \right|^2} + {\left| {{\rm{\vec a}} \times {\rm{\hat k}}} \right|^2}\) = 2 (x² + y² + z²) = \(2{\left| {{\rm{\vec a}}} \right|^2}\)

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) ⇔ \(\frac{{{\rm{\;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} + \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\) (a > b)
• The sum of the focal distance of any point on an ellipse is constant and equal to the length of the major axis of the ellipse. ⇔ PS + PS' = 2a = Major axis

Focus: (± ae, 0)

Eccentricity: \({\rm{e}} = {\rm{\;}}\sqrt {1 - {\rm{\;}}\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2}}}} \) ⇔ a2e² = a2 - b2

Distance between Focus  = 2ae

Calculation:

Given: PS + PS^' = 10

⇒ 2a = 10

∴ a = 5

Distance between Focus = 8

⇒ 2ae = 8

⇒ ae = 4

Squaring both sides, we get

⇒ a²e² = 16

⇒ (a² - b²) = 16

⇒ b² = a² - 16 = 25 - 16 = 9

∴ b = 3

The area of the path he encloses = Area of ellipse = πab = π × 5 × 3 = 15π

Concept:

Let A(x1, y1) and B(x2, y2)

\({\rm{Distance\;between\;A\;and\;B\;}} = \sqrt {{\rm{\;}}{{\left( {{{\rm{x}}_2} - {\rm{\;}}{{\rm{x}}_1}} \right)}^2} + {\rm{\;}}{{\left( {{{\rm{y}}_2} - {\rm{\;}}{{\rm{y}}_1}} \right)}^2}} \)

Calculation:

We have to find the distance of the point (1, 3) from the line 2x + 3y = 6   ....

(1) Other line is parallel to 4x + y = 4

So the slope is same.

It can be written as y = −4x + 4

So, slope is -4.

Given point = (1, 3)

Equation of line passing through (1, 3) and slope -4 is y – 3 = −4(x − 1)

⇒ y −3 = −4x + 4

⇒ 4x + y = 7           …. (2) Solving the equations 1 and 2, we get

⇒ x = 3/2 and y = 1

Point is (3/2, 1)

Distance between the points (1, 3) and (3/2,1) is = \(\sqrt {{{\left( {\frac{3}{2} - 1} \right)}^2} + \;{{\left( {1 - 3} \right)}^2}} = \;\frac{{\sqrt {17} }}{2}\;\)

Concept:

Scaler triple product of the vectors:

The scaler triple product of the vectors \({\rm{\bar A}} = {{\rm{x}}_1}\hat i + \;{y_1}\hat j + {\rm{\;}}{{\rm{z}}_1}\hat k,\;\bar B = \;{x_2}\hat i + \;{y_2}\hat j + {z_2}\hat k\) and \({\rm{\bar C}} = {\rm{\;}}{{\rm{x}}_3}\hat i + \;{y_3}\hat j + {\rm{\;}}{{\rm{z}}_3}\hat k\) is given by:

\({\rm{\bar A}} \cdot \left[ {{\rm{\bar B}} \times {\rm{\bar C}}} \right] = {\rm{\;}}\left| {\begin{array}{*{20}{c}} {{{\rm{x}}_1}}&{{{\rm{y}}_1}}&{{{\rm{z}}_1}}\\ {{{\rm{x}}_2}}&{{{\rm{y}}_2}}&{{{\rm{z}}_2}}\\ {{{\rm{x}}_3}}&{{{\rm{y}}_3}}&{{{\rm{z}}_3}} \end{array}} \right|\)

Coplaner vectors:

Three vectors \({\rm{\bar A}} = {{\rm{x}}_1}\hat i + \;{y_1}\hat j + {\rm{\;}}{{\rm{z}}_1}\hat k,\;\bar B = \;{x_2}\hat i + \;{y_2}\hat j + {z_2}\hat k\) and \({\rm{\bar C}} = {\rm{\;}}{{\rm{x}}_3}\hat i + \;{y_3}\hat j + {\rm{\;}}{{\rm{z}}_3}\hat k\) are said to be coplaner if the scaler triple product \({\bf{\bar A}} \cdot \left[ {{\bf{\bar B}} \times {\bf{\bar C}}} \right] = 0.\)

 

Solution:

Let the given vectors be \({\rm{\bar A}} = {\rm{a}}\hat i + \;\hat j + {\rm{\;}}\hat k,\;\bar B = \;\hat i + \;b\hat j + \hat k\) and \({\rm{\bar C}} = {\rm{\;}}\hat i + \;\hat j + {\rm{c}}\hat k\).

It is given that the vectors are coplaner therefore the scaler triple product \({\rm{\bar A}} \cdot \left[ {{\rm{\bar B}} \times {\rm{\bar C}}} \right] = 0.\)

Therefore,

\(\left| {\begin{array}{*{20}{c}} {\rm{a}}&1&1\\ 1&{\rm{b}}&1\\ 1&1&{\rm{c}} \end{array}} \right| = 0\)

Perform the coloumn operation C₁ – C₂ as follows:

\(\left| {\begin{array}{*{20}{c}} {{\rm{a}} - 1}&1&1\\ {1 - {\rm{b}}}&{\rm{b}}&1\\ 0&1&{\rm{c}} \end{array}} \right| = 0\)

Now perform another coloum operation C₂ – C₃ as follows:

\(\left| {\begin{array}{*{20}{c}} {{\rm{a}} - 1}&0&1\\ {1 - {\rm{b}}}&{{\rm{b}} - 1}&1\\ 0&{1 - {\rm{c}}}&{\rm{c}} \end{array}} \right| = 0\)

Take (1 - a)(1 - b)(1 - c) common. Note that it is given that a,b,c ≠ 0 therefore this action is jstified.

\(\left( {1 - {\rm{a}}} \right)\left( {1 - {\rm{b}}} \right)\left( {1 - {\rm{c}}} \right)\left| {\begin{array}{*{20}{c}} { - 1}&0&{\frac{1}{{1 - {\rm{a}}}}}\\ 1&{ - 1}&{\frac{1}{{1 - {\rm{b}}}}}\\ 0&1&{\frac{{\rm{c}}}{{1 - {\rm{c}}}}} \end{array}} \right| = 0\)

Since a, b, c ≠ 0 therefore (1 - a)(1 - b)(1 - c) ≠ 0.Therefore the determinant has to be zero.

\( - 1\left( {\frac{{\rm-{c}}}{{1 - {\rm{c}}}} - {\rm{}}\frac{1}{{1 - {\rm{b}}}}} \right) + {\rm{\;}}\frac{1}{{1 - {\rm{a}}}} = 0\)

\(\frac{1}{{1 - {\rm{b}}}} + {\rm{\;}}\frac{{\rm{c}}}{{1 - {\rm{c}}}} + {\rm{\;}}\frac{1}{{1 - {\rm{a}}}} = 0\)

Simplify the above expression as follows:

\(\frac{1}{{1 - {\rm{b}}}} + {\rm{\;}}\frac{1}{{1 - {\rm{c}}}} - {\rm{\;}}1 + {\rm{\;}}\frac{1}{{1 - {\rm{a}}}} = 0\)

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

Therefore, \(\frac{1}{{1 - {\rm{b}}}} + {\rm{\;}}\frac{1}{{1 - {\rm{c}}}} + {\rm{\;}}\frac{1}{{1 - {\rm{a}}}} = 1\).

Concept:

Equation of line passing through two points

• Let, points A(x₁, y₁, z₁ ) and B(x₂, y₂, z₂) lies on a line

 

Then equation of line: \(\frac{{{\bf{x}} - {{\bf{x}}_1}}}{{{{\bf{x}}_2} - {{\bf{x}}_1}}} = \frac{{{\bf{y}} - {{\bf{y}}_1}}}{{{{\bf{y}}_2} - {{\bf{y}}_1}}} = \frac{{{\bf{z}} - {{\bf{z}}_1}}}{{{{\bf{z}}_2} - {{\bf{z}}_1}}}\)

 

Calculation:

Given points: (-3, 4, -8) and (5, -6, 4)

Equation of line passing through points (-3, 4, -8) and (5, -6, 4)

\(\frac{{{\rm{x}} - \left( { - 3} \right)}}{{5 - \left( { - 3} \right)}} = \frac{{{\rm{y}} - 4}}{{ - 6 - 4}} = \frac{{{\rm{z}} - \left( { - 8} \right)}}{{4 - \left( { - 8} \right)}}\)

\(\Rightarrow \frac{{{\rm{x}} + 3}}{8} = \frac{{{\rm{y}} - 4}}{{ - 10}} = \frac{{{\rm{z}} + 8}}{{12}}\)

Let, \(\frac{{{\rm{x}} + 3}}{8} = \frac{{{\rm{y}} - 4}}{{ - 10}} = \frac{{{\rm{z}} + 8}}{{12}} = {\rm{\lambda }}\)

So, x + 3 = 8λ ⇒ x = 8λ - 3

 y - 4 = -10λ ⇒ y = -10λ + 4

z + 8 = 12λ ⇒ z = 12λ - 8     ---(1)

Now given that the line intersects the XY plane so, z = 0,

From (1),

z =12λ - 8

⇒ 12λ - 8 = 0

⇒ λ = 8/12 = 2/3

∴x = 8λ - 3

⇒ x = 8(2/3) - 3

\(= \frac{{16}}{3} - 3\)

\(= \frac{7}{3}\)

Also, y = -10λ + 4

\(\Rightarrow {\rm{y}} = - 10\left( {\frac{2}{3}} \right) + 4\)

\(= - \frac{{20}}{3} + 4\)

\(= - \frac{8}{3}\)

So the point of intersection \(\left( {\frac{7}{3}, - \frac{8}{3},0} \right)\)

Hence, option (1) is correct

Concept:

1. Angle between the lines given the direction ratios:

• The angle between the lines with direction ratios \(\left\langle {{a_1},\;{b_1},\;{c_1}} \right\rangle \) and \(\left\langle {{a_2},\;{b_2},\;{c_2}} \right\rangle \) is given by: \(\cos {\bf{\theta }} = \;\frac{{{{\bf{a}}_1}{{\bf{a}}_2} + \;{{\bf{b}}_1}{{\bf{b}}_2} + {{\bf{c}}_1}{{\bf{c}}_2}\;}}{{\sqrt {{\bf{a}}_{1\;}^2 + \;{\bf{b}}_1^2 + \;{\bf{c}}_1^2} \sqrt {{\bf{a}}_{2\;}^2 + \;{\bf{b}}_2^2 + \;{\bf{c}}_2^2} }}\)

Solution:

It is given that the angle between the lines whose direction ratios are \(\left\langle {2, - 1,\;2} \right\rangle \) And \(\left\langle {x,\;3,\;5\;} \right\rangle\) is \(\frac{\pi }{4}\).

Therefore from the formula for the angle between the lines we get:

\(\cos \frac{{\rm{\pi }}}{4} = {\rm{}}\frac{{2{\rm{x}} - 3 + 10}}{{\sqrt {4 + 1 + 4} \sqrt {{{\rm{x}}^2} + 9 + 25} }}\)

\(\frac{1}{{\sqrt 2 }} = {\rm{\;}}\frac{{2{\rm{x}} + 7}}{{3\sqrt {{{\rm{x}}^2} + 34} }}\)

\(\frac{1}{2} = \frac{{{{\left( {2{\rm{x}} + 7} \right)}^2}}}{{9\left( {{{\rm{x}}^2} + 34} \right)}}\)     ....squaring both sides

Now simplify the above equation to form a quadratic equation as follows:

\(9{{\rm{x}}^2} + {\rm{\;}}306 = {\rm{\;}}8{{\rm{x}}^2} + 56{\rm{x}} + 98\)

\({{\rm{x}}^2} - 56{\rm{x}} + 208 = 0\)

\(\left( {{\rm{x}} - 52} \right)\left( {{\rm{x}} - 4} \right) = 0\)

x = 52, 4

Therefore, the minimum value is 4.

Concept:

1. Position of a point with respect to a curve:

• For any curve F(x, y) = k in the xy-plane with k is a real number, then the point (a, b) lies on the curve if F(a, b) = k.
• For any curve F(x, y) = k in the xy-plane with k is a real number, then the point (a, b) lies outside the curve if F(a, b) > k.
• For any curve F(x, y) = k in the xy-plane with k is a real number, then the point (a, b) lies inside the curve if F(a, b) < k.

2. Standard equation of an ellipse and its foci:

• The standard equation of an ellipse is of the form \(\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} + \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1\). If a² > b² then the ellipse is horizontal and vertical otherwise.
• The foci of the ellipse is given by (±c, 0) if the ellipse is horizontal and it is given by (0, ±c) if the ellipse is vertical.

 

Solution:

Observe that the given equation is of the form F(x, y) = k, where F(x, y) = 2x² + 7y² and k = 20.

Therefore, to find the position of the given point (1, 2) with respect to ellipse substitute x = 1 and y = 2 in the equation of the ellipse..

Therefore,

\(2{\left( 1 \right)^2} + {\rm{\;}}7{\left( 2 \right)^2} = 30\)

But 30 > 20 therefore we have F(1, 2) > k and therefore the point (1, 2) lies outside the ellipse.

Concept:

• Slope of line if angle is known: Slope (m) = tan (θ)
• Equation of straight line: y - y₁ = m (x - x₁),

Here, m = slope and (x₁, y₁) is given point

 

Calculation:

Here, a straight line cuts negative Y-axis at 5 units

So, the line is passing through point (0, -5)

And the line makes an angle 120° with X-axis

So, slope of line (m) = tan (120°) = -√3                 (Slope (m) = tan (θ))

Now, equation of straight line passing through point (0, -5) having slope -√3

y - (-5) = -√3 (x - 0)

⇒ y + 5 = -√3 x

i.e., y + √3 x + 5 = 0

Hence, option (1) is correct

Concept:

Let L₃ is the line passing through the intersection of lines L₁ and L₂

Then equation of line L₃ is given by L₁ + (λ)L₂ = 0

 

Calculation:

Equation of line passing through the intersection of lines,

2x - 3y + 7 = 0 and 7x + 4y + 2 = 0 is given by,

2x - 3y + 7 + λ (7x + 4y + 2) = 0                     (1)

Now, this line is also passing through the point (2, 3),

So, it will satisfy the equation of line 2x - 3y + 7 + λ (7x + 4y + 2) = 0

\(\begin{array}{l} \therefore 2\left( 2 \right) - 3\left( 3 \right) + 7 + {\rm{\lambda }}\left( {7\left( 2 \right) + 4\left( 3 \right) + 2} \right) = 0\\ \Rightarrow 4 - 9 + 7 = - {\rm{\lambda }}\left( {14 + 12 + 2} \right) \end{array}\)

\(\Rightarrow - {\rm{\lambda }} = \frac{2}{{28}}\)

\(\Rightarrow {\rm{\lambda }} = - \frac{1}{{14}}\)

Now, putting \({\rm{\lambda }} = - \frac{1}{{14}}\) in (1), we get equation of line,

\(2{\rm{x\;}} - {\rm{\;}}3{\rm{y\;}} + {\rm{\;}}7{\rm{\;}} + \left( { - \frac{1}{{14}}} \right)\left( {7{\rm{x\;}} + {\rm{\;}}4{\rm{y\;}} + {\rm{\;}}2} \right) = {\rm{\;}}0\)

Multiply by 14,

\(\begin{array}{l} \Rightarrow 28{\rm{x}} - 42{\rm{y}} + 98 - \left( {7{\rm{x\;}} + {\rm{\;}}4{\rm{y\;}} + {\rm{\;}}2} \right) = 0\\ \Rightarrow 28{\rm{x}} - 7{\rm{x}} - 42{\rm{y}} - 4{\rm{y}} + 98 - 2 = 0\\ \Rightarrow 21{\rm{x}} - 46{\rm{y}} + 96 = 0 \end{array}\)

Hence, option (2) is correct

Concept:

1. The length of the latus rectum of an ellipse:

• For an ellipse with x-axis as its major axis with equation \(\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} + \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1\) where a > b the length of the latus rectum is given by: \(l = \;\frac{{2{{\bf{b}}^2}}}{{\bf{a}}}\)

2. The relation between major axis, minor axis and eccentricity of an ellipse:

• For an ellipse with x-axis as its major axis with equation \(\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} + \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1\) where a > b and eccentricity e the following relation always holds: \({{\rm{b}}^2} = {{\rm{a}}^2}\left( {1 - {{\rm{e}}^2}} \right)\)

 

Solution:

It is given that the length of latus rectum l = 4 and eccentricity \(e = \frac{3}{4}\).

Let the equation of the ellipse be \(\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} + \frac{{{{\rm{y}}^2}}}{{{{\rm{b}}^2}}} = 1\) where a > b.

For the given ellipse we have already discovered that the relation \({b^2} = {a^2}\left( {1 - {e^2}} \right)\) holds.

Therefore, the equation of the ellipse is \(\frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}} + \frac{{{{\rm{y}}^2}}}{{{a^2}\left( {1 - {e^2}} \right){\rm{\;}}}} = 1\).

Given that \(e = \;\frac{3}{4}\) therefore \(1 - {e^2} = \;\frac{7}{{16}}\).

Also we have the length of latus rectum is \(l = 4\) 

Therefore, \(4 = \;\frac{{2{b^2}}}{a}\)

Substituting \({b^2} = {a^2}\left( {1 - {e^2}} \right)\) in the above equation we get \(4 = \frac{{2{a^2}\left( {1 - {e^2}} \right)\;}}{a}\).

Therefore,

\(2 = \frac{{7a}}{{16}}\)

\(a = \frac{{32}}{7}\)

Therefore, \({a^2} = \frac{{1024}}{{49}}\) and \({b^2} = \;\frac{{1024}}{{49}}\left( {\frac{7}{{16}}} \right) = \;\frac{{64}}{7}\).

Thus, the equation of the ellipse is \(\frac{{49{x^2}}}{{1024}} + \frac{{7{y^2}}}{{64}} = 1\).

Concept:

1. The general equation of a circle:

• The general equation of a circle with centre (-g, -h) is x² + y² + 2gx + 2fy + c = 0.

2. Points on the curve:

• For a curve F(x, y) = k if a point (a, b) lies on the curve then F (a, b) = 0.

Solution:

Let the equation of the circle be F(x, y) = x² + y² + 2gx + 2fy + c.

Since the point (1, 0) lies on the circle, we can write F (1, 0) = 0.

Therefore,

1 + 2g + c = 0     ...1)

Similarly, the point (0, -6) also lies on the circle therefore, F (0, -6) = 0.

Therefore,

36 - 12f + c = 0     ...2)

Similarly, the point (3, 4) also lies on the circle therefore F (3, 4) = 0.

9 + 16 + 6g + 8f + c = 0

25 + 6g + 8f + c = 0     ...3)

Solving the above three simultaneous equations we get:

\(g = - \frac{{71}}{4},\;f = \;\frac{{47}}{8},\;c = \;\frac{{69}}{2}\)

Therefore, the equation of the circle is given by:

\({x^2} + \;{y^2} - \;2\left( {\frac{{71}}{4}} \right)x + 2\left( {\frac{{47}}{8}} \right)y + \;\frac{{69}}{2} = 0\)

\({x^2} + \;{y^2} - \;\left( {\frac{{71}}{2}} \right)x + \left( {\frac{{47}}{4}} \right)y + \;\frac{{69}}{2} = 0\)

\(4{x^2} + \;4{y^2} - \;142x + 47y + \;138 = 0\)

Therefore, the equation of the circle passing through the points (1, 0), (0, -6) and (3, 4) is

\(4{x^2} + \;4{y^2} - \;142x + 47y + \;138 = 0\).

Shortcut Trick

Substitute the given points in the options one by one in the options to see which of the equations it satisfies. The option for which all the points lie on the curve is the correct option.

Concept:

• Consider a plane cut at (p, 0, 0) on X-axis, (0, q, 0) on Y-axis, and (0, 0, s) on Z-axis

Then,

Intercept form of the equation of a plane

\(\frac{{\bf{x}}}{{\bf{p}}} + \frac{{\bf{y}}}{{\bf{q}}} + \frac{{\bf{z}}}{{\bf{s}}} = 1\)

• The general equation of a sphere is: \({\left( {{\rm{x}} - {\rm{a}}} \right)^2} + {\left( {{\rm{y}} - {\rm{b}}} \right)^2} + {\left( {{\rm{z}} - {\rm{c}}} \right)^2} = {{\rm{r}}^2}\)

Where (a, b, c) = center of the sphere, r = radius, and x, y, and z are the coordinates of the points on the surface of the sphere.

 

Calculation:

Let, (p, q, s) = Center of sphere

\(\therefore {\rm{equation\;of\;a\;sphere}} = {\left( {{\rm{x}} - {\rm{p}}} \right)^2} + {\left( {{\rm{y}} - {\rm{q}}} \right)^2} + {\left( {{\rm{z}} - {\rm{s}}} \right)^2} = {{\rm{r}}^2}\)

Now, \({\rm{radius\;of\;sphere\;}}\left( {\rm{r}} \right) = \sqrt {{{\left( {{\rm{p}} - 0} \right)}^2} + {{\left( {{\rm{q}} - 0} \right)}^2} + {{\left( {{\rm{s}} - 0} \right)}^2}} \)  (distance from origin)

\(\begin{array}{l} \therefore {{\rm{r}}^2} = {{\rm{p}}^2} + {{\rm{q}}^2} + {{\rm{s}}^2}\\ \therefore {\rm{equation\;of\;a\;sphere}} = {\left( {{\rm{x}} - {\rm{p}}} \right)^2} + {\left( {{\rm{y}} - {\rm{q}}} \right)^2} + {\left( {{\rm{z}} - {\rm{s}}} \right)^2} = {{\rm{p}}^2} + {{\rm{q}}^2} + {{\rm{s}}^2}\\ \Rightarrow {{\rm{x}}^2} - 2{\rm{px}} + {{\rm{p}}^2} + {{\rm{y}}^2} - 2{\rm{qy}} + {{\rm{q}}^2} + {{\rm{z}}^2} - 2{\rm{zs}} + {{\rm{s}}^2} = {{\rm{p}}^2} + {{\rm{q}}^2} + {{\rm{s}}^2}\\ \Rightarrow {{\rm{x}}^2} + {{\rm{y}}^2} + {{\rm{z}}^2} - 2{\rm{px}} - 2{\rm{qy}} - 2{\rm{zs}} = 0 \Rightarrow \end{array}\)

Now, we will find point of intersection on X- axis, so y = 0, z = 0,

\(\begin{array}{l} {{\rm{x}}^2} + {0^2} + {0^2} - 2{\rm{px}} - 2{\rm{q}}\left( 0 \right) - 2\left( 0 \right){\rm{s}} = 0\\ \Rightarrow {{\rm{x}}^2} - 2{\rm{px}} = 0 \end{array}\)

⇒ x(x - 2p) = 0

∴ x = 0, OR x = 2p

Plane cut at point (2p, 0, 0) on X-axis

Similarly, we will find point of intersection on Y- axis, so x = 0, z = 0,

\(\begin{array}{l} {0^2} + {{\rm{y}}^2} + {0^2} - 2{\rm{p}}\left( 0 \right) - 2{\rm{qy}} - 2\left( 0 \right){\rm{s}} = 0\\ \Rightarrow {{\rm{y}}^2} - 2{\rm{q}} = 0 \end{array}\)

∴ y = 0, y = 2q

Plane cut at point (0, 2q, 0) on Y-axis

Also, we will find point of intersection on Z- axis, so x = 0, y = 0,

\(\begin{array}{l} {0^2} + {0^2} + {{\rm{z}}^2} - 2{\rm{p}}\left( 0 \right) - 2{\rm{q}}\left( 0 \right) - 2{\rm{zs}} = 0\\ \Rightarrow {{\rm{z}}^2} - 2{\rm{s}} = 0 \end{array}\)

∴ z = 0, z = 2s

Plane cut at point (0, 0, 2s) on Z-axis

So, equation of plane will be \(\frac{{\rm{x}}}{{2{\rm{p}}}} + \frac{{\rm{y}}}{{2{\rm{q}}}} + \frac{{\rm{z}}}{{2{\rm{s}}}} = 1\)

\(\Rightarrow \frac{1}{2}\left( {\frac{{\rm{x}}}{{\rm{p}}} + \frac{{\rm{y}}}{{\rm{q}}} + \frac{{\rm{z}}}{{\rm{s}}}} \right) = 1\)

\(\Rightarrow \frac{{\rm{x}}}{{\rm{p}}} + \frac{{\rm{y}}}{{\rm{q}}} + \frac{{\rm{z}}}{{\rm{s}}} = 2\)

But plane passing through fixed point (a, b, c)

∴ x = a, y = b, z = c

So, \(\frac{{\rm{a}}}{{\rm{p}}} + \frac{{\rm{b}}}{{\rm{q}}} + \frac{{\rm{c}}}{{\rm{s}}} = 2\)

Now, for the locus of center of sphere p = x, q = y, s = z

⇒ \(\frac{{\rm{a}}}{{\rm{x}}} + \frac{{\rm{b}}}{{\rm{y}}} + \frac{{\rm{c}}}{{\rm{z}}} = 2\)

Hence, option (3) is correct

Concept:

• If P₃ is the plane passing through the line of intersection of planes P₁ and P₂

Then the equation of plane P₃ is given by P₁ + (λ)P₂ = 0

• Let, \({\rm{\bar a}} = {{\rm{a}}_1}{\rm{x}} + {{\rm{b}}_1}{\rm{y}} + {\rm{c_1}}\) and \({\rm{\bar b}} = {{\rm{a}}_2}{\rm{x}} + {{\rm{b}}_2}{\rm{y}} + {\rm{c_2}}\)

Then dot product: \({\rm{\bar a}}.{\rm{\bar b}} = {{\rm{a}}_1}{\rm{a_2}} + {{\rm{b}}_1}{\rm{b_2}} + {{\rm{c}}_1}{\rm{c_2}} \)

• If two planes (let’s say P₁ and P₂) are perpendicular, then the dot product is zero

i.e., P₁∙P₂ = 0

Calculation:

Let P₁: x + y + z - 1= 0

P₂: 2x +3y + 4z-7 = 0

Then, equation of plane passing through the line of intersection of P₁ and P₂ is

x + y + z - 1 + λ (2x + 3y + 4z - 7) = 0                           (1)

\(\Rightarrow {\rm{x}}\left( {1 + 2{\rm{\lambda }}} \right) + {\rm{y}}\left( {1 + 3{\rm{\lambda }}} \right) + {\rm{z}}\left( {1 + 4{\rm{\lambda }}} \right) - 1 - 7{\rm{\lambda }} = 0\)

It is perpendicular to the plane x - 5y + 3z = 5

\(\begin{array}{l} \therefore \left[ {{\rm{x}}\left( {1 + 2{\rm{\lambda }}} \right) + {\rm{y}}\left( {1 + 3{\rm{\lambda }}} \right) + {\rm{z}}\left( {1 + 4{\rm{\lambda }}} \right) - 1 - 7{\rm{\lambda }}} \right] \cdot \left[ {{\rm{x\;}} - {\rm{\;}}5{\rm{y\;}} + {\rm{\;}}3{\rm{z}} - {\rm{\;}}5} \right] = 0\\ \Rightarrow 1\left( {1 + 2{\rm{\lambda }}} \right) - 5\left( {1 + 3{\rm{\lambda }}} \right) + 3\left( {1 + 4{\rm{\lambda }}} \right) = 0\\ \Rightarrow 1 + 2{\rm{\lambda }} - 5 - 15{\rm{\lambda }} + 3 + 12{\rm{\lambda }} = 0\\ \Rightarrow - {\rm{\lambda }} - 1 = 0\\ \Rightarrow {\rm{\lambda }} = - 1 \end{array}\)

Putting \({\rm{\lambda }} = - 1\) in (1), we get the equation of required plane,

\(\begin{array}{l} {\rm{x\;}} + {\rm{\;y\;}} + {\rm{\;z\;}} - {\rm{\;}}1{\rm{\;}} + \left( { - 1} \right)\left( {2{\rm{x\;}} + {\rm{\;}}3{\rm{y\;}} + {\rm{\;}}4{\rm{z\;}} - {\rm{\;}}7} \right) = {\rm{\;}}0\\ \Rightarrow {\rm{x}} - 2{\rm{x}} + {\rm{y}} - 3{\rm{y}} + {\rm{z}} - 4{\rm{z}} - 1 + 7 = 0\\ \Rightarrow - {\rm{x}} - 2{\rm{y}} - 3{\rm{z}} + 6 = 0\\ \Rightarrow {\rm{x}} + 2{\rm{y}} + 3{\rm{z}} - 6 = 0 \end{array}\)

Hence, option (1) is correct