Quantitative Aptitude PYQs

FORMULA USED:

sec² x - tan² x = 1

\(\frac{d}{dx}tan(x)=sec^2(x)\)

\(\smallint \frac{{dx}}{{1 + {x^2}}} = {\tan ^{ - 1}}x + C\)

where C is a constant

CALCULATION:

Let \(I = ∫ \dfrac{dx}{sec^2({tan}^{-1}x)}\)

As we know that,

sec2 x - tan2 x = 1

⇒ \(I = ∫ \dfrac{dx}{1 + tan^2({tan}^{-1}x)}\)

⇒ \(I = ∫ \dfrac{dx}{1 + [tan({tan}^{-1}x)]^2}\)

⇒ \(I = ∫ \dfrac{dx}{1 + x^2}\)

As we know,

\(\smallint \frac{{dx}}{{1 + {x^2}}} = {\tan ^{ - 1}}x + C\) where C is constant.

⇒ \(I = ∫ \dfrac{dx}{sec^2({tan}^{-1}x)} = tan^{-1} x + C\)

Hence, option 2 is correct.

Alternate Method We have,

\(I = ∫ \dfrac{dx}{sec^2({tan}^{-1}x)}\) ....

(1) Consider, tan^-1(x) = θ .... (2) Hence, x = tan θ

⇒ dx = sec² (θ) dθ

From equation (1),

\(I = ∫ \dfrac{sec^2 (θ) dθ}{sec^2({θ})}\)

or, I = ∫dθ = θ + c .... (3) Hence, from equation (1) & (3),

I = tan-1 x + c

CONCEPT:

Second Derivative Test: Let f be a function defined on an interval I.

• Calculate f’(x)
• Solve f’(x) = 0 and find the roots of f'(x) = 0. Suppose x = c is the root of f’(x) = 0.
• Calculate f’’(x) and put x = c to get the value of f’’(c).
• If f’’(c) < 0 then x = c is a point of local maxima.
• If f’’(c) > 0 then x = c is a point of local minima.
• If f’’(c) = 0 then we need to use the first derivative test.

Note:

Maxima and Minima on a closed Interval:

Let f(x) be a given function defined on [a, b].Let the local minimum value of f(x) be m and let the local maximum value of f(x) be M. Then

Minimum value of f(x) on [a, b] is the smallest of m, f(a) and f(b)

Maximum value of f(x) on [a, b] is the greatest of M, f(a) and f(b)

CALCULATION:

Given: x + y = 20 and P = xy

⇒ P = x ⋅ (20 - x) = 20x - x²

⇒ P(x) = 20x - x²

⇒ P'(x) = 20 - 2x

If P'(x) = 0 ⇒ 20 - 2x = 0 ⇒ x = 10

By substituting x = 10 in the equation x + y = 20 we get, y = 10

⇒ P''(x) = - 2

⇒ P''(x = 10) = - 2 < 0

So, x = 10 is the point of maxima

So, the maximum value of P = xy = 10 × 10 = 100

Hence, option 1 is correct.

CONCEPT:

• \(\frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x\)
• \(\frac{{d\left( {\cos x} \right)}}{{dx}} = ; - \sin x\)
• \(\frac{{d\left( {\ln x} \right)}}{{dx}} = \frac{1}{x}\)

CALCULATION:

Let f(x) = sin(ln x) + cos(ln x)

As we know that, \(\frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x, \frac{{d\left( {\cos x} \right)}}{{dx}} = ; - \sin x \ and \ \frac{{d\left( {\ln x} \right)}}{{dx}} = \frac{1}{x}\)

⇒ \(f'\left( x \right) = \frac{{\cos \left( {\ln x} \right)}}{x} - \frac{{\sin \left( {\ln x} \right)}}{x}\)

⇒ \(f'\left( e \right) = \frac{{\cos \left( {\ln e} \right)}}{e} - \frac{{\sin \left( {\ln e} \right)}}{e}\)

As we know that, ln(e) = 1

⇒ \(f'(e) = \frac{cos1 - sin1}{e}\)

Hence, option 1 is correct.

CONCEPT:

• \(\frac{{d\left( {{e^x}} \right)}}{{dx}} = {e^x}\)
• \(\frac{{d\left( {\sin x} \right)}}{{dx}} = \cos x\)
• \(\frac{{d\left( {\cos x} \right)}}{{dx}} = ; - \sin x\)

Parametric Differentiation:

If x = f(t), y = g(t), where t is a parameter, then \(\frac{{dx}}{{dy}} = \frac{{f'\left( t \right)}}{{g'\left( t \right)}}\)

CALCULATION:

Given: x = et cost and y = et sint

First let's find out dx/dt

⇒ \(\frac{dx}{dt} =e^t cos t -e^t sin t\)

Now let's find out dy/dt

⇒ \(\frac{dy}{dt} = e^t sint + e^t cos t\)

⇒ \(\frac{dx}{dy} =\frac{e^t cos t \ - \ e^t sin t}{e^t sint \ + \ e^t cos t}\)

⇒ \({\left[ {\frac{{dy}}{{dx}}} \right]_{t= 0}} = \frac{1 - 0}{0+1} = 1\)

Hence, correct option is 2.

CONCEPT:

Second Derivative Test: Let f be a function defined on an interval I.

• Calculate f’(x)
• Solve f’(x) = 0 and find the roots of f'(x) = 0. Suppose x = c is the root of f’(x) = 0.
• Calculate f’’(x) and put x = c to get the value of f’’(c).
• If f’’(c) < 0 then x = c is a point of local maxima.
• If f’’(c) > 0 then x = c is a point of local minima.
• If f’’(c) = 0 then we need to use the first derivative test.

Note:

Maxima and Minima on a closed Interval:

Let f(x) be a given function defined on [a, b].Let the local minimum value of f(x) be m and let the local maximum value of f(x) be M. Then

Minimum value of f(x) on [a, b] is the smallest of m, f(a) and f(b)

Maximum value of f(x) on [a, b] is the greatest of M, f(a) and f(b)

CALCULATION:

Let f(x) = sin 2x ⋅ cos 2x

⇒ f'(x) = 2 cos² 2x - 2 sin² 2x

⇒ f'(x) = 2 ⋅ (cos² 2x - sin² 2x)

As we know that, cos 2x = cos² x - sin² x

⇒ f'(x) = 2cos 4x

If f'(x) = 0 then 2cos 4x = 0 ⇒ x = π/8

⇒ f''(x) = - 8 sin4x

⇒ f''(x) = - 8 < 0

So, x = π/8 is the point of maxima.

So, the maximum value of f(x) = sin 2x ⋅ cos 2x is given by f(π/8) = sin (π/4) ⋅ cos(π/4) = 1/2

Hence, correct option is 1.

CONCEPT:

The three points A (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) are said to be collinear if the area of Δ ABC is zero i.e \(\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\ {{x_2}}&{{y_2}}&1\ {{x_3}}&{{y_3}}&1 \end{array}} \right| = 0\)

CALCULATION:

Given: (p, p - 3), (q + 3, q) and (6, 3) are three points in a 2D plane.

Statement 1: The points lie on a straight line.

As we know that, if three points A (x1, y1), B (x2, y2) and C (x3, y3) are said to be collinear if the area of Δ ABC is zero i.e \(\left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1\ {{x_2}}&{{y_2}}&1\ {{x_3}}&{{y_3}}&1 \end{array}} \right| = 0\)

Let A = (p, p - 3), B = (q + 3, q) and C = (6, 3)

⇒ \(\left| {\begin{array}{*{20}{c}} {{p}}&{{p-3}}&1\ {{q+3}}&{{q}}&1\ {{6}}&{{3}}&1 \end{array}} \right| = p (q - 3) - (p-3)(q+3 - 6) + 1(3q+9-6q) \)

⇒ \(\left| {\begin{array}{*{20}{c}} {{p}}&{{p-3}}&1\ {{q+3}}&{{q}}&1\ {{6}}&{{3}}&1 \end{array}} \right| = 0\)

So, as we can see that area of triangle ABC = 0 for any value of p and q.

Hence, statement 1 is true.

Statement 2: The points always lie in the first quadrant only for any value of p and q.

As we can see that, for any value of p and q it is not necessary that the points lies in the first quadrant only.

Hence, statement 2 is false.

Hence, the correct option is 1.

Concept:

Let θ be the angle between two lines of slope m₁ and m₂, then the acute angle between the lines is given by:

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

Calculations:

Consider, the slope of the line (x - 2 = 0) is m₁

So, m₁ = ∞ .

And, the slope of the line (√3x - y - 2 = 0) is m₂

So, m₂ = √3.

Now, the angle between the given lines is θ.

⇒ \(\tan θ = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1} \cdot \ {m_2}}}} \right|\)

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

⇒ \(tan \ θ = \frac{1}{\sqrt 3}\)

⇒ θ = 30°

Hence, the correct option is 2.

Alternate Method

For line 1:x – 2 = 0Hence, x = 2

Since it is a vertical line, the slope is undefined.

Hence, θ₁ = 90°

For line 2:

√3x - y - 2 = 0

Compare this equation with, y = mx + c

It becomes, y = √3x + 2

Hence, m = √3 = tan θ₂

Hence, θ₂ = \(tan^{-1}{\sqrt 3}\) = 60°

Now, the graph of intersection between two lines can be drawn as

Hence, acute angle between the lines = θ₁ – θ₂ = 90° – 60° = 30°

Concept:

The equation of a line passing through the point (x₁, y₁) and having the slope ‘m’ is given as:

y – y₁ = m ⋅ (x – x₁)

Calculations:

Given: The point of intersection of diagonals of a square ABCD is at the origin and one of its vertices is at A(4, 2).

So, the diagonal AC passes through the origin

As we know that, the slope of the line joining the points (x₁, y₁) and (x₂, y₂) is: \(m = \frac{{{y_2}; - ;{y_1}}}{{{x_2} - {x_1}}}\)

The slope of line AC is given by

\(\frac{2\ -\ 0}{4\ -\ 0}\ =\ \frac{1}{2}\)

In square ABCD, the diagonals AC and BD are perpendicular to each other.

⇒ Slope of AC × slope of BD  = - 1.

So, the slope of BD is - 2.

As we know that, the equation of a line passing through the point (x1, y1) and having the slope ‘m’ is given as:

y – y1 = m ⋅ (x – x1)

The equation of BD whose slope is - 2 and passes through origin is given by:

y - 0 = (-2) ⋅ (x - 0)

⇒ 2x + y = 0. 

Hence, the correct option is 1.

CONCEPT:

The eccentricity of a hyperbola of the form \(\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1\) is given by: \(e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}} \)

CALCULATION:

Given: The point (3tan θ, 2sec θ) lies on the hyperbola.

As we can see that, if we substitute x = 3tan θ and y = 2sec θ in the expression \(\frac{y^2}{4}-\frac{x^2}{9} \) we get

⇒ \(\frac{y^2}{4}-\frac{x^2}{9} = sec^2 \ \theta - tan^2 \ \theta = 1\)

So, we can say that, the point (3tan θ, 2sec θ) lies on the hyperbola \(\frac{y^2}{4}-\frac{x^2}{9} = 1\)

As we know that, eccentricity of the hyperbola of the form \(\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1\) is given by: \(e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}} \)

Here, a² = 4 and b² = 9

⇒ \(e = \sqrt {1 + \frac{{{9}}}{{{4}}}} = \frac{\sqrt {13}}{2} \)

Hence, correct option is 4.

CONCEPT:

Eccentricity:

The constant ratio of distance of point lying on conic from the focus to its perpendicular distance from directrix is called the eccentricity of a conic section and is denoted by e.

• For an ellipse: e < 1
• For a parabola: e = 1
• For a hyperbola: e > 1
• For a circle: e = 0
• For a pair of straight lines: e = ∞

CALCULATION:

As we know that, eccentricity of a circle is 0 i.e e = 0 for a circle.

So, statement 1 is correct.

We also know that, eccentricity of a parabola is equal to 1 i.e e = 1 for a parabola.

So, statement 2 is also correct.

We also know that, eccentricity of an ellipse is always less than 1 i.e e < 1 for an ellipse.

So, statement 3 is also correct.

Hence, correct option is 4.

CONCEPT:

The angle θ between two lines whose direction ratios are proportional to a₁, b₁, c₁ and a₂, b₂, c₂ respectively is given by: \(\cos θ = \left| {\frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} ;\sqrt {a_2^2 + b_2^2 + c_2^2} }}} \right|\)

CALCULATION:

Here, we have to find the angle between the two lines having direction ratios (6, 3, 6) and (3, 3, 0).

Here, a₁ = 6, b₁ = 3, c₁ = 6, a₂ = 3, b₂ = 3 and c₂ = 0

As we know that, if θ is the angle between two lines having direction ratios proportional to a1, b1, c1 and a2, b2, c₂ is given by: \(\cos θ = \left| {\frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} ;\sqrt {a_2^2 + b_2^2 + c_2^2} }}} \right|\)

⇒ \(\cos θ = \left| {\frac{{6 \cdot 3 + 3 \cdot 3 +6 \cdot 0}}{{\sqrt {6^2 + 3^2 + 6^2} ;\sqrt {3^2 + 3^2 + 0^2} }}} \right|\)

⇒ \(cos θ = \frac{1}{\sqrt 2}\)

⇒ θ = π/4

Hence, correct option is 2.

Concept:

If a, b, c are the direction ratios of a line passing through the point (x₁, y₁, z₁), then the equation of the line is given by: \(\frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}\)

If a, b, c are the direction ratios of a line then the direction cosine of the line is given by: \(\rm l= \frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}} m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}},n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

Calculation

Given: l, m, n are the direction cosines of the line x - 1 = 2(y + 3) = 1 - z

The given equation of lines can be re-written as \(\frac{{x - 1}}{1} = \frac{{y + 3}}{\frac{1}{2}} = \frac{{z - 1}}{-1}\)

So, by comparing the equation \(\frac{{x - 1}}{1} = \frac{{y + 3}}{\frac{1}{2}} = \frac{{z - 1}}{-1}\) with \(\frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}\) we get

⇒ a = 1, b = 1/2 and c = - 1

As we know, if a, b, c are the direction ratios of a line, then the direction cosine of the line is given by: \(\rm l= \frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}} m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}},n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

The direction cosine of the given line is:

\(\Rightarrow \rm l= \frac{2}{3}, m=\frac{1}{3},n=\frac{-2}{3}\)

⇒ l4 + m4 + n4  = \(\frac{16}{81}+\frac{1}{81}+\frac{16}{81} =\frac{33}{81}=\frac{11}{27}\) 

Hence, the correct option is 2.

Concept:

The projection of any point on (a, b, c) on x -axis is (a, 0, 0), y-axis is (0, b, 0) and z-axis is (0, 0, c).

Calculation:

The projection of A(1, 7, -5) on y-axis is (0, 7, 0) and projection of B(-3, 4, -2) on y-axis is (0, 4, 0).

Let C = (0 ,7, 0), D = (0, 4, 0) and O = (0, 0, 0) is the origin

⇒ OC = 7 and OD = 4

Let the projection of line segment AB is m = OD - OC.

⇒ m = OD - OC = 7 - 4 = 3

Hence, correct option is 3.

CONCEPT:

• The direction ratios of the line joining the points (x₁, y₁, z₁) and (x₂, y₂, z₂) is given by: a = x₂ - x₁, b = y₂ - y₁ and c = z₂ - z₁
• If a, b, c are the direction ration ratios of a line passing through the point (x₁, y₁, z₁), then the equation of line is given by: \(\frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}\)

CALCULATION:

Given: The line joining the points (k, 1, 3) and (1, -2, k + 1) also passes through the point (15, 2, -4)

As we know that, the direction ratios of the line joining the points (x1, y1, z1) and (x2, y2, z2) is given by: a = x2 - x1, b = y2 - y1 and c = z2 - z₁

So, the direction ratios of the line joining the points (k, 1, 3) and (1, -2, k + 1) is: a = 1 - k, b = - 3 and c = k - 2

As we know, if a, b, and c are the direction ratio ratios of a line passing through the point (x1, y1, z1), then the equation of a line is given by: \(\frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}\)

So, the equation of the line with direction ratios a, b, c and passing through the point (k, 1, 3) is given by: \(\frac{{x - k}}{1 - k} = \frac{{y - 1}}{- 3} = \frac{{z - 3}}{k - 2}\)

∵ It is given that, the line represented by \(\frac{{x - k}}{1 - k} = \frac{{y - 1}}{- 3} = \frac{{z - 3}}{k - 2}\) also passes through the point (15, 2, -4)

So, substitute x = 15, y = 2 and z = - 4 in the equation \(\frac{{x - k}}{1 - k} = \frac{{y - 1}}{- 3} = \frac{{z - 3}}{k - 2}\)

⇒ \(\frac{{15 - k}}{1 - k} = \frac{{2 - 1}}{- 3} = \frac{{-4 - 3}}{k - 2}\)

⇒ \(\frac{{15 - k}}{1 - k} = \frac{{1}}{- 3} \)

⇒ - 45 + 3k = 1 - k

⇒ k = 23/2

⇒ \(\frac{{1}}{- 3} = \frac{{-7}}{k - 2}\)

⇒ k - 2 = 21

⇒ k = 23

So, there are two possible values of k as shown above.

Hence, correct option is 3

CONCEPT:

Perpendicular distance of a plane ax + by + cz + d = 0 from a point P (x₁, y₁, z₁) is given by: \(d = \left| {\frac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\)

CALCULATION:

Let A(x, y, z) be the foot of the perpendicular drawn from the origin to the plane x + y + z = 3.

As we know that, the perpendicular distance of a plane ax + by + cz + d = 0 from a point P (x1, y1, z1) is given by: \(d = \left| {\frac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\)

So, the distance between the origin and the plane x + y + z - 3 = 0 is given by: \(d = \left| {\frac{{0 + 0 + 0 - 3}}{{\sqrt {{1^2} + {1^2} + {1^2}} }}} \right| = \frac{3}{\sqrt 3} = \sqrt 3\)

So, this means that the length of the line joining the points origin and A is √3

As we can see that from the given options, if A = (1, 1, 1) then the distance between the points origin and A is √3

Hence, correct option is 3.

CONCEPT:

The scalar product of two vectors \(\vec a \ and \ \vec b \)is given by \(\vec a \cdot ;\vec b = \left| {\vec a} \right| \times \left| {\vec b} \right|\cos θ \)

CALCULATION:

Given: Vector \(\vec r=a \hat i+b \hat j\) is equally inclined to both x and y axes and the magnitude of the vector is 2 units.

i.e \(|\vec r| = 2\)

⇒ \(\sqrt {a^2 + b^2} = 2\)

⇒ a² + b² = 4 --------(1)

∵ The vector \(\vec r=a \hat i+b \hat j\) is equally inclined to both x and y axes

Let θ be the angle between the vector \(\vec r=a \hat i+b \hat j\) and both the x and y axes.

⇒ \(cos \ \theta = \frac{a}{2} = \frac{b}{2}\)

⇒ a = b

So, by substituting a = b in equation (1), we get

⇒ 2b² = 4 ⇒ b = √2

So, a = b = √2

Hence, correct option is 3

CONCEPT:

• The scalar triple product of three vectors is zero if any two of them are equal
• If \(\vec a \) is perpendicular to the vectors \(\vec b\) then \(\vec a \cdot \vec b = 0\)

CALCULATION:

Given:  \(\vec c=\vec a+\vec b\), where \(|\vec a|=|\vec b|\ne0\)

Statement 1: \(\vec c\) is perpendicular to \((\vec a-\vec b).\)

First let's find out \(\vec c \cdot (\vec a-\vec b) = (\vec a + \vec b) \cdot (\vec a-\vec b)\)

⇒ \((\vec a + \vec b) \cdot (\vec a-\vec b) = |\vec a|^2 - \vec a \cdot \vec b + \vec b \cdot \vec a - |\vec b|^2\)

As we know that, \(\vec a \cdot \vec b = \vec b \cdot \vec a\)

⇒ \((\vec a + \vec b) \cdot (\vec a-\vec b) = |\vec a|^2 - |\vec b|^2\)

∵ It is given that \(|\vec a|=|\vec b|\ne0\)

⇒ \((\vec a + \vec b) \cdot (\vec a-\vec b) = |\vec a|^2 - |\vec b|^2 = 0\)

⇒ \(\vec c \cdot (\vec a-\vec b) = 0\)

Hence, statement 1 is true.

Statement 2: \(\vec c\) is perpendicular to \(\vec a \times \vec b.\)

First let's find out \(\vec c \cdot (\vec a \times \vec b) = (\vec a + \vec b) \cdot (\vec a \times \vec b)\)

⇒ \((\vec a + \vec b) \cdot (\vec a \times \vec b) = \vec a \cdot (\vec a \times \vec b) + \vec b \cdot (\vec a \times \vec b)\)

As we know that, the scalar triple product of three vectors is zero if any two of them are equal

⇒ \((\vec a + \vec b) \cdot (\vec a \times \vec b) = 0\)

Hence, statement 2 is true.

Hence, the correct option is 3.

CONCEPT:

The scalar product of two vectors \(\vec a \ and \ \vec b \)is given by \(\vec a \cdot ;\vec b = \left| {\vec a} \right| \times \left| {\vec b} \right|\cos θ \)

If the vectors \(\vec a \ and \ \vec b \) are perpendicular then \(\vec a \cdot ;\vec b = 0\)

CALCULATION:

Given: \(\vec a :and: \vec b\) are two vectors such that \(|\vec a + \vec b|= |\vec a - \vec b|=4\)

⇒ \(|\vec a + \vec b|^2= |\vec a - \vec b|^2\)

⇒ \((\vec a + \vec b) \cdot (\vec a + \vec b) = (\vec a - \vec b) \cdot (\vec a - \vec b)\)

⇒ \(|\vec a|^2 + \vec a \cdot \vec b + \vec b \cdot \vec a + |\vec b|^2 = |\vec a|^2 - \vec a \cdot \vec b - \vec b \cdot \vec a + |\vec b|^2\)

∵ \(\vec a \cdot \vec b = \vec b \cdot \vec a\)

⇒ \(|\vec a|^2 + 2\vec a \cdot \vec b + |\vec b|^2 = |\vec a|^2 - 2\vec a \cdot \vec b + |\vec b|^2\)

⇒ \(4\vec a \cdot \vec b = 0\)

⇒ \(\vec a \cdot \vec b = 0\)

So, \(\vec a \) must be perpendicular to \(\vec b.\)

Hence, correct option is 3.

CONCEPT:

Properties of Scalar Triple Product

• [a b c] = [b c a] = [c a b]
• [a b c] = - [b a c] = - [c b a] = - [a c b]
• [(a + b) c d] = [a c d] + [b c d]
• [λa, b c] = λ [a b c]
• Three non-zero vectors \(\vec a,;\vec b;and;\vec c\) are coplanar if and only if [a b c] = 0

CALCULATION:

Given: \(\vec a, \vec b:and : \vec c\) are coplanar i.e \([\vec a \ \vec b \ \vec c] = 0\)

⇒ \((2\vec a\times 3\vec b)\cdot4\vec c = [2\vec a \ 3\vec b \ 4\vec c]\) and \((5\vec b\times 3\vec c)\cdot6\vec a = [5\vec b \ 3\vec c \ 6\vec a]\)

⇒ \((2\vec a\times 3\vec b)\cdot4\vec c+(5\vec b\times 3\vec c)\cdot6\vec a = [2\vec a \ 3\vec b \ 4\vec c] + [5\vec b \ 3\vec c \ 6\vec a]\)

As we know that, [a b c] = [b c a] = [c a b]

⇒ \((2\vec a\times 3\vec b)\cdot4\vec c+(5\vec b\times 3\vec c)\cdot6\vec a = [2\vec a \ 3\vec b \ 4\vec c] + [6\vec a \ 5\vec b \ 3\vec c]\)

As we know that, [λa, b c] = λ [a b c]

⇒ \( [2\vec a \ 3\vec b \ 4\vec c] + [6\vec a \ 5\vec b \ 3\vec c] = 24 [\vec a \ \vec b \ \vec c] + 90 [\vec a \ \vec b \ \vec c]\)

As we know that, vectors \(\vec a,;\vec b;and;\vec c\) are coplanar if and only if [a b c] = 0

⇒ \( [2\vec a \ 3\vec b \ 4\vec c] + [6\vec a \ 5\vec b \ 3\vec c] = 0\)

⇒ \((2\vec a\times 3\vec b)\cdot4\vec c+(5\vec b\times 3\vec c)\cdot6\vec a = 0\)

Hence, correct option is  3.

Concept:

The cross product of two vectors \(\vec a \ and \ \vec b \)is given by \(\vec a \times \vec b = \left| {\vec a} \right| \cdot \left| {\vec b} \right|\sin θ ;\hat n\) and \(|\vec a \times \vec b| = \left| {\vec a} \right| \cdot \left| {\vec b} \right|\sin θ ;\)

The scalar product of two vectors \(\vec a \ and \ \vec b \)is given by \(\vec a \cdot ;\vec b = \left| {\vec a} \right| \times \left| {\vec b} \right|\cos θ \)

If \(\vec a\) is a unit vector then \(|\vec a| = 1\)

Calculations:

Statement 1: The cross product of two unit vectors is always a unit vector.

Let \(\vec a\) and \(\vec b\) are two unit vectors.

i.e \(|\vec a| = 1 \ and \ |\vec b| = 1\)

As we know that, the cross product of two vectors \(\vec a \ and \ \vec b \)is given by \(\vec a \times \vec b = \left| {\vec a} \right| \cdot \left| {\vec b} \right|\sin θ ;\hat n\) and \(|\vec a \times \vec b| = \left| {\vec a} \right| \cdot \left| {\vec b} \right|\sin θ ;\)

⇒ \(|\vec a \times \vec b| = \left| {\vec a} \right| \cdot \left| {\vec b} \right|\sin θ ; = sin θ \)

The range of sin θ is [-1, 1]

So, it is not necessarily true that the cross product of two unit vectors is always a unit vector.

Hence, statement 1 is false.

Statement 2: The dot product of two unit vectors is always unity.

Let \(\vec a\) and \(\vec b\) are two unit vectors.

i.e \(|\vec a| = 1 \ and \ |\vec b| = 1\)

As we know that, the scalar product of two vectors \(\vec a \ and \ \vec b \)is given by \(\vec a \cdot ;\vec b = \left| {\vec a} \right| \times \left| {\vec b} \right|\cos θ \) 

⇒ \(|\vec a \cdot ;\vec b |= cos \ θ \)

The range of cos θ is [-1, 1].

So, it is not necessarily true that the dot product of two unit vectors is always a unit vector.

Hence, statement 2 is false.

Statement 3: The magnitude of sum of two unit vectors is always greater than the magnitude of their difference.

Let \(\vec a \ = \hat i \ and \ \ \vec b = \hat j\) 

As we can see that, the vectors  \(\vec a\) and \(\vec b\) are two unit vectors

⇒  \(|\hat i + \hat j| = \sqrt 2\) and \(|\hat i - \hat j| = \sqrt 2\)

⇒ \(|\vec a + \vec b| = |\vec a - \vec b|\)

So, statement 3 is also false.

Hence, the correct option is 4.