Quantitative Aptitude PYQs

Concept:

1). The area of the triangle formed by three vertices that are collinear is zero.

2). Area of a triangle formed by (x1​,y1​), (x2​,y2​), and (x3​,y3​) is given by,

     Area = \(\displaystyle \frac{1}{2}\)​[x1​(y2 ​- y3​) + x2​(y3 ​− y1​) + x3​(y1 ​− y2​)]

Calculation:

Given,

Points with coordinates (-5, 0), (5p2, 10p) and (5q2, 10q) are collinear

∴  Area of △ABC=0

We know that,

Area of a triangle formed by (x₁​,y₁​), (x₂​,y₂​), and (x₃​,y₃​) is given by,

Area = \(\displaystyle \frac{1}{2}\)​[x₁​(y₂ ​- y₃​) + x₂​(y₃ ​− y₁​) + x₃​(y₁ ​− y₂​)]

∴ Area of △ABC = \(\displaystyle \frac{1}{2}\)[-5(10p − 10q) + 5p²(10q − 0) + 5q²(0 − 10p)]

⇒ 0 = -50p + 50q + 50p2q - 50q2p

⇒ 0 = 50(q - p) + 50(p2q - q2p)

⇒ 50(p - q) = 50(p2q - q2p)

(p - q) = (p2q - q2p)

(p - q)  = pq(p - q) 

⇒ pq = 1           [p ≠ q]

∴ The value of pq = 1

Concept:

The equation of line in intercept form is

\(\displaystyle \frac{x}{a}+\frac{y}{b}=1\), where a and b are the intercept made on the x and y-axis.

Calculation:

The equation of line in intercept form is

⇒ \(\displaystyle \frac{x}{a}+\frac{y}{b}=1\)

Given, the line cuts off equal intercepts. So a = b

⇒ \(\displaystyle \frac{x}{a}+\frac{y}{a}=1\)

⇒ x + y = a      ------(i)

It passes through (1, -2).

So 1 – 2 = a

⇒ a = -1

⇒ x + y = -1        [from (i)]

⇒ x + y + 1 = 0

∴ x + y + 1 = 0 is the required equation.

Concept:

The general equation for a circle is (x - h)² + (y - k)² = r², where ( h, k ) is the center and r is the radius.

Calculation:

The circle which touches both the axes in the first quadrant and the line y - 2 = 0 is shown in the figure.

From the above figure, we can say that the diameter of the circle is 2.

⇒ r = 1 and the center of the circle is (1, 1).

We know that the equation of the circle is (x - h)2 + (y - k)2 = r2, where (h, k) is the center and r is the radius.

⇒ (x - 1)² + (y - 1)2 = 12

⇒ x2 - 2x + 1 + y2 - 2y + 1 = 12

⇒ x2 + y2 - 2x - 2y + 2 = 1

⇒ x2 + y2 - 2x - 2y + 1 = 0

∴ The equation of the circle is x2  + y2  - 2x - 2y + 1 = 0.

Concept:

Section of a right circular cone by a plane parallel to a generator of the cone is a parabola.

It is a locus of a point, which moves so that distance from a fixed point (FP) (focus) is equal to the distance from a fixed line (directrix)

From definition, FP = MP

⇒ \(\displaystyle \sqrt{(x-a)^2+(y−0)^2}​=\frac{∣x+a∣​}{1} \)

⇒ (x - a)² + y² = (x + a)²

⇒ y² = 4ax

This is the standard equation of Parabola.

Calculation:

The focus of the parabola is F (-3,0) and its directrix is the line x = 3 i.e., x - 3 = 0

Let P (x,y) be any point in the plane of directrix and focus, and MP be the perpendicular

distance from P to the directrix, then P lies on the parabola if FP = MP

⇒\(\displaystyle \sqrt{(x+3)^2+(y−0)^2}​=\frac{∣x-3∣​}{1} \)

Squaring both sides, we get,

⇒ x² + 6x + 9 + y² = x² - 6x + 9

⇒ 6x + y² = - 6x

⇒ y² = -12x

∴ y2 = -12x is the required equation of the parabola.

Concept:

The general equation of an ellipse is written as: 

\(\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)  where a and b are the lengths of the semi-major and semi-minor axes respectively.

Distance between foci is = 2ae, where e = \(\displaystyle \sqrt{1-\frac{b^2}{a^2}} \) 

Calculation:

Given:

The equation of the ellipse is x2 + 2y2 = 1

⇒ \(\displaystyle \frac{x^2}{1}+\frac{y^2}{\frac{1}{2}}=1\)

⇒ a² = 1, b² = \(\displaystyle \frac{1}{2}\)

e = \(\displaystyle \sqrt{1-\frac{b^2}{a^2}} \) = \(\displaystyle \sqrt{1-\frac{1}{2}} \) = \(\displaystyle \frac{1}{\sqrt2}\)

Distance between foci is = 2ae = 2 × 1 × \(\displaystyle \frac{1}{\sqrt2}\) = \(\sqrt 2\)

∴ Distance between foci is \(\sqrt 2\)

Concept:

The perimeter of the triangle = sum of all sides

Area of triangle = (1/2)base × height

Calculation:

Since it is not mentioned, Δ ABC is which type of triangle. So, to make calculation easy

Let us assume, Δ ABC is a right-angle triangle where ∠A is 90°.

Let a = 5, b = 4, and c = 3 (Pythagorean triple)

Using the above concept, Perimeter

p = 3 + 4 + 5

⇒ p = 12  

Area of triangle

q = (1/2) 3 × 4

⇒ q = 6

Hence, the required value

\(\Rightarrow p(p-2a) \tan \frac{A}{2}\) = 12(12 - 2 × 5)tan (90°/2) [∵ ∠A = 90°]

⇒ p(p − 2a)tan(A/2)= 24        [∵ tan 45° = 1

From option 4

4q = 4 × 6 = 24

∴ 4q is the correct answer.

Alternate MethodConcept:

The perimeter of the triangle = sum of all sides = a + b + c

Area of scalene triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\), 

Half-angle formulae: \(\displaystyle tan(\frac{A}{2})=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \)

Where 2s = a + b + c and a, b, c are the sides of the triangle.

Calculation:

If p be the perimeter of the triangle,

p = a + b + c                ------(1)

q = \(\sqrt{s(s-a)(s-b)(s-c)}\)        ------(2)

where a, b, c are the sides of the triangle.

Now, 

\(\displaystyle p(p-2a) \tan \left(\frac{A}{2}\right)\) = (a + b+ c) (a + b+ c - 2a) \(\displaystyle \tan \left(\frac{A}{2}\right) \)

⇒ (a + b+ c) (a + b+ c - 2a) \(\displaystyle \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \)

Multiplying and dividing \(\sqrt{s(s - a)}\), we get, 

⇒ (a + b+ c) (a + b+ c - 2a) \(\displaystyle \sqrt{\frac{(s-b)(s-c)}{s(s-a)}×\frac{s(s-a)}{ s(s-a)}} \)

⇒  (a + b+ c) (a + b+ c - 2a) \(\displaystyle \sqrt{\frac{s(s-a)(s-b)(s-c)}{s^2(s-a)^2}} \)

⇒ (a + b+ c) (a + b+ c - 2a) \(\displaystyle \frac{\sqrt{s(s-a)(s-b)(s-c)}}{s(s-a)}\)     

From equation (1) & (2)

⇒ 2s (b + c - a) \(\displaystyle \frac{q}{s(s-a)}\)     [∵ b + c - a = 2(s - a)]    

⇒  2s × 2(s - a) \(\displaystyle \frac{q}{s(s-a)}\)

⇒  4q

∴  \(\displaystyle p(p-2a) \tan \left(\frac{A}{2}\right)\) = 4q

Concept:

1. A positive slope means y increases as x increases (visually, the line moves up as you go from left to right).
2. A negative slope means y decreases as x increases (visually, the line moves down as you go from left to right).

Calculation:

Given:

x + 2y + 2 = 0      ------(i)

2x - 3y - 3 = 0      ------(ii)

By solving (i) and (ii) we get the point of intersection of the lines.

So x = 0 and y = -1

The point of intersection is (0, -1)

Given that the line makes equal intercepts with axes. So slope (m) can be tan θ = 1

or -1 but it is in the fourth quadrant so, it is positive.

Then the equation of the line is y - y₁ = m(x - x₁)

⇒ y - (-1) = 1(x - 0)

⇒ y + 1 = x

⇒ x - y = 1

From this equation, the intercept on the x and y axes are 1 and -1 respectively.

∴ The sum of the absolute values of the intercepts = 2.

Concept:

1. Standard Equation of a line y = mx + c, where m is the slope of the line.
2. Slopes of parallel lines are equal. The parallel lines are equally inclined with the positive x-axis and hence the slope of parallel lines are equal. If the slopes of two parallel lines are represented as m₁, m₂ then we have m₁ = m₂.

Calculation:

Given:

Lines ax + by + c = 0 and bx + ay + c = 0

Express the two lines in standard format that is y = mx + c

ax + by + c = 0 ⇒ by = -ax - c ⇒ \(\displaystyle y=(\frac{-a}{b})x-\frac{c}{b}\)      ------(i)

bx + ay + c = 0 ⇒ ay = -bx - c ⇒ \(\displaystyle y=(\frac{-b}{a})x-\frac{c}{a}\)      ------(ii)

Given both lines are parallel implies their slopes are equal, that is

⇒ \(\displaystyle \frac{-a}{b}=\frac{-b}{a}\)

⇒ b² = a²

⇒ a2 - b2 = 0

∴ a2 - b2 = 0 is correct.

Concept:

The distance between two points in a two-dimensional plane is given by \(\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Calculation:

Given:

The equation of the line is x + y = p. 

Let the line cut the coordinate axes at A and B and if C(h, k) be the midpoints of AB, then

Now the intercept made by the line on the coordinate axes is p.

Point C(h, k) is equidistant from the points A(p, 0) and B(0, p)

As C is the midpoint, AC = BC

⇒ \(\displaystyle \sqrt{(h-p)^2+(k-0)^2}=\sqrt{(h-0)^2+(k-p)^2}\)

Squaring both sides, we get,

⇒ (h - p)2 + k2 = h2 + (k - p)2

⇒ h² + p2 - 2hp + k2 = h2 + k2 + p2 - 2pk

⇒ - 2hp = - 2pk

⇒ h = k

Now, to obtain the locus of the mid-point replace (h, k) with (x, y),

⇒ x = y

∴ x - y = 0 is correct.

Concept:

The distance between two points in a two-dimensional plane is given by \(\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Calculation:

Given:

Point (x, y) is equidistant from the points (2a, 0) and (0, 3a)

\(\displaystyle d=\sqrt{(2a-x)^2+(0-y)^2}=\sqrt{(0-x)^2+(3a-y)^2}\)

Squaring both sides, we get,

⇒ (2a - x)² + y² = x² + (3a - y)2

⇒ 4a² + x² - 4ax + y2 = x2 + 9a2 + y2 - 6ay

⇒ 4a2 - 4ax = 9a2 - 6ay

⇒  5a2 - 6ay + 4ax = 0

⇒ 5a - 6y + 4x = 0

⇒ 4x - 6y + 5a = 0

∴ 4x - 6y + 5a = 0 is correct.

Concept:

1). The general form of the equation of the sphere is x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 where u, v, w and d are constant.

2). The equation of the plane whose intercepts are a, b, c on the x, y, z axes respectively is 

     \(\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\), where a, b, c ≠  0

Calculation:

The given equation of plane is 6x + ky + 3z - 12 = 0

In intercept form  \(\displaystyle \frac{x}{2}+\frac{y}{\frac{12}{k}}+\frac{z}{4}=1\)

So the coordinates of A, B, C are (2,0,0), (0,\(\frac{12}{k}\),0) and (0,0,4) respectively.

Let the equation of the sphere be x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0

As the sphere passes through the origin, it satisfies point (0, 0, 0) ⇒ d = 0.

⇒  x² + y² + z² + 2ux + 2vy + 2wz = 0    ------(1)

Equation (1) passes through the point A, B, C.

At A (2,0,0) we get, 4 + 2u.2 = 0 ⇒ u = -1

At B (0,\(\frac{12}{k}\),0) we get, \(\frac{144}{k^2}\) + 2v.\(\frac{12}{k}\) = 0 ⇒ v = -\(\frac{6}{k}\)

At C (0,0,4) we get, 16 + 2w.4 = 0 ⇒ w = -2

So the required equation of the sphere is

x2 + y2 + z2 + 2.(-1).x + 2.(-\(\frac{6}{k}\)).y + 2.(-2).z = 0

⇒ x2 + y2 + z2 - 2x + (-\(\frac{12}{k}\)).y - 4z = 0      ------(2)

Comparing x2 + y2 + z2 - 2x - 3y - 4z = 0  with equation (1) we get, k = 4.

∴ The value of k is 4.

Concept:

1). The general form of the equation of the sphere is x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 where u, v, w and d are constant.

2). Centre of  the sphere is given by (-u, -v, -w)

3). Radius of the sphere is given by r = \(\displaystyle \sqrt{{u^2+v^2+w^2-d}} \)

4). The perpendicular distance of a point (x₁, y₁, z₁) from plane ax + by + cz = d is given by \(\displaystyle p=\mid \frac{ax_1+by_1+cz_1-d}{\sqrt{{a^2+b^2+c^2}}}\mid \)

Calculation:

The given equation of the sphere is x2 + y2 + z2 - 2x - 3y - 4z = 0.

Centre of  the sphere is given by (-u, -v, -w) ⇒ (1, \(\frac{3}{2}\), 2)

The given equation of the plane is 6x + 4y + 3z - 12 = 0

If p is the perpendicular distance from the center of the sphere to the plane,

Then \(\displaystyle p=\mid \frac{ax_1+by_1+cz_1-d}{\sqrt{{a^2+b^2+c^2}}}\mid \)

⇒ \(\displaystyle p=\mid \frac{6\times1+4\times\frac{3}{2}+3\times2-12}{\sqrt{{6^2+4^2+3^2}}}\mid \)

⇒ \(\displaystyle p=\mid \frac{6}{\sqrt{{61}}}\mid \)

⇒ p = 0.768

∴ 0.5 < p < 1

Concept: 

1). The general form of the equation of the sphere is x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 where u, v, w and d are constant.

2). Centre of  the sphere is given by (-u, -v, -w)

3). Equation of line passing through two points (x1, y1, z1) and (x2, y2, z2) is given by \(\displaystyle \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\)

Calculation:

The given equation of the sphere is x2 + y2 + z2 - 2x - 3y - 4z = 0.

Centre of  the sphere is given by (-u, -v, -w) ⇒ (1, \(\frac{3}{2}\), 2)

Here the two points are (0, 0, 0) and (1, \(\frac{3}{2}\), 2).

Equation of line passing through two points (0, 0, 0) and (1, \(\frac{3}{2}\), 2) is given by,

\(\displaystyle \frac{x-0}{1-0}=\frac{y-0}{\frac{3}{2}-0}=\frac{z-0}{2-0}\)

⇒ x = \(\displaystyle \frac{2y}{3}\) = \(\displaystyle \frac{z}{2}\)

⇒ 6x = 4y = 3z

∴ The required equation of the line is 6x = 4y = 3z.

Concept:

The equation to the plane passing through P(x₁, y₁, z₁) and having direction ratios (a, b, c) for its normal is

a(x - x₁) + b(y - y₁) + c (z - z₁) = 0.

Calculation:

Since the plane \(\displaystyle \frac{2x}{k} + \frac{2y}{3} + \frac{z}{3} = 2\) pass through the point (2, 3, -6);

it satisfies the given equation.

⇒ \(\displaystyle \frac{2\times 2}{k} + \frac{2\times3}{3} + \frac{(-6)}{3} = 2\)

⇒ k = 2

The given equation of the plane is \(\displaystyle \frac{2x}{k} + \frac{2y}{3} + \frac{z}{3} = 2\)

Putting the value of k = 2, we get,

\(\displaystyle \frac{2x}{2} + \frac{2y}{3} + \frac{z}{3} = 2\)

⇒ 3x + 2y + z = 6

The direction ratios of normal to the plane are <3, 2, 1>

∴ The direction ratios of normal to the plane are <3, 2, 1>

Concept:

The equation of the plane whose intercepts are a, b, c on the x, y, z axes, respectively, is

 \(\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)  (a, b, c ≠  0)

Calculation:

Since the plane \(\displaystyle \frac{2x}{k} + \frac{2y}{3} + \frac{z}{3} = 2\) pass through the point (2, 3, -6);

So, it satisfies the given equation.

So, \(\displaystyle \frac{2\times 2}{k} + \frac{2\times3}{3} + \frac{(-6)}{3} = 2\)

⇒ k = 2

The given equation of the plane is \(\displaystyle \frac{2x}{k} + \frac{2y}{3} + \frac{z}{3} = 2\)

Putting the value of k = 2, we get,

\(\displaystyle \frac{2x}{2} + \frac{2y}{3} + \frac{z}{3} = 2\)

⇒ 3x + 2y + z = 6

The equation of the plane in intercepts form,

\(\displaystyle \frac{3x}{6}+\frac{2y}{6}+\frac{z}{6}=1​​\)

⇒  \(\displaystyle \frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1\)

2, 3, and 6 are the intercepts made by the plane on the coordinate axes, respectively.

So, Sum of the intercept = 2 + 3 + 6 = 11.

∴ The value of (p + q + r) is 11.

Concept:

For two vectors \(\vec m \ and \ \vec n \) to be collinear,​ \(\vec m; = ;λ \vec n\) where λ is a scalar.

Calculation:

Given that, the vectors \(4\hat i + \hat j - 3\hat k\) & \(p\hat i + q\hat j - 2\hat k\) are collinear.

Since two vectors \(\vec m \ and \ \vec n \) are collinear then \(\vec m; = ;λ \vec n\) where λ is a scalar.

⇒ \(4\hat i + \hat j - 3\hat k;\ = λ × (;p\hat i + q\hat j - 2\hat k)\)

⇒ \(4\hat i + 1\hat j - 3\hat k;\ = λ p \hat i + λq \hat j - 2λ \hat k\)

⇒ λp = 4,  λq = 1 and -2λ = -3

⇒  λ = 3/2 

So, by substituting λ = 3/2 in  λp = 4 and λq = 1, we get

⇒ (3/2)p = 4 and (3/2)q = 1

⇒ p = 8/3 and q  = 2/3

∴  \(\frac{8}{3}, \frac{2}{3}\)is the correct answer.

Concept:

Centroid: The point at which the three medians of the triangle intersect is known as the centroid of a triangle.

If the three vertices of the triangle are A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃), then the

 \(\displaystyle G(x,y)=\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)

Calculation:

Given:

A, B, and C are vertices of the triangle

Position vectors of A, B, and C be  \(\vec a,\vec b,\vec c\).

G is the centroid of △ABC

∴ G = \(\displaystyle \frac{\vec a + \vec b + \vec c}{3} \)

⇒ \(\overrightarrow{AG}\) =  \(\displaystyle \frac{(\vec a+\vec b+\vec c)}{3} - \vec{a}\)

⇒ \(\overrightarrow{AG}\) = \(\displaystyle \frac{(\vec a+\vec b+\vec c)-3\vec{a}}{3}\)

⇒\(\overrightarrow{AG}\) = \(\displaystyle\frac{\vec b + \vec c -2\vec a}{3}\)

∴​ \(\overrightarrow{AG}\) = \(\displaystyle\frac{\vec b + \vec c -2\vec a}{3}\)

Concept:

One algebraic property of real numbers is the distributive law. The distributive law for the real numbers says: "For all real numbers x, y, and z, \(x.( y+ z)=x. y+ x.z\)

The vector dot product is distributive over addition. In general: \(\vec a.( \vec b+ \vec c)=\vec a. \vec b+ \vec a.\vec c\)  

The vector cross product is distributive over addition. In general: \(\vec a × (\vec b + \vec c) = \vec a × \vec b + \vec a × \vec c\)

Associative property: (p × q) × r = p × (q × r) (where p, q, and r are any three natural/whole numbers)

Calculation:

Let

\(\displaystyle \vec a = a_x \overline i+a_y \overline j+a_z \overline k\vec b = b_x \overline i+b_y \overline j+b_z \overline k\vec c = c_x \overline i+c_y \overline j+c_z \overline k\)

Statement I: Dot product over vector addition is distributive

We have to prove \(\vec a.( \vec b+ \vec c)=\vec a. \vec b+ \vec a.\vec c\)

• \(\displaystyle \vec a.( \vec b+ \vec c)=(a_x ̂ i+a_y ̂ j+a_z ̂ k).[(b_x ̂ i+b_y ̂ j+b_z ̂ k)+(c_x ̂ i+c_y ̂ j+c_z ̂ k)] \)
• \(\displaystyle \vec a.( \vec b+ \vec c)=(a_x ̂ i+a_y ̂ j+a_z ̂ k).[(b_x+c_x)̂ i+(b_y+c_y)̂ j+(b_z +c_z )̂ k] \)
• \(\vec a.( \vec b+ \vec c)=\) a_x (b_x + c_x) + a_y (b_y + c_y) + a_z (b_z + c_z)
• \(\vec a.( \vec b+ \vec c)=\) ax bx + ax cx + ay by + ay cy + az bz + az cz...................................

(1) - \(\displaystyle \vec a. \vec b+ \vec a.\vec c=(a_x ̂ i+a_y ̂ j+a_z ̂ k).(b_x ̂ i+b_y ̂ j+b_z ̂ k)+(a_x ̂ i+a_y ̂ j+a_z ̂ k).(c_x ̂ i+c_y ̂ j+c_z ̂ k)\)

• \(\displaystyle \vec a. \vec b+ \vec a.\vec c=(a_x.b_x+a_y.b_y +a_z.b_z)+(a_x.c_x +a_y.c_y+a_z.c_z)\)
• \(\vec a. \vec b+ \vec a.\vec c=\) ax bx + ax cx + ay by + ay cy + az bz + az cz................................... (2) - From equation (1) and (2)
• ∴ \(\vec a.( \vec b+ \vec c)=\vec a. \vec b+ \vec a.\vec c\)

Statement II: Cross product over vector addition is distributive

We have to prove \(\vec a × (\vec b + \vec c) = \vec a × \vec b + \vec a × \vec c\)

• \(\displaystyle \vec a\times(\vec b+ \vec c)=(a_x ̂ i+a_y ̂ j+a_z ̂ k)\times[(b_x ̂ i+b_y ̂ j+b_z ̂ k)+(c_x ̂ i+c_y ̂ j+c_z ̂ k)] \)
• ​\(\displaystyle \vec a\times( \vec b+ \vec c)=(a_x ̂ i+a_y ̂ j+a_z ̂ k)\times[(b_x+c_x)̂ i+(b_y+c_y)̂ j+(b_z +c_z )̂ k] \)​
• \(\vec a\times( \vec b+ \vec c)=\begin{bmatrix} \overline i & \overline j & \overline k \[0.3em] a_x & a_y & a_z \[0.3em] b_x+c_x &b_y+c_y & b_z+c_z \end{bmatrix}\)
• \(\vec a\times( \vec b+ \vec c)=\) ̂̂î [ay (bz + cz) - az (by + cy)] - ĵ [ax (bz + cz) - az (bx + cx)] + k̂ [ax (by + cy) - ay (bx + cx)] ............(3)
• \(\displaystyle \vec a\times \vec b+ \vec a\times\vec c=(a_x ̂ i+a_y ̂ j+a_z ̂ k)\times(b_x ̂ i+b_y ̂ j+b_z ̂ k)+(a_x ̂ i+a_y ̂ j+a_z ̂ k)\times(c_x ̂ i+c_y ̂ j+c_z ̂ k)\)
• \(\displaystyle \vec a\times \vec b+ \vec a\times\vec c=\begin{bmatrix} ̂ i & ̂ j & ̂ k \[0.3em] a_x & a_y & a_z \[0.3em] b_x &b_y& b_z \end{bmatrix}+\begin{bmatrix} ̂ i & ̂ j & ̂ k \[0.3em] a_x & a_y & a_z \[0.3em] c_x &c_y& c_z \end{bmatrix}\)
• \((\vec a\times \vec b+\vec a\times \vec c)=\) ̂̂î [aybz - azby)] - ĵ (axbz - a_zbx)] + k̂ [ax by - aybx] + î [aycz - azcy)] - ĵ (axcz - azcx)] + k̂ [ax cy - aycx]
• \((\vec a\times \vec b+\vec a\times \vec c)=\) î [ay (bz + cz) - az (by + cy)] - ĵ [ax (bz + cz) - az (bx + cx)] + k̂ [ax (by + cy) - ay (bx + cx)] ..............(4)
• ∴ \(\vec a × (\vec b + \vec c) = \vec a × \vec b + \vec a × \vec c\)

Statement III: Cross product of vectors is associative

• Consider two non-zero perpendicular vectors, a and b.
• We have (a × a) × b = 0 × b = 0
• However, a × b is perpendicular to a and is not the zero vector, so
• a × (a × b) ≠ 0
• (a × a) × b ≠ a × (a × b)
• Cross product of vectors is not associative

∴ Only Statements I and II are correct.

Concept:

Properties of the cross product of vector

1. a × b is a vector that is perpendicular to both a and b.
2. The direction of a × b is determined by the right-hand rule. (This means that if we curl the fingers of the right hand from a to b, then the thumb points in the direction of (a × b)).
3. The magnitude (or length) of the vector a × b, written as ∥a × b∥
4. b × a = − a × b and  a × a = 0

 

Calculation:

Statement I:  \(\vec a\) is unique if  \(\vec b\) and is given

Given that 

\(\vec{a}× \vec{b} = \vec{c} \)

If \(\displaystyle \vec b\ and \ \vec c\) are given,

Let b = ĵ & c = k̂ &

Case: 1 Let a =  (î + ĵ)

\(\vec{a}× \vec{b} = \vec{c} \) = (î + ĵ) × ĵ

\(\vec{a}× \vec{b} = \vec{c} \) = î × ĵ + ĵ × ĵ   

\(\vec{a}× \vec{b} = \vec{c} \) = k̂ + 0 

\(\vec{a}× \vec{b} = \vec{c} \) = k̂

Case: 2   Let a = î    

\(\vec{a}× \vec{b} = \vec{c} \) = î × ĵ = k̂  

Hence, a can be either î or (î + ĵ)

Statement II:  \(\vec c\) is unique if \(\vec a\) and is given

\(\vec{a}× \vec{b}\) represent vector with magnitude |\(\vec{a}× \vec{b}\)| & direction ⊥ to both vector

a & vector b.

∴ only statement II is correct.

Concept:

1. The scalar product of two vectors a and b of magnitude |a| and |b| is given as |a||b| cos θ, where θ represents the angle between the vectors a and b taken in the direction of the vectors.
2. We can express the scalar product as \(\mid\vec a\mid.\mid\vec b\mid\)=|a||b| cosθ, where |a| and |b| represent the magnitude of the vectors a and b while cos θ denotes the cosine of the angle between both the vectors and a.b indicates the dot product of the two vectors.

Calculation:

Given:

\(\vec a\) and \(\vec b\) be two unit vectors such that \(|\vec a - \vec b|<2\) and 2θ is the angle between \(\vec a\) and \(\vec b\)

⇒ \(\mid\vec a\mid= \mid\vec b\mid=1\)

We are given \(|\vec a - \vec b|<2.\)

Squaring both sides, we get, 

⇒ \(|\vec a - \vec b|^2<2^2\)

⇒ \((\vec a - \vec b)(\vec a - \vec b)<4\)

⇒ \(\mid\vec a\mid^2+ \mid\vec b\mid^2-2\ \vec a.\vec b<4\)

⇒ 1 + 1 - 2 \(\mid\vec a\mid. \mid\vec b\mid\ cos2θ<4\)

⇒ 1 + 1 - 2​ . 1 . 1 . cos2θ < 4

⇒ 2 - 2.cos2θ < 4

⇒ 1 - cos2θ < 2

⇒ 2 sin²θ < 2

⇒ sin2θ < 1

∴ -1≤ sin θ < 1 is correct.