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Concept:

If 2 forces are \(\rm \vec {a}\) and \(\rm \vec {b}\)

Resultant of  \(\rm \vec {a}\) and \(\rm \vec {b}\):

\(\rm \left|\vec a+\vec b\right| = \sqrt{a^2+b^2+2ab\cosθ}\)

Where a, b are magnitude of vectors \(\rm \vec a \text{ and } \vec b\) and θ is angle between them.

Calculation:

Given:

Resultant of forces F₁ and F₂ = \(\rm \left|\vec F_1+\vec F_2\right|\)

⇒ \(\rm \sqrt{(F_1^2+F_2^2)}=\sqrt{F_1^2+F_2^2+2F_1F_2\cosθ}\)

⇒ \(\rm F_1^2+F_2^2=F_1^2+F_2^2+2F_1F_2\cosθ\)

⇒ \(\rm 0=2F_1F_2\cosθ\)

⇒ \(\rm \cos θ=0\)

⇒ θ = 90o

Concept:

• The cross product of vector to itself = 0
• The cross product of collinear vectors = 0
• The dot product of collinear vectors = Product of their Magnitudes
• For dot product \(\rm (\vec P + \vec Q) \cdot \vec R = (\vec P\cdot \vec R) +(\vec Q\cdot \vec R)\)
• For cross product \(\rm (\vec P + \vec Q) \times\vec R = (\vec P\times\vec R) + (\vec Q\times\vec R) \)

 

Calculation:

Given:

\(\rm \vec{a}+\vec{b}+\vec{c}\) is collinear with \(\rm \vec d\)

⇒ \(\rm (\vec{a}+\vec{b}+\vec{c})\times\vec d = 0\)

⇒ \(\rm (\vec{a}+\vec{b}+\vec{c})\times\vec d +\vec d\times\vec d = 0\) (∵ \(\rm \vec d\times\vec d = 0\))

⇒ \(\rm (\vec{a}+\vec{b}+\vec{c}+\vec d)\times\vec d = 0\)                                     ...(i)

Also given:

\(\rm \vec{b}+\vec{c}+\vec{d}\) is collinear with \(\rm \vec{a}\)

⇒ \(\rm (\vec{b}+\vec{c}+\vec d)\times\vec{a} = 0\)

⇒ \(\rm \vec{a}\times\vec a+(\vec{b}+\vec{c}+\vec d)\times\vec a= 0\) (∵ \(\rm \vec a\times\vec a= 0\))

⇒ \(\rm (\vec{a}+\vec{b}+\vec{c}+\vec d)\times\vec a = 0\)                                     ...(ii)

On adding (i) and (ii):

⇒ \(\rm (\vec{a}+\vec{b}+\vec{c}+\vec d)\times\vec a +(\vec{a}+\vec{b}+\vec{c}+\vec d)\times\vec d= 0\) 

⇒ \(\rm (\vec{a}+\vec{b}+\vec{c}+\vec d)\times(\vec a+\vec d) = 0\)                            ...(iii)

On subtracting (i) from (ii):

⇒ \(\rm (\vec{a}+\vec{b}+\vec{c}+\vec d)\times\vec a -(\vec{a}+\vec{b}+\vec{c}+\vec d)\times\vec d= 0\)

⇒ \(\rm (\vec{a}+\vec{b}+\vec{c}+\vec d)\times(\vec a-\vec d) = 0\)                            ...(iv)

From (iii) and (iv); it is clear that \(\rm (\vec{a}+\vec{b}+\vec{c}+\vec d)\) is collinear to \(\rm (\vec a+\vec d)\text{ and }(\vec a-\vec d)\) Both Option 2 and 3 are correct

Concept:

The work done \(\rm W =(\vec{F_r})\cdot(\vec{dx})\) 

where \(\rm \vec{F_r}\) is the resultant force, and  \(\rm \vec{dx}\) is the displacement 

The unit vector in the direction of a \(\rm \vec P= \hat P={\vec P\over\left|\vec P\right|}\) 

Calculation:

The initial position of the particle (A) = 2i - j - 3k

The initial position of the particle (B) = 5i - j + k

Displacement \(\rm \vec{dx}\) = B - A = (5i - j + k) - (2i - j - 3k) = 3i + 4k

First force F₁ = magnitude of the 1^st force × unit vector in the given direction

⇒ \(\rm \vec {F_1}=5\times{6i+2j+3k\over\left|6i+2j+3k\right|}\) 

⇒ \(\boldsymbol{\rm \vec {F_1}={30i+10j+15k\over7}}\) 

Second force F₂ = magnitude of the 2^nd force × unit vector in the given direction

⇒ \(\rm \vec {F_2}=3\times{3i-2j+6k\over\left|3i-2j+6k\right|}\) 

⇒ \(\boldsymbol{\rm \vec {F_2}={9i-6j+18k\over7}}\) 

Third force F₃ = magnitude of the 3^rd force × unit vector in the given direction

⇒ \(\rm \vec {F_3}=1\times{2i-3j-6k\over\left|2i-3j-6k\right|}\) 

⇒ \(\boldsymbol{\rm \vec {F_3}={2i-3j-6k\over7}}\) 

Resultant Force \(\rm \vec{F_r} =\vec{F_1}+\vec{F_2}+\vec{F_3}\) 

⇒ \(\rm \vec{F_r} ={30i+10j+15k\over7}+{9i-6j+18k\over7}+{2i-3j-6k\over7}\)

⇒ \(\boldsymbol{\rm \vec{F_r} ={41i+j+27k\over7}}\) 

The work done by the forces:

\(\rm W =(\vec{F_r})\cdot(\vec{dx})\)

⇒ \(\rm W =\left({41i+j+27k\over7}\right)\cdot(3i+4k)\) 

⇒ \(\rm W =\left({123+0+108\over7}\right)={231\over7}\) 

⇒ \(\boldsymbol{\rm W =33}\) units

Concept:

If a point P divides a line joining points A(x₁, y₁, z₁) and B(x₂, y₂, z₂) in a ratio of m:n, then

\(\rm x_P ={nx_1 + mx_2\over n+m}\), \(\rm y_P ={ny_1 + my_2\over n+m}\) and \(\rm z_P ={nz_1 + mz_2\over n+m}\)

Calculation:

Assuming,

\(\rm \vec a =x_ai+y_aj+z_ak\)

\(\rm \vec b =x_bi+y_bj+z_bk\)

Vector of point p is 

\(\rm \vec p =x_pi+y_pj+z_pk\)

∵ \(\rm x_P ={nx_1 + mx_2\over n+m}\), \(\rm y_P ={ny_1 + my_2\over n+m}\) and \(\rm z_P ={nz_1 + mz_2\over n+m}\)

⇒ \(\rm \vec p ={nx_a + mx_b\over n+m}i+{ny_a + my_b\over n+m}j+{nz_a + mz_b\over n+m}k\)

⇒ \(\rm \vec p ={n(x_ai + y_aj+z_ak)+m(x_bi+y_bj+z_bk)\over n+m}\)

⇒ \(\boldsymbol{\rm \vec p ={n\vec a+m\vec b\over n+m}}\)

Cocept:

If a vector A is collinear to other vector B, it can be written as:

\(\rm \vec A = λ × \vec B\) where λ is any scalar number

Calculation:

Given \(\rm \vec{a}+2\vec{b}\) is collinear with \(\rm \vec {c}\) 

⇒ \(\rm \vec{a}+2\vec{b}\) = λ₁ × \(\rm \vec {c}\)                      ...(i)

Also given \(\rm \vec{b}+3\vec{c}\) is collinear with \(\rm \vec {a}\)

⇒ \(\rm \vec{b}+3\vec{c}\) = λ₂ × \(\rm \vec {a}\)                     ...(ii)

Subtracting 2×(ii) from (i)

⇒ \(\rm \vec{a}-6\vec{c} = λ_1 \vec c-2λ_2 \vec a\)

∴ λ₁ = - 6, λ₂ = \(\rm {-1\over2}\) 

From (ii)

⇒ \(\rm \vec{b}+3\vec{c}\)= \(\rm {-1\over2}\)(\rm \vec {a}\)

⇒ \(\rm 2\vec{b}+6\vec{c}\)= \(\rm -\vec {a}\)

⇒ \(\rm \vec a+ 2\vec{b}+6\vec{c} = 0\)

⇒ \(\boldsymbol{\rm \left|\vec a+ 2\vec{b}+6\vec{c}\right| = 0}\)

Concept:

• The vector \(\rm \vec {AB}\) joining two points with position vectors \(\rm \vec A\) and \(\rm \vec B\) is given by \(\rm \vec {AB}=\vec B-\vec A\).
• The length of a vector \(\rm \vec A\) = a₁î + a₂ĵ + a₃k̂ is given by \(\rm |\vec A|=\sqrt{a_1^2+a_2^2+a_3^2}\).

 

Calculation:

Let's say that the vertices of the triangle are:

\(\rm \vec A\) = î - 2ĵ + 2k̂

\(\rm \vec B\) = 2î + ĵ - k̂

\(\rm \vec C\) = 3î - ĵ + 2k̂

∴ \(\rm \vec {AB}=\vec B-\vec A\) = (2î + ĵ - k̂) - (î - 2ĵ + 2k̂) = î + 3ĵ - 3k̂

\(\rm |\vec{AB}|=\sqrt{1^2+3^2+(-3)^2}=\sqrt{19}\)

\(\rm \vec {BC}=\vec C-\vec B\) = (3î - ĵ + 2k̂) - (2î + ĵ - k̂) = î - 2ĵ + 3k̂

\(\rm |\vec{BC}|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{14}\)

\(\rm \vec {CA}=\vec A-\vec C\) = (î - 2ĵ + 2k̂) - (3î - ĵ + 2k̂) = -2î - ĵ + 0k̂

\(\rm |\vec{CA}|=\sqrt{(-2)^2+(-1)^2+0^2}=\sqrt5\)

Since, AB² = 19 = BC² + CA² = 14 + 5, the triangle is right angled at C.

Concept:

For two vectors \(\rm \vec A\) and \(\rm \vec B\) at an angle θ to each other:

• Dot Product is defined as \(\rm \vec A.\vec B=|\vec A||\vec B|\cos \theta\).
• Cross Product is defined as \(\rm \vec A\times \vec B=\vec n|\vec A||\vec B|\sin \theta\) where \(\rm \vec n\) is the unit vector perpendicular to the plane containing \(\rm \vec A\) and \(\rm \vec B\).

 

For three vectors \(\rm \vec A\), \(\rm \vec B\) and \(\rm \vec C\):

• Triple Cross Product: is defined as: \(\rm \vec A\times(\vec B\times\vec C)=(\vec A.\vec C)\vec B-(\vec A.\vec B)\vec C\).
• Triple Scalar Product (Box Product): is defined as \(\rm [\vec A\ \vec B\ \vec C]=\vec A.(\vec B\times\vec C)=\begin{vmatrix} \rm a_1 & \rm a_2 & \rm a_3 \ \rm b_1 & \rm b_2 & \rm b_3 \rm c_1 & \rm c_2 & \rm c_3 \end{vmatrix}\).

 

Volume of a parallelepiped, with vectors \(\rm \vec a\), \(\rm \vec b\) and \(\rm \vec c\) as its sides, is given by the box product of the three vectors.

• Volume = \(\rm [\vec a\ \vec b\ \vec c]\).

 

Calculation:

The sides of the parallelepiped are:

\(\rm \vec a\) = 2î + 3ĵ + 4k̂

\(\rm \vec b\) = î + αĵ + 2k̂

\(\rm \vec c\) = î + 2ĵ + αk̂

∴ Volume = \(\rm \rm [\vec a\ \vec b\ \vec c]=\begin{vmatrix} 2 & 3 & 4 \ 1 & \alpha & 2 \ 1 & 2 & \alpha \end{vmatrix}\) = 15

⇒ 2(α² - 4) + 3(2 - α) + 4(2 - α) = 15

⇒ 2α2 - 8 + 6 - 3α + 8 - 4α = 15

⇒ 2α2 - 7α - 9 = 0

⇒ 2α2 - 9α + 2α - 9 = 0

⇒ α(2α - 9) + (2α - 9) = 0

⇒ (2α - 9)(α + 1) = 0

⇒ 2α - 9 = 0 OR α + 1 = 0

⇒ α = \(\dfrac92\) OR α = -1.

Concept:

The solution to the quadratic equation ax2 + bx + c = 0 is given by: \(\rm x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).

• sin (-θ) = -sin θ.
• sin θ = sin (2nπ + θ).

 

Calculation:

sin2 x - sin x - 2 = 0

Using the formula for the roots of a quadratic equation: 

⇒ \(\rm \sin x = \frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-2)}}{2(1)}\)

⇒ \(\rm \sin x = \dfrac{1 +\sqrt{9}}{2}\) OR \(\rm \sin x = \dfrac{1 -\sqrt{9}}{2}\)

⇒ sin x = 2 OR sin x = -1.

∵ -1 ≤ sin x ≤ 1, ∴ sin x = 2 is not possible.

sin x = -1 = \(\rm -\sin\dfrac{\pi}{2}=\sin\left(-\dfrac{\pi}{2}\right)=\sin\left(2n\pi-\dfrac{\pi}{2}\right)\), n ∈ Z.

For n = 0, x = \(-\dfrac{\pi}{2}\).

For n = 1, x = \(\dfrac{3\pi}{2}\).

For n = 2, x = \(\dfrac{7\pi}{2}\).

The only value of x on the interval 0 ≤ x < 2π is x = \(\dfrac{3\pi}{2}\).

Additional Information

• Period of sin θ is 2π. ⇒ sin θ = sin (2nπ + θ).
• Period of cos θ is 2π. ⇒ cos θ = cos (2nπ + θ).
• Period of tan θ is π. ⇒ tan θ = tan (nπ + θ).

Concept:

Trigonometric Identities:

• tan (A + B) = \(\rm \frac{\tan A+\tan B}{1-\tan A \tan B}\).
• tan (nπ + θ) = tan θ.
• tan (-θ) = -tan θ

 

Calculations:

It is given that \(\rm \frac{\tan x}{2} = \frac{\tan y}{3} = \frac{\tan z}{5}\) = k (say).

∴ tan x = 2k, tan y = 3k and tan z = 5k.

It is also given that x + y + z = π.

⇒ x + y = π - z

⇒ tan (x + y) = tan [π + (-z)]

⇒ \(\rm \frac{\tan x+\tan y}{1-\tan x \tan x}=\tan (-z)=-\tan z\)

⇒ tan x + tan y = - tan z + tan x tan y tan z

⇒ tan x + tan y + tan z = tan x tan y tan z

Substituting the values in terms of k from the above result, we get:

2k + 3k + 5k = (2k)(3k)(5k)

⇒ 10k = 30k³

⇒ \(\rm k^2=\frac13\).

Now, tan2 x + tan2 y + tan2 z

= (2k)² + (3k)² + (5k)²

= 4k² + 9k² + 25k²

= 38k²

= \(\rm \dfrac{38}3\).

 

Additional Information

Trigonometric Ratios for Allied Angles: 

• sin (-θ) = -sin θ.
• cos (-θ) = cos θ.
• sin (nπ + θ) = (-1)n sin θ.
• cos (nπ + θ) = (-1)n cos θ.
• \(\rm \sin \left [(2n+1)\frac{\pi}{2}+\theta \right ]\) = (-1)n cos θ.
• \(\rm \cos \left [(2n+1)\frac{\pi}{2}+\theta \right ]\) = (-1)n (-sin θ).

Concept:

Trigonometric Identities:​

• cos (A ± B) = cos A cos B ∓ sin A sin B.
• 2 cos A cos B = cos (A - B) + cos (A + B).
• 2 sin A sin B = cos (A - B) - cos (A + B).

 

Trigonometric Values:

• cos 60° = \(\rm \dfrac{1}{2}\).
• cos 36° = \(\rm \dfrac{\sqrt5+1}{4}\).

 

Trigonometric Ratios for Allied Angles: 

• sin (-θ) = -sin θ.
• cos (-θ) = cos θ.
• sin (nπ + θ) = (-1)ⁿ sin θ.
• cos (nπ + θ) = (-1)ⁿ cos θ.
• \(\rm \sin \left [(2n+1)\dfrac{\pi}{2}+\theta \right ]\) = (-1)ⁿ cos θ.
• \(\rm \cos \left [(2n+1)\dfrac{\pi}{2}+\theta \right ]\) = (-1)ⁿ (-sin θ).

 

Calculation:

sin 12° sin 48° sin 54°

= \(\rm \dfrac{2\sin 48^{\circ}\sin 12^{\circ}}{2} \sin 54^{\circ}\)

= \(\rm \left [\dfrac{\cos (48^{\circ}-12^{\circ})-\cos (48^{\circ}+12^{\circ})}{2} \right ] \sin (90^{\circ}-36^{\circ})\)

= \(\rm \left (\dfrac{\cos 36^{\circ}-\cos 60^{\circ}}{2} \right ) \cos 36^{\circ}\)

= \(\rm \left ( \dfrac{1}{2} \right )\left (\dfrac{\sqrt5+1}{4}-\dfrac{1}{2} \right )\left ( \dfrac{\sqrt5+1}{4} \right )\)

= \(\rm \left ( \dfrac{1}{2} \right )\left (\dfrac{\sqrt5-1}{4} \right )\left ( \dfrac{\sqrt5+1}{4} \right )\)

= \(\rm \left ( \dfrac{1}{2} \right )\left (\dfrac{5-1}{16} \right )\)

= \(\rm \dfrac{1}{8}\).

 

Additional Information

• sin 18° = \(\rm \dfrac{\sqrt5-1}{4}\).
• cos 36° = \(\rm \dfrac{\sqrt5+1}{4}\).
• sin 36° = \(\rm \sqrt{\dfrac{5-\sqrt5}{8}}\).
• cos 18° = \(\rm \sqrt{\dfrac{5+\sqrt5}{8}}\).