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Suryansh Tiwari

Suryansh Tiwari

Physics Enthusiast / Physics Tutor

2–2 Vector Operations

3 min readDec 7, 2021

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1. Resolution of vectors:

The process of splitting a given vector

into its components is called resolution of the vector.

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Eg1: Find the x and y components of a vector having a magnitude of 12 and making an angle of 30° with the positive x-axis.

Solution:

The given vector is V= 12, and it makes an angle θ = 30º.

The x component of the vector = Vx = VCosθ = 12.Cos30º = 12.(√3/2) = 6√3.

The y component of the vector = Vy = VSinθ = 12.Sin30º = 12.(1/2) = 6.

Therefore, the x component of the vector is equal to 6√3 and y component is equal to 6.​​​​​​

2. Magnitude of a vector:

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Press enter or click to view image in full size

3. Unit vector along the given vector:

A unit vector along the given vector can be found by dividing the vector by it’s magnitude.

Eg1: Find the unit vector in the direction of vector a⃗ = 2î + 3ĵ + k̂.

Solution :

4. DOT/SCALAR product of Vectors:

The scalar product or dot product of two vectors P and Q is defined as the product of magnitudes of the two vectors and the cosine of the angle θ between the two vectors.

|P.Q|
= |P||Q|cos θ

P = mî+nĵ+ok̂, Q = xî+yĵ+zk̂

|P.Q| = (mx + ny + ok)

. . . . .

Eg1: If f=10N, s=5m, θ=60° Find Work done.

Solution :

W =|f.s|=|f||s|cos θ = 10×5×cos 60° =50×½ =25 Nm (J)

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. . . . .

Eg2: If f=(2î-7ĵ+3k̂)N, v=(5î+ĵ+4k̂)m/s calculate P(power).

Solution :

P = f.v = (2×5) + (-7×1) + (3×4) = 10–7+12 = 15 Nm/s (W)

5. CROSS/VECTOR product of Vectors:

The vector product or cross product of two vectors ( P
and Q
) is equal to the product of magnitudes of the two vectors and sine of the smaller angle (θ) between the two vectors.

|P×Q|
= |P||Q|sin θ

P = mî+nĵ+ok̂, Q = xî+yĵ+zk̂

R = |P×Q| = (nz-oy)î - (mz-ox)ĵ +(my-nx)k̂

Eg1: If r=5√2 m, f=10N, θ=45° Find τorque(τ).

Solution :

τ =|r×f|=|r||f|sin θ = 5√2×10×sin 45° =5√2×10×1/√2 =50 Nm

. . . . .

Eg2: If r = (2î-7ĵ+3k̂)m, ω = (5î+ĵ+4k̂) rad/s calculate ν(velocity).

Solution :

ν = |r×ω| = (nz-oy)î -(mz-ox)ĵ + (my-nx)k̂ = (-28+3)î - (8-15)ĵ +(2-(-7))k̂ = (-25î+7ĵ+9k̂)m/s

. . . . .

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6. Area of parallelogram using cross product of vectors: Area = |P×Q|

The magnitude of cross product of two

vectors is numerically equal to the area of a parallelogram whose adjacent sides represent the two vectors.

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Press enter or click to view image in full size

7. Angle between two vectors using dot product

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Eg1: Find the angle between a=2î+ĵ-3k̂ & b=3î-2ĵ-k̂.

Solution :

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Suryansh Tiwari
Suryansh Tiwari

Written by Suryansh Tiwari

Physics Enthusiast / Physics Tutor