Vector | Numericals For Class 11 Students

A Vector Quantity or Vector is a Physical Quantity that have both magnitude and direction. 
Symbol for denoting Vector are generally written as a letter with an arrow on top pointing to the right, or as a bolded letter without an arrow on top.

\begin{equation} \vec{F} \text{ for denoting force vector} \\ \vec{v} \text{ for denoting velocity vector} \\ \textbf{Or} \\ \textbf{f} \text{ for denoting force vector} \\ \textbf{v} \text{ for denoting velocity vector} \\ \end{equation}

Vector Quantities include force, velocity, acceleration, and momentum. The magnitude of a vector denotes how little or large it is, and its direction defines the direction in which it points or is being applied.

Like if we say there’s 10 newton force acting on top of a box, that means there’s a force of magnitude 10 newton pushing the box from top pointing towards the ground.

In order to succeed, in Physics or in general Science or IIT JEE Mains/Advanced Exams you need to have complete understanding of Vectors and be able to draw them, multiply them etc.

That’s why below I have documented some good numericals which can be helpful for Class 11 students who are just starting out in their PCM journey.

Numerical 1

A vector has component along the X-axis equal to 25 unit and along the Y-axis equal to 60 unit.

Find the magnitude and direction of the vector.

Solution

If we’re given two vectors let’s say vector A and B then magnitude of their resultant vector will be square root of A2 + B2 + 2AB Cosθ

Let’s suppose that vector along X-axis is vector A and that along Y-axis is vector B.

A = 25 unit
B = 60 unit
θ = 90° (as angle between X-axis and Y-axis is 90°)

Thus magnitude of resultant vector will be square root of 252 + 602 + 2 × 25 × 60 × Cos90°
= 625 + 3600 + 3000 × Cos90°

Value of Cos90° = 0
= 625 + 3600 + 3000 × 0

= 625 + 3600 = 4225

Magnitude of resultant vector will be square root = √4225 = 65 unit

Similarly if we’re given two vectors say vector A and B then direction of resultant vector will be at an angle α with respect to vector B and α = Tan-1(B/A)

A = 25 unit
B = 60 unit

α = Tan-1(60/25)

Thus if there’s a 25 unit vector along X-axis and 60 unit vector along Y-axis then resultant vector will be of magnitude 65 unit and it’s direction will be Tan-1(60/25) with respect to X-axis.

Numerical 2

The sum of three vectors is shown in below diagram.
Find out magnitudes of the vectors OB and OC, if OA is resultant vector.

Solution

From the diagram it’s clear that resultant vector OA doesn’t have any component along X-axis.

Thus sum of X-components of vectors OB and OC should be zero

X-component of vector OB = OB
X-component of vector OC = OC Cos45°

OB + OC Cos45° = 0

Similarly from the digram itself, its pretty much clear that Y-component of resultant vector have total magnitude 5.

Thus sum of Y-components of vectors OB and OC should be zero

Y-component of vector OB = 0
Y-component of vector OC = OC Sin45°

⇒ 0 + OC Sin45° = 5

Replacing Sin45° = 1/√2

0 + OC/√2 = 5

OC = 5√2

Putting this value in equation OB + OC Cos45° = 0
⇒ OB + 5√2 Cos45° = 0

Replacing Cos45° = 1/√2

OB + 5√2(1/√2) = 0

OB + 5 = 0

OB = – 5

As OB is just magnitude of vector, thus it should be positive. Therefore OB = 5

So magnitudes of vectors OC, OB is 5√2 and 5 respectively.

Numerical 3

Write the unit vector in the direction of vector

\begin{equation} \text{Write the unit vector in the direction of } \vec{A} = 5\vec{i} + \vec{j} – 2\vec{k} \end{equation}

Solution

Unit vector of any vector can be calculated by just dividing given vector with it’s magnitude.

\begin{equation} \text{Unit vector of } \vec{A} = \frac{\vec{A}}{|\vec{A}|} \\ \text{If } \vec{A} = a\vec{i} + b\vec{j} + c\vec{k} \text{ then } \\ |\vec{A}| = \sqrt{a^{2} + b^{2} + c^{2}} \\ \implies \text{Unit vector of } \vec{A} = \frac{\vec{A}}{\sqrt{a^{2} + b^{2} + c^{2}}} \\ \text{So Unit vector of } \vec{A} = 5\vec{i} + \vec{j} – 2\vec{k} \\ = \frac{5\vec{i} + \vec{j} – 2\vec{k}}{\sqrt{5^{2} + 1^{2} + (-2)^{2}}} \\ = \frac{5\vec{i} + \vec{j} – 2\vec{k}}{\sqrt{25 + 1 + 4}} \\ = \frac{5\vec{i} + \vec{j} – 2\vec{k}}{\sqrt{30}} \\ = \frac{5}{\sqrt{30}}\vec{i} + \frac{1}{\sqrt{30}}\vec{j} – \frac{2}{\sqrt{30}}\vec{k} \end{equation}

Numerical 4

If |a + b| = |a – b| then show that a ⊥ b (angle between a and b is 90). Here a, b are both vectors.

Solution

Dot product of two vectors is zero if and only if they both are at an angle of 90° with each other.
Thus if |a + b| = |a – b| and both a, b are vectors then both of these vectors will be at an angle of 90° with each other.

Numerical 5

if a = 2i + 3j + 4k and b = 4i + 3j + 2k then find the angle between vectors a and b

Solution

Numerical 6

Solution

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