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Vectors questions are a great tool for students trying to comprehend complex mathematical concepts and three-dimensional spaces. These questions offer a comprehensive overview of the topic and practising them will enhance your understanding. To check your answers, go through the detailed explanations for each question. For more information on vectors, click here.

1. Calculate the magnitude of the vector:

\(\begin{array}{l}\vec{a}= 4\hat{i}-3\hat{j}+7\hat{k}\end{array} \)

Solution:

Given vector:

\(\begin{array}{l}\vec{a}= 4\hat{i}-3\hat{j}+7\hat{k}\end{array} \)

Here, x = 4, y = -3 and z = 7

The formula to find the magnitude of vector A is:

\(\begin{array}{l}|A| = \sqrt{x^{2}+y^{2}+z^{2}}\end{array} \)

Substituting the values of x, y and z in the formula, we get

\(\begin{array}{l}|A| = \sqrt{4^{2}+(-3)^{2}+7^{2}}\end{array} \)

\(\begin{array}{l}|A| = \sqrt{16 + 9 + 49}\end{array} \)

|A| = √74

Hence, the magnitude of the given vector is √74.

2. Compute the magnitude of the vector: 6i – 5j + 3k.

Solution:

Let the given vector be A.

I.e., A = 6i – 5j + 3k

Here, the components x, y and z of vector A are 6, -5, and 3, respectively.

Now, substitute the values in the magnitude of a vector formula:

\(\begin{array}{l}|A| = \sqrt{x^{2}+y^{2}+z^{2}}\end{array} \)

Hence, we get

\(\begin{array}{l}|A| = \sqrt{(6)^{2}+(-5)^{2}+(3)^{2}}\end{array} \)

\(\begin{array}{l}|A| = \sqrt{36 + 25 + 9}\end{array} \)

|A| = √70

Hence, the magnitude of the vector 6i – 5j + 3k is √70.

The dot product formula is used to calculate the angle between two vectors. If a and b are two vectors and θ is the angle between them, the dot product of two vectors is:

a·b = |a||b| cosθ

To find the value of the angle, we use the following formula:

θ = cos -1 [(a·b)/|a||b|]

Also read: Angle Between Two Vectors .

3. Calculate the angle between two vectors:

\(\begin{array}{l}\hat{i}-3\hat{j}+4\hat{k}\end{array} \)

\(\begin{array}{l}4\hat{i}-3\hat{j}+\hat{k}\end{array} \)

Solution:

Let,

\(\begin{array}{l}\vec{a} = 1\hat{i}-3\hat{j}+4\hat{k}\end{array} \)

\(\begin{array}{l}\vec{b} = 4\hat{i}-3\hat{j}+1\hat{k}\end{array} \)

We know that the formula to find the angle between two vectors is given by:

\(\begin{array}{l}\vec{a}.\vec{b} = |\vec{a}||\vec{b}|cos \theta \end{array} \)

Let the above equation be (1)

Here, θ is the angle between two vectors.

\(\begin{array}{l}\vec{a}.\vec{b} = (1\hat{i}-3\hat{j}+4\hat{k}).(4\hat{i}-3\hat{j}+1\hat{k})\end{array} \)

Now, take the dot product for two vectors.

= 1(4) + (-3)(-3) + 4(1)

= 4 + 9 + 4

= 17

\(\begin{array}{l}\vec{a}.\vec{b} = 17\end{array} \)

Now, find the magnitude of two vectors:

\(\begin{array}{l}|\vec{a}| = \sqrt{(1)^{2}+ (-3)^{2} + (4)^{2}}\end{array} \)

\(\begin{array}{l}|\vec{a}| = \sqrt{1+9+16}\end{array} \)

\(\begin{array}{l}|\vec{a}| = \sqrt{26}\end{array} \)

Similarly,

\(\begin{array}{l}|\vec{b}| = \sqrt{(4)^{2}+ (-3)^{2} + (1)^{2}}\end{array} \)

\(\begin{array}{l}|\vec{b}| = \sqrt{16+9+1}\end{array} \)

\(\begin{array}{l}|\vec{b}| = \sqrt{26}\end{array} \)

Now, substituting the derived values in equation (1), we get

\(\begin{array}{l}17 = \sqrt{26}\times \sqrt{26}\times cos \theta \end{array} \)

\(\begin{array}{l}17 = 26 \times cos \theta \end{array} \)

Thus,

\(\begin{array}{l}cos \theta = \frac{17}{26}\end{array} \)

Therefore,

\(\begin{array}{l}\theta = cos^{-1}\frac{17}{26}\end{array} \)

4. Calculate the projection on the vector

\(\begin{array}{l}\hat{i}+ 4\hat{j}+8\hat{k}\end{array} \)

on the vector

\(\begin{array}{l}8\hat{i}-\hat{j}+9\hat{k}\end{array} \)

Solution:

Let,

\(\begin{array}{l}\vec{a} = 1\hat{i} + 4\hat{j}+8\hat{k} \end{array} \)

\(\begin{array}{l}\vec{b} = 8\hat{i} -1\hat{j}+9\hat{k} \end{array} \)

The formula to find the projection of vector on the other vector is given by:

\(\begin{array}{l}Projection\ of\ \vec{a}\ on\ \vec{b}= \frac{1}{|\vec{b}|}(\vec{a}.\vec{b})\end{array} \)

Thus,

\(\begin{array}{l}\vec{a}.\vec{b} = (1\times 8)+(4\times -1)+(8\times 9)\end{array} \)

\(\begin{array}{l}\vec{a}.\vec{b} = 8 +(-4) + 72\end{array} \)

\(\begin{array}{l}\vec{a}.\vec{b} = 76\end{array} \)

Now, find the magnitude:

\(\begin{array}{l}|\vec{b}| = \sqrt{8^{2}+(-1)^{2}+9^{2}}\end{array} \)

\(\begin{array}{l}|\vec{b}| = \sqrt{64 + 1 + 81}\end{array} \)

\(\begin{array}{l}|\vec{b}| = \sqrt{146}\end{array} \)

Now, substitute the obtained values in the formula, we get

\(\begin{array}{l}Projection\ of\ \vec{a}\ on\ \vec{b}= \frac{1}{\sqrt{146}}(76)\end{array} \)

5. A boy walks 5 kilometers east, then 4 kilometers in a direction 40 degrees west of north, before coming to a halt. Determine the boy’s distance from his starting position.

Solution:

Let's say O and B are the starting and final positions of the boy as shown in the diagram below.

The boy's position can be represented as follows:

\(\begin{array}{l}\vec{OA} = -5\hat{i}\end{array} \)

\(\begin{array}{l}\vec{AB} = \hat{i}|\vec{AB}|cos 40^{\circ} + \hat{j}|\vec{AB}|sin 40^{\circ}\end{array} \)

\(\begin{array}{l}\vec{AB} = \hat{i}4\times \frac{4}{5} + \hat{j}4\times \frac{3}{5}\end{array} \)

\(\begin{array}{l}\vec{AB} =\frac{16}{5}\hat{i} + \frac{12}{5}\hat{j}\end{array} \)

Using the triangle law of vector addition, we can write:

\(\begin{array}{l}\vec{OB} = \vec{OA} + \vec{AB}\end{array} \)

Substituting the obtained values in the formula, we get

\(\begin{array}{l}\vec{OB} = -5\hat{i} + \left ( \frac{16}{5}\hat{i} + \frac{12}{5}\hat{j} \right )\end{array} \)

\(\begin{array}{l}\vec{OB} = \left ( -5 + \frac{16}{5} \right )\hat{i}+ \frac{12}{5}\hat{j}\end{array} \)

\(\begin{array}{l}\vec{OB} = \left ( \frac{-25+16}{5} \right )\hat{i}+ \frac{12}{5}\hat{j}\end{array} \)

\(\begin{array}{l}\vec{OB} = \frac{-9}{5}\hat{i}+ \frac{12}{5}\hat{j}\end{array} \)

Therefore, the boy’s distance from his starting position is:

\(\begin{array}{l}\frac{-9}{5}\hat{i}+ \frac{12}{5}\hat{j}\end{array} \)

6. Add the given vectors and find their sum:

\(\begin{array}{l}\vec{a}=\hat{i}-3\hat{j}+2\hat{k}\end{array} \)

\(\begin{array}{l}\vec{b}=-3\hat{i}+5\hat{j}+6\hat{k}\end{array} \)

\(\begin{array}{l}\vec{c}=\hat{i}-7\hat{j}-8\hat{k}\end{array} \)

Solution:

Given vectors are:

\(\begin{array}{l}\vec{a}=\hat{i}-3\hat{j}+2\hat{k}\end{array} \)

\(\begin{array}{l}\vec{b}=-3\hat{i}+5\hat{j}+6\hat{k}\end{array} \)

\(\begin{array}{l}\vec{c}=\hat{i}-7\hat{j}-8\hat{k}\end{array} \)

Finding the sum of three vectors:

\(\begin{array}{l}\vec{a}+\vec{b}+\vec{c} = (1-3+1)\hat{i} + (-3+5-7)\hat{j} + (2+6-8)\hat{k}\end{array} \)

\(\begin{array}{l}\vec{a}+\vec{b}+\vec{c} = -1\hat{i} -5\hat{j} -\hat{k}\end{array} \)

Therefore,

\(\begin{array}{l}\vec{a}+\vec{b}+\vec{c} = -\hat{i} -5\hat{j} -\hat{k}\end{array} \)

Also, read : Unit Vectors .

7. Determine the unit vector in the direction of vector:

\(\begin{array}{l}\vec{a}=3\hat{i}+4\hat{j}+\hat{k}\end{array} \)

Solution:

Given vector:

\(\begin{array}{l}\vec{a}=3\hat{i}+4\hat{j}+\hat{k}\end{array} \)

Finding Magnitude of the vector:

\(\begin{array}{l}|\vec{a}| = \sqrt{3^{2} + 4^{2}+1^{2}}\end{array} \)

\(\begin{array}{l}|\vec{a}| = \sqrt{9+16+1}\end{array} \)

\(\begin{array}{l}|\vec{a}| = \sqrt{26}\end{array} \)

Thus, the formula to find the unit vector in the direction of given vector is:

Now, substituting the values in the formula, we get

\(\begin{array}{l}\hat{a} = \frac{1}{\sqrt{26}}(3\hat{i}+4\hat{j}+1\hat{k})\end{array} \)

Therefore,

\(\begin{array}{l}\hat{a} = \frac{3}{\sqrt{26}}\hat{i}+\frac{4}{\sqrt{26}}\hat{j}+\frac{1}{\sqrt{26}}\hat{k}\end{array} \)

8. Prove that the given point A, B, C are collinear using vector method:

A(7,−8,−2), B(3,−4,2) and C(5,−6,1).

Solution:

Given points are A(7,−8,−2), B(3,−4,2) and C(5,−6,1).

Hence,

\(\begin{array}{l}\vec{AB} = (3-7)\hat{i} + (-4+8)\hat{j} + (2+2)\hat{k} = -4\hat{i} + 4\hat{j} + 4\hat{k}\end{array} \)

\(\begin{array}{l}\vec{BC} = (5-3)\hat{i} + (-6+4)\hat{j} + (1-2)\hat{k} = 2\hat{i} -2\hat{j}-1\hat{k}\end{array} \)

\(\begin{array}{l}\vec{AC} = (5-7)\hat{i} + (-6+8)\hat{j} + (1+2)\hat{k} = -2\hat{i} +2\hat{j}+3\hat{k}\end{array} \)

Finding Magnitudes:

\(\begin{array}{l}|\vec{AB}| = \sqrt{16 + 16 + 16} = \sqrt{48} = 4\sqrt{3}\end{array} \)

\(\begin{array}{l}|\vec{BC}| = \sqrt{4 + 4 + 1} = \sqrt{9} = 3\end{array} \)

\(\begin{array}{l}|\vec{AC}| = \sqrt{4 + 4 + 9} = \sqrt{17}\end{array} \)

Hence, clearly we can say that

\(\begin{array}{l}|\vec{BC}| + |\vec{AC}| = |\vec{AB}|\end{array} \)

Hence, the three points A(7,−8,−2), B(3,−4,2) and C(5,−6,1) are collinear.

9. Determine the vector joining the points A(3,4,1) and B(-2, -3, -5) directed from A to B.

Solution:

Given points are A(3,4,1) and B(-2, -3, -5).

Hence, the vectors joining the points P and Q is given by:

\(\begin{array}{l}\overrightarrow{AB} = (-2-3)\hat{i} + (-3-4)\hat{j}+(-5-1)\hat{k}\end{array} \)

\(\begin{array}{l}\overrightarrow{AB} = -5\hat{i} -7\hat{j}-6\hat{k}\end{array} \)

Practice Questions

1. Find the position vector of the midpoint of the vector that joins the points P(3, 4, 5) and Q(5, 2, -3).
2. Calculate the unit vector in the direction of the vector:
   \(\begin{array}{l}\vec{a}=\hat{i}+\hat{j}+3\hat{k}\end{array} \)
3. Determine the projection on the vector
   \(\begin{array}{l}3\hat{i}+ 4\hat{j}+3\hat{k}\end{array} \)
   on the vector
   \(\begin{array}{l}2\hat{i}+3\hat{j}+2\hat{k}\end{array} \)

Key Properties of Vectors:

1. Equal Vectors
   Two vectors are equal if they have the same length (magnitude) and point in the same direction, even if they start from different points.
    
2. Zero Vector
   A vector with no length and no direction. It is written as 0.It acts like adding nothing to a vector.
3. Negative Vector
   The negative of a vector has the same length but opposite direction.
4. Addition of Vectors
   You can add two vectors using the head-to-tail method or by adding their components.
5. Commutative Property
   A + B = B + A
   (You can add vectors in any order.)
6. Associative Property
   (A + B) + C = A + (B + C)
   (Grouping doesn’t affect the result.)
7. Scalar Multiplication
   Multiplying a vector by a number (scalar) changes its length, but not its direction (unless multiplied by a negative number, which reverses the direction).

Applications of Vectors

Vectors are very useful in many areas of life and science because they help describe things that have both magnitude (size) and direction. Let's look at some common and practical applications of vectors:

1. Physics and Engineering

In physics, vectors are used to describe quantities like force, velocity, acceleration, and displacement. For example, when you push an object, the force you apply has both a direction and strength, which is exactly what a vector represents. Engineers also use vectors to design bridges, machines, and buildings by analyzing forces acting on different parts.

2. Navigation and Maps

Vectors help in navigation for ships, airplanes, and even in GPS systems. They are used to show direction and distance. If a plane needs to travel from one city to another, its path and speed can be shown using a vector.

3. Computer Graphics and Animation

In computer games and animations, vectors are used to move characters and objects smoothly. They help in showing motion, rotation, and direction of different elements on the screen.

4. Robotics

Robots use vectors to calculate movement and direction. Whether it's a robotic arm or an automated vehicle, vectors help them move precisely in the right direction.

5. Sports and Mechanics

In sports like football or basketball, players and coaches use vector ideas (even without realizing it) to plan movement, passes, and shots. In mechanics, vectors explain how forces act on machines and tools.

6. Astronomy and Space Science

Scientists use vectors to track the motion of planets, satellites, and rockets. The direction and speed of their movement are studied using vector mathematics.

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