Vectors questions and answers may help students grasp the idea more effectively. Vectors are important in higher-level mathematical concepts and three-dimensional spaces. Students can use these questions to get a thorough summary of the topics and practice answering them to enhance their understanding. To verify your answers, check the whole explanations for each question. To know more about vectors, click here.
Vectors Definition:
Geometrical entities with both magnitude and direction are known as vectors. A vector is presented graphically with an arrow pointing in the vector’s direction, and the magnitude of the vector is determined by its length. Vectors are thus represented by arrows and have both initial and terminal points. The magnitude of the vector is the length of the line connecting points A and B, and the direction of vector AB is the orientation of the displacement from point A to point B. Spatial vectors and Euclidean vectors are other names for vectors. Vectors are used in Maths, engineering, physics, and various other professions. |
Vectors Questions with Solutions
Magnitude of Vectors
To find the magnitude of a vector, first find the sum of the squares of the vector components and finally take the square root to the obtained value. If A is the vector with the components (x, y, z), then the magnitude of vector A is found using the formula,
\(\begin{array}{l}|A| = \sqrt{x^{2}+y^{2}+z^{2}}\end{array} \)
A vector’s magnitude is a scalar value.
Also, read: Magnitude of Vectors. |
1. Determine the magnitude of the vector: \(\begin{array}{l}\vec{a}= 3\hat{i}-2\hat{j}+6\hat{k}\end{array} \)
Solution:
Given vector:
\(\begin{array}{l}\vec{a}= 3\hat{i}-2\hat{j}+6\hat{k}\end{array} \)
Here, x = 3, y = -2 and z = 6
As we know, the formula to find the magnitude of vector A is given by:
\(\begin{array}{l}|A| = \sqrt{x^{2}+y^{2}+z^{2}}\end{array} \)
Now, substitute the values of x, y and z in the formula, we get
\(\begin{array}{l}|A| = \sqrt{3^{2}+(-2)^{2}+6^{2}}\end{array} \)
\(\begin{array}{l}|A| = \sqrt{9 + 4 + 36}\end{array} \)
|A| = √49
|A| = 7
Hence, the magnitude of the given vector is 7.
2. Compute the vector’s magnitude: 5i – 4j + 2k.
Solution:
Let the given vector be A.
I.e., A = 5i – 4j + 2k
Here, the components x, y and z of vector A are 5, -4, and 2, 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
\(\begin{array}{l}|A| = \sqrt{(5)^{2}+(-4)^{2}+(2)^{2}}\end{array} \)
\(\begin{array}{l}|A| = \sqrt{25 + 16 + 4}\end{array} \)
|A| = √45
Hence, the magnitude of the vector 5i – 4j + 2k is √45.
| Angle Between Two Vectors
The dot product formula can be used to calculate the angle between two vectors. Let’s say there are two vectors a and b, and the angle between them is θ. Hence, the dot product of two vectors is:
a·b = |a||b| cosθ
Now, the value of the angle must be determined. The direction of two vectors is also indicated by the angle between them. The following formula can be used to evaluate θ:
θ = cos-1[(a·b)/|a||b|]
Also read: Angle Between Two Vectors. |
3. Calculate the angle between two vectors:
\(\begin{array}{l}\hat{i}-2\hat{j}+3\hat{k}\end{array} \)
\(\begin{array}{l}3\hat{i}-2\hat{j}+\hat{k}\end{array} \)
Solution:
Let,
\(\begin{array}{l}\vec{a} = 1\hat{i}-2\hat{j}+3\hat{k}\end{array} \)
\(\begin{array}{l}\vec{b} = 3\hat{i}-2\hat{j}+1\hat{k}\end{array} \)
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.
\(\begin{array}{l}\vec{a}.\vec{b} = (1\hat{i}-2\hat{j}+3\hat{k}).(3\hat{i}-2\hat{j}+1\hat{k})\end{array} \)
Now, take the dot product for two vectors.
= 1(3) + (-2)(-2) + 3(1)
= 3 + 4 + 3
= 10
\(\begin{array}{l}\vec{a}.\vec{b} = 10\end{array} \)
Now, find the magnitude of two vectors:
\(\begin{array}{l}|\vec{a}| = \sqrt{(1)^{2}+ (-2)^{2} + (3)^{2}}\end{array} \)
\(\begin{array}{l}|\vec{a}| = \sqrt{1+4+9}\end{array} \)
\(\begin{array}{l}|\vec{a}| = \sqrt{14}\end{array} \)
Similarly,
\(\begin{array}{l}|\vec{b}| = \sqrt{(3)^{2}+ (-2)^{2} + (1)^{2}}\end{array} \)
\(\begin{array}{l}|\vec{b}| = \sqrt{9+4+1}\end{array} \)
\(\begin{array}{l}|\vec{b}| = \sqrt{14}\end{array} \)
Now, substituting the derived values in equation (1), we get
\(\begin{array}{l}10 = \sqrt{14}\times \sqrt{14}\times cos \theta \end{array} \)
\(\begin{array}{l}10 = 14 \times cos \theta \end{array} \)
Thus,
\(\begin{array}{l}cos \theta = \frac{10}{14}\end{array} \)
Therefore,
\(\begin{array}{l}\theta = cos^{-1}\frac{5}{7}\end{array} \)
4. Calculate the projection on the vector \(\begin{array}{l}\hat{i}+ 3\hat{j}+7\hat{k}\end{array} \)
on the vector \(\begin{array}{l}7\hat{i}-\hat{j}+8\hat{k}\end{array} \)
Solution:
Let,
\(\begin{array}{l}\vec{a} = 1\hat{i} + 3\hat{j}+7\hat{k} \end{array} \)
\(\begin{array}{l}\vec{b} = 7\hat{i} -1\hat{j}+8\hat{k} \end{array} \)
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,
\(\begin{array}{l}\vec{a}.\vec{b} = (1\times 7)+(3\times -1)+(7\times 8)\end{array} \)
\(\begin{array}{l}\vec{a}.\vec{b} = 7 +(-3) + 56\end{array} \)
\(\begin{array}{l}\vec{a}.\vec{b} = 60\end{array} \)
Now, find the magnitude:
\(\begin{array}{l}|\vec{b}| = \sqrt{7^{2}+(-1)^{2}+8^{2}}\end{array} \)
\(\begin{array}{l}|\vec{b}| = \sqrt{49 + 1 + 64}\end{array} \)
\(\begin{array}{l}|\vec{b}| = \sqrt{114}\end{array} \)
Now, substitute the obtained values in the formula, we get
\(\begin{array}{l}Projection\ of\ \vec{a}\ on\ \vec{b}= \frac{1}{\sqrt{114}}(60)\end{array} \)
5. A girl walks 4 kilometers west, then 3 kilometers in a direction 30 degrees east of north, before coming to a halt. Determine the girl’s distance from her starting position.
Solution:
According to the given conditions, let O and B are the starting and final positions of the girl as shown in the below diagram.
Thus, the position of girl can be shown as follows:
\(\begin{array}{l}\vec{OA} = -4\hat{i}\end{array} \)
\(\begin{array}{l}\vec{AB} = \hat{i}|\vec{AB}|cos 60^{\circ} + \hat{j}|\vec{AB}|sin 60^{\circ}\end{array} \)
\(\begin{array}{l}\vec{AB} = \hat{i}3\times \frac{1}{2} + \hat{j}3\times \frac{\sqrt{3}}{2}\end{array} \)
\(\begin{array}{l}\vec{AB} =\frac{3}{2}\hat{i} + \frac{3\sqrt{3}}{2}\hat{j}\end{array} \)
Now, using the triangle law of vector addition, we can write:
\(\begin{array}{l}\vec{OB} = \vec{OA} + \vec{AB}\end{array} \)
Now, substitute the obtained values in the formula, we get
\(\begin{array}{l}\vec{OB} = -4\hat{i} + \left ( \frac{3}{2}\hat{i} + \frac{3\sqrt{3}}{2}\hat{j} \right )\end{array} \)
\(\begin{array}{l}\vec{OB} = \left ( -4 + \frac{3}{2} \right )\hat{i}+ \frac{3\sqrt{3}}{2}\hat{j}\end{array} \)
\(\begin{array}{l}\vec{OB} = \left ( \frac{-8+3}{2} \right )\hat{i}+ \frac{3\sqrt{3}}{2}\hat{j}\end{array} \)
\(\begin{array}{l}\vec{OB} = \frac{-5}{2}\hat{i}+ \frac{3\sqrt{3}}{2}\hat{j}\end{array} \)
Therefore, the girl’s distance from her starting position is:
\(\begin{array}{l}\frac{-5}{2}\hat{i}+ \frac{3\sqrt{3}}{2}\hat{j}\end{array} \)
6. Add the given vectors and find their sum:
\(\begin{array}{l}\vec{a}=\hat{i}-2\hat{j}+\hat{k}\end{array} \)
\(\begin{array}{l}\vec{b}=-2\hat{i}+4\hat{j}+5\hat{k}\end{array} \)
\(\begin{array}{l}\vec{c}=\hat{i}-6\hat{j}-7\hat{k}\end{array} \)
Solution:
Given vectors are:
\(\begin{array}{l}\vec{a}=\hat{i}-2\hat{j}+\hat{k}\end{array} \)
\(\begin{array}{l}\vec{b}=-2\hat{i}+4\hat{j}+5\hat{k}\end{array} \)
\(\begin{array}{l}\vec{c}=\hat{i}-6\hat{j}-7\hat{k}\end{array} \)
Finding the sum of three vectors:
\(\begin{array}{l}\vec{a}+\vec{b}+\vec{c} = (1-2+1)\hat{i} + (-2+4-6)\hat{j} + (1+5-7)\hat{k}\end{array} \)
\(\begin{array}{l}\vec{a}+\vec{b}+\vec{c} = 0\hat{i} -4\hat{j} -\hat{k}\end{array} \)
Therefore,
\(\begin{array}{l}\vec{a}+\vec{b}+\vec{c} = -4\hat{j} -\hat{k}\end{array} \)
7. Determine the unit vector in the direction of vector:
\(\begin{array}{l}\vec{a}=2\hat{i}+3\hat{j}+\hat{k}\end{array} \)
Solution:
Given vector:
\(\begin{array}{l}\vec{a}=2\hat{i}+3\hat{j}+\hat{k}\end{array} \)
Finding Magnitude of the vector:
\(\begin{array}{l}|\vec{a}| = \sqrt{2^{2} + 3^{2}+1^{2}}\end{array} \)
\(\begin{array}{l}|\vec{a}| = \sqrt{4+9+1}\end{array} \)
\(\begin{array}{l}|\vec{a}| = \sqrt{14}\end{array} \)
Thus, the formula to find the unit vector in the direction of given vector is:
Now, substituting the values in the formula, we get
\(\begin{array}{l}\hat{a} = \frac{1}{\sqrt{14}}(2\hat{i}+3\hat{j}+1\hat{k})\end{array} \)
Therefore,
\(\begin{array}{l}\hat{a} = \frac{2}{\sqrt{14}}\hat{i}+\frac{3}{\sqrt{14}}\hat{j}+\frac{1}{\sqrt{14}}\hat{k}\end{array} \)
Also, read: Unit Vectors.
8. Prove that the given point A, B, C are collinear using vector method:
A(6,−7,−1), B(2,−3,1) and C(4,−5,0).
Solution:
Given points are A(6,−7,−1), B(2,−3,1) and C(4,−5,0).
Hence,
\(\begin{array}{l}\vec{AB} = (2-6)\hat{i} + (-3+7)\hat{j} + (1+1)\hat{k} = -4\hat{i} + 4\hat{j} + 2\hat{k}\end{array} \)
\(\begin{array}{l}\vec{BC} = (4-2)\hat{i} + (-5+3)\hat{j} + (0-1)\hat{k} = 2\hat{i} -2\hat{j}-1\hat{k}\end{array} \)
\(\begin{array}{l}\vec{AC} = (4-6)\hat{i} + (-5+7)\hat{j} + (0+1)\hat{k} = -2\hat{i} +2\hat{j}+1\hat{k}\end{array} \)
Finding Magnitudes:
\(\begin{array}{l}|\vec{AB}| = \sqrt{16 + 16 + 4} = \sqrt{36} = 6\end{array} \)
\(\begin{array}{l}|\vec{BC}| = \sqrt{4 + 4 + 1} = \sqrt{9} = 3\end{array} \)
\(\begin{array}{l}|\vec{AC}| = \sqrt{4 + 4 + 1} = \sqrt{9} = 3\end{array} \)
Hence, clearly we can say that
\(\begin{array}{l}|\vec{BC}| + |\vec{AC}| = |\vec{AB}|\end{array} \)
Hence, the three points A(6,−7,−1), B(2,−3,1) and C(4,−5,0) are collinear.
9. Show that the parallelogram on the same base and between the same parallels are equal in area using the vector method.
Solution:
Let PQRS and PQR’S’ be two parallelograms on the same base PQ and between the same parallels l and m
Thus, the vector area of parallelogram PQRS is given by:
\(\begin{array}{l}PQRS = \overrightarrow{PQ}\times \overrightarrow{PS}\end{array} \)
\(\begin{array}{l}PQRS = \overrightarrow{PQ}\times (\overrightarrow{PS’} + \overrightarrow{S’S})\end{array} \)
\(\begin{array}{l}PQRS = \overrightarrow{PQ}\times \overrightarrow{PS’} + \overrightarrow{PQ}\times \overrightarrow{S’S} \end{array} \)
Since, vector AB and vector S’S are parallel, we can write:
\(\begin{array}{l}PQRS = \overrightarrow{PQ}\times \overrightarrow{PS’} + 0\end{array} \)
\(\begin{array}{l}PQRS = \overrightarrow{PQ}\times \overrightarrow{PS’}\end{array} \)
Therefore, the vector area of parallelogram PQRS = vector area of parallelogram PQR’S’.
Hence, the parallelogram on the same base and between the same parallels are equal in area using the vector method is proved.
10. Determine the vector joining the points A(2,3,0) and B(-1, -2, -4) directed from A to B.
Solution:
Given points are A(2,3,0) and B(-1, -2, -4).
Hence, the vectors joining the points P and Q is given by:
\(\begin{array}{l}\overrightarrow{AB} = (-1-2)\hat{i} + (-2-3)\hat{j}+(-4-0)\hat{k}\end{array} \)
\(\begin{array}{l}\overrightarrow{AB} = -3\hat{i} -5\hat{j}-4\hat{k}\end{array} \)
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Practice Questions
- Determine the position vector of the midpoint of the vector that joins the points P(2, 3, 4) and Q(4, 1, -2).
- Calculate the unit vector in the vector’s direction:
\(\begin{array}{l}\vec{a}=\hat{i}+hat{j}+2\hat{k}\end{array} \)
- Determine the projection on the vector
\(\begin{array}{l}2\hat{i}+ 3\hat{j}+2\hat{k}\end{array} \)
on the vector \(\begin{array}{l}\hat{i}+2\hat{j}+1\hat{k}\end{array} \)
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