Vector Algebra: Chapter 10 Links
Current page: Vector Algebra 10.2 NCERT Solutions. Go to other exercises and content from this chapter :
10.1 | 10.3 | 10.4 | Misc. | MCQs
Vector Algebra Exercise 10.2 introduces the basic operations on vectors and their representation in three dimensions. In this exercise, you will learn how to add and subtract vectors, multiply a vector by a scalar, express vectors in component form, find their magnitude and unit vectors, determine direction ratios and direction cosines, and apply the section formula to divide a line segment internally or externally.
\displaystyle \vec r=x\hat i+y\hat j+z\hat k;\; |\vec r|=\sqrt{x^2+y^2+z^2};\; \hat r=\frac{\vec r}{|\vec r|}
The main idea is to represent vectors using their components so that operations such as addition, subtraction, scalar multiplication and finding magnitudes become straightforward. You will also learn how to obtain the vector joining two points and use the section formula to locate points that divide a line segment in a given ratio.
By practising these questions, you will develop a strong understanding of vector operations and their geometric interpretation. These concepts form the foundation of Vector Algebra and are important for the CBSE board examinations as well as CUET.
Key Concepts
Before solving Exercise 10.2, it is important to understand the basic operations on vectors, their component form, magnitude, unit vectors, vector joining two points, section formula, and direction ratios and direction cosines.
1. Component Form of a Vector
Any vector in three dimensions can be written in component form as
\displaystyle \vec r=x\hat i+y\hat j+z\hat k
where x, y and z are the scalar components of the vector along the x-, y– and z-axes, respectively.
2. Magnitude and Unit Vector
If \displaystyle \vec a=a_1\hat i+a_2\hat j+a_3\hat k
then its magnitude is
\displaystyle |\vec a|=\sqrt{a_1^2+a_2^2+a_3^2}
The unit vector in the direction of \vec a is given by
\displaystyle \hat a=\frac{\vec a}{|\vec a|}
3. Addition, Subtraction and Scalar Multiplication
Vector Addition Laws
- Triangle Law: If two vectors are represented by two sides of a triangle taken in order, then their resultant is represented by the third side taken in the opposite order.
- Parallelogram Law: If two vectors are represented by two adjacent sides of a parallelogram drawn from the same point, then their resultant is represented by the diagonal passing through that point.
If \displaystyle \vec a=a_1\hat i+a_2\hat j+a_3\hat k,\qquad \vec b=b_1\hat i+b_2\hat j+b_3\hat k
then \displaystyle \vec a+\vec b=(a_1+b_1)\hat i+(a_2+b_2)\hat j+(a_3+b_3)\hat k
\displaystyle \vec a-\vec b=(a_1-b_1)\hat i+(a_2-b_2)\hat j+(a_3-b_3)\hat k
\displaystyle \lambda\vec a=(\lambda a_1)\hat i+(\lambda a_2)\hat j+(\lambda a_3)\hat k
Also, two vectors are equal if and only if their corresponding components are equal.
\displaystyle \vec a=\vec b\iff a_1=b_1,\;a_2=b_2,\;a_3=b_3
4. Scalar Multiplication Properties
- For any vectors \vec a,\vec b and scalars k,m:
- k\vec a+m\vec a=(k+m)\vec a
- k(m\vec a)=(km)\vec a
- k(\vec a+\vec b)=k\vec a+k\vec b
5. Vector Joining Two Points
If \displaystyle P(x_1,y_1,z_1) and \displaystyle Q(x_2,y_2,z_2) are any two points, then the vector directed from P to Q is
\displaystyle \overrightarrow{PQ}=(x_2-x_1)\hat i+(y_2-y_1)\hat j+(z_2-z_1)\hat k
Its magnitude (length) is given by the distance formula, i.e.
\displaystyle |\overrightarrow{PQ}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}
6. Section Formula
If a point R divides the line segment joining the position vectors \vec a and \vec b internally in the ratio m:n, then its position vector is
\displaystyle \vec r=\frac{m\vec b+n\vec a}{m+n}
If the division is external, then
\displaystyle \vec r=\frac{m\vec b-n\vec a}{m-n},\qquad (m\ne n)
Putting m=n=1 in the internal section formula gives the midpoint.
\displaystyle \vec r=\frac{\vec a+\vec b}{2}
7. Direction Ratios and Direction Cosines
For the vector
\displaystyle \vec a=a_1\hat i+a_2\hat j+a_3\hat k
the direction ratios are simply its components, namely a, b and c.
The corresponding direction cosines are
\displaystyle \left(\frac{a}{|\vec r|},\frac{b}{|\vec r|},\frac{c}{|\vec r|}\right)
8. Collinear Vectors
- Two vectors are collinear if one is a non-zero scalar multiple of the other.
- If \vec a=a_1\hat i+a_2\hat j+a_3\hat k and \vec b=b_1\hat i+b_2\hat j+b_3\hat k, then the vectors are collinear if
\displaystyle \frac{b_1}{a_1}=\frac{b_2}{a_2}=\frac{b_3}{a_3}
(where the denominators are non-zero)
9. Important Points
- Add and subtract vectors by combining their corresponding components. Multiply a vector by multiplying each component by the scalar.
- Always calculate the magnitude before finding the unit vector.
- To obtain \overrightarrow{PQ}, subtract the coordinates of P (initial point) from those of Q (terminal point).
- Use the correct section formula depending on whether the division is internal or external.
- The midpoint is obtained by taking the average of the corresponding coordinates.
- Two vectors are equal only when all their corresponding components are equal.
- Direction ratios are the components of a vector, while direction cosines are obtained by dividing each component by the magnitude.
- If l,\;m,\;n are the direction cosines of a vector, then l\hat i+m\hat j+n\hat k is the unit vector in the direction of that vector.
Quick Reference Table
| Component Form: \displaystyle \vec r=x\hat i+y\hat j+z\hat k |
| Scalar Multiplication: \displaystyle k\vec a+m\vec a=(k+m)\vec a,\quad k(m\vec a)=(km)\vec a,\quad k(\vec a+\vec b)=k\vec a+k\vec b |
| Magnitude: \displaystyle |\vec a|=\sqrt{a_1^2+a_2^2+a_3^2} |
| Unit Vector: \displaystyle \hat a=\frac{\vec a}{|\vec a|} |
| Equal Vectors: Corresponding components must be equal, i.e., \displaystyle a_1=b_1,\;a_2=b_2,\;a_3=b_3. |
| Collinear Vectors: \displaystyle \vec b=\lambda\vec a\iff\frac{b_1}{a_1}=\frac{b_2}{a_2}=\frac{b_3}{a_3} (where the denominators are non-zero). |
| Vector Joining Two Points: \displaystyle \overrightarrow{PQ}=(x_2-x_1)\hat i+(y_2-y_1)\hat j+(z_2-z_1)\hat k |
| Magnitude of \displaystyle \overrightarrow{PQ}: \displaystyle |\overrightarrow{PQ}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2} |
| Internal Section Formula: \displaystyle \vec r=\frac{m\vec b+n\vec a}{m+n} |
| External Section Formula: \displaystyle \vec r=\frac{m\vec b-n\vec a}{m-n},\quad(m\ne n) |
| Midpoint Formula: \displaystyle \vec r=\frac{\vec a+\vec b}{2} |
| Direction Ratios (DRs): For \displaystyle \vec a=a_1\hat i+a_2\hat j+a_3\hat k, the direction ratios are a_1,\;a_2,\;a_3. |
| Direction Cosines (DCs): \displaystyle \left(\frac{a_1}{|\vec a|},\frac{a_2}{|\vec a|},\frac{a_3}{|\vec a|}\right) |
Before You Begin… Don’t rush into calculations. First identify the concept involved, then choose the appropriate formula and solve the question step by step.
Let us now solve all the NCERT questions of Vector Algebra 10.2 step by step.
Question 1: Magnitude of a Vector
1. Compute the magnitude of the following vectors.
(i) \vec a=\hat i+\hat j+\hat k
(ii) \vec b=2\hat i-7\hat j-3\hat k
(iii) \vec c=\frac1{\sqrt3}\hat i+\frac1{\sqrt3}\hat j-\frac1{\sqrt3}\hat k
Solution
Think First… If \vec a=a_1\hat i+a_2\hat j+a_3\hat k, then its magnitude is |\vec a|=\sqrt{a_1^2+a_2^2+a_3^2}.
- (i)
|\vec a|=\sqrt{1^2+1^2+1^2}=\sqrt3 - (ii)
|\vec b|=\sqrt{2^2+(-7)^2+(-3)^2}=\sqrt{4+49+9}=\sqrt{62} - (iii)
|\vec c|=\sqrt{\left(\frac1{\sqrt3}\right)^2+\left(\frac1{\sqrt3}\right)^2+\left(-\frac1{\sqrt3}\right)^2}=\sqrt{\frac13+\frac13+\frac13}=\sqrt1=1
Hence,
- |\vec a|=\sqrt3
- |\vec b|=\sqrt{62}
- |\vec c|=1
Question 2: Vector Algebra 10.2
2. Write two different vectors having same magnitude.
Solution
Think First… Two vectors have the same magnitude if their lengths are equal. Their directions may be the same or different.
There are infinitely many possible answers. One such pair of vectors is:
- \vec a=\hat i+\hat j
- \vec b=\hat i-\hat j
Since
|\vec a|=\sqrt{1^2+1^2}=\sqrt2
|\vec b|=\sqrt{1^2+(-1)^2}=\sqrt2
Hence, \vec a=\hat i+\hat j and \vec b=\hat i-\hat j are two different vectors having the same magnitude.
You may already be following Maths Better for the NCERT Solutions of the previous Class 12 Maths chapters, namely:
- Matrices
- Determinants
- Relations and Functions
- Inverse Trigonometric Functions
- Continuity and Differentiability
- Application of Derivatives
- Integrals
- Application of Integrals
- Differential Equations
Likewise, Vector Algebra 10.2 introduces important concepts such as the magnitude, unit vector, direction cosines, collinear vectors and the section formula. With regular practice, you will learn to apply these concepts confidently while solving vector problems. These ideas also form the foundation for many topics in Three-Dimensional Geometry. Now, let’s proceed to the next question.
Question 3: Vector Algebra 10.2
3. Write two different vectors having same direction.
Solution
Think First… Two vectors have the same direction if one is a positive scalar multiple of the other.
There are infinitely many possible answers. One such pair of vectors is:
- \vec a=\hat i+\hat j
- \vec b=2\hat i+2\hat j
Since \vec b=2\vec a, the vectors have the same direction.
Hence, \vec a=\hat i+\hat j and \vec b=2\hat i+2\hat j are two different vectors having the same direction.
Question 4: Equal Vectors
4. Find the values of x and y so that the vectors 2\hat i+3\hat j and x\hat i+y\hat j are equal.
Solution
Think First… Two vectors are equal if their corresponding components are equal.
Given, 2\hat i+3\hat j=x\hat i+y\hat j
Comparing the corresponding components, we get
- x=2
- y=3
Hence, x=2 and y=3
Maths Better Tip… Two vectors are equal only when they have the same magnitude and the same direction. Equivalently, if the vectors are given in component form, their corresponding components must be equal. Having the same magnitude alone or the same direction alone is not sufficient for two vectors to be equal.
Question 5: Vector Joining Two Points
5. Find the scalar and vector components of the vector with initial point (2,1) and terminal point (-5,7).
Solution
Think First… The vector from point P(x_1,y_1) to Q(x_2,y_2) is obtained by subtracting the coordinates of the initial point from those of the terminal point.
Let the initial point be P(2,1) and the terminal point be Q(-5,7).
The vector \overrightarrow{PQ} is
\overrightarrow{PQ}=(-5-2)\hat i+(7-1)\hat j=-7\hat i+6\hat j
Therefore,
- Scalar components: -7,\;6
- Vector components: -7\hat i,\;6\hat j
Hence, the scalar components are -7,\;6 and the vector components are -7\hat i,\;6\hat j
Question 6: Vector Algebra 10.2
6. Find the sum of the vectors \vec a=\hat i-2\hat j+\hat k, \vec b=-2\hat i+4\hat j+5\hat k and \vec c=\hat i-6\hat j-7\hat k.
Solution
Think First… To add vectors, add their corresponding \hat i, \hat j and \hat k components separately.
We have, \vec a=\hat i-2\hat j+\hat k,\; \vec b=-2\hat i+4\hat j+5\hat k,\; \vec c=\hat i-6\hat j-7\hat k
Therefore, \vec a+\vec b+\vec c
=(1-2+1)\hat i+(-2+4-6)\hat j+(1+5-7)\hat k
=0\hat i-4\hat j-\hat k=-4\hat j-\hat k
Hence, \vec a+\vec b+\vec c=-4\hat j-\hat k
Question 7: Unit Vector
7. Find the unit vector in the direction of the vector \vec a=\hat i+\hat j+2\hat k.
Solution
Think First… A unit vector in the direction of a vector is obtained by dividing the vector by its magnitude.
Given, \vec a=\hat i+\hat j+2\hat k
The magnitude of \vec a is
|\vec a|=\sqrt{1^2+1^2+2^2}=\sqrt6
Therefore, the required unit vector:
We get, \hat a=\frac{\vec a}{|\vec a|}
=\frac{\hat i+\hat j+2\hat k}{\sqrt6}
=\frac{1}{\sqrt6}\hat i+\frac{1}{\sqrt6}\hat j+\frac{2}{\sqrt6}\hat k
By now, you have already completed the NCERT Solutions for Matrices, Determinants, Relations and Functions, Inverse Trigonometric Functions, Continuity and Differentiability, Application of Derivatives, Integrals, Application of Integrals and Differential Equations. Now, we have started Vector Algebra, where every exercise is explained with detailed, step-by-step solutions to help you build strong concepts and prepare confidently for your CBSE board examinations.
Many of these questions are also available in video format on my YouTube Channel, @MathsBetter, where each solution is explained in a simple and student-friendly manner. Now, let’s proceed to the next question.
Question 8: Vector Algebra 10.2
8. Find the unit vector in the direction of vector \overrightarrow{PQ}, where P(1,2,3) and Q(4,5,6) are the given points.
Solution
Think First… First find the vector joining the two points. Then divide it by its magnitude to obtain the required unit vector.
We have, P(1,2,3),\; Q(4,5,6)
The vector joining P to Q is
\overrightarrow{PQ}=(4-1)\hat i+(5-2)\hat j+(6-3)\hat k=3\hat i+3\hat j+3\hat k
Its magnitude is
|\overrightarrow{PQ}|=\sqrt{3^2+3^2+3^2}=\sqrt{27}=3\sqrt3
Therefore, the required unit vector is
\widehat{PQ}=\frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}=\frac{3\hat i+3\hat j+3\hat k}{3\sqrt3}=\frac{1}{\sqrt3}\hat i+\frac{1}{\sqrt3}\hat j+\frac{1}{\sqrt3}\hat k
Hence, the required unit vector is \dfrac{1}{\sqrt3}\hat i+\dfrac{1}{\sqrt3}\hat j+\dfrac{1}{\sqrt3}\hat k
Question 9: Vector Algebra 10.2
9. For given vectors \vec a=2\hat i-\hat j+2\hat k and \vec b=-\hat i+\hat j-\hat k, find the unit vector in the direction of the vector \vec a+\vec b.
Solution
Think First… First find the required vector. Then divide it by its magnitude to obtain the unit vector in its direction.
Given, \vec a=2\hat i-\hat j+2\hat k,\; \vec b=-\hat i+\hat j-\hat k
Therefore,
\vec a+\vec b=(2-1)\hat i+(-1+1)\hat j+(2-1)\hat k=\hat i+\hat k
Its magnitude is
|\vec a+\vec b|=\sqrt{1^2+0^2+1^2}=\sqrt2
Hence, the required unit vector is
\frac{\vec a+\vec b}{|\vec a+\vec b|}=\frac{\hat i+\hat k}{\sqrt2}=\frac{1}{\sqrt2}\hat i+\frac{1}{\sqrt2}\hat k
Question 10: Vector Algebra
10. Find a vector in the direction of vector 5\hat i-\hat j+2\hat k which has magnitude 8 units.
Solution
Think First… First find the unit vector in the given direction. Then multiply it by the required magnitude.
Let \vec a=5\hat i-\hat j+2\hat k
Its magnitude is
|\vec a|=\sqrt{5^2+(-1)^2+2^2}=\sqrt{30}
Therefore, the unit vector in the direction of \vec a is
\hat a=\frac{\vec a}{|\vec a|}=\frac{5\hat i-\hat j+2\hat k}{\sqrt{30}}=\frac{5}{\sqrt{30}}\hat i-\frac{1}{\sqrt{30}}\hat j+\frac{2}{\sqrt{30}}\hat k
Hence, the required vector of magnitude 8 units is
8\hat a=\frac{40}{\sqrt{30}}\hat i-\frac{8}{\sqrt{30}}\hat j+\frac{16}{\sqrt{30}}\hat k
Maths Better Tip! To find a vector of a given magnitude in the direction of another vector, first find its unit vector. Then multiply the unit vector by the required magnitude.
- The unit vector in the direction of a vector is given by \hat a=\frac{\vec a}{|\vec a|}.
- If a vector of magnitude \lambda is required in the same direction, multiply the unit vector by \lambda. Thus, the required vector is \lambda\hat a=\lambda\left(\frac{\vec a}{|\vec a|}\right).
Remember, a unit vector has magnitude 1. Multiplying it by any positive number changes only its magnitude, not its direction.
Question 11: Collinear Vectors
11. Show that the vectors 2\hat i-3\hat j+4\hat k and -4\hat i+6\hat j-8\hat k are collinear.
Solution
Think First… Two vectors are collinear if one is a scalar multiple of the other.
Let
\vec a=2\hat i-3\hat j+4\hat k,\qquad \vec b=-4\hat i+6\hat j-8\hat k
We observe that \vec b=-2\vec a
Thus, one vector is a scalar multiple of the other.
Hence, the vectors 2\hat i-3\hat j+4\hat k and -4\hat i+6\hat j-8\hat k are collinear.
Alternative Method Two vectors are collinear if their corresponding components are proportional.
For the given vectors,
\frac{-4}{2}=\frac{6}{-3}=\frac{-8}{4}=-2
Since the corresponding components are proportional, the given vectors are collinear.
Maths Better Tip! Two non-zero vectors are collinear if one is a scalar multiple of the other. In other words, there exists a scalar \lambda such that \vec b=\lambda\vec a.
- If \vec b=\lambda\vec a for some scalar \lambda, then the vectors are collinear.
- Equivalently, if \vec a=a_1\hat i+a_2\hat j+a_3\hat k and \vec b=b_1\hat i+b_2\hat j+b_3\hat k, then the vectors are collinear if their corresponding components are proportional, i.e., \displaystyle \frac{b_1}{a_1}=\frac{b_2}{a_2}=\frac{b_3}{a_3}=\lambda (where the ratios are defined).
Both methods are equivalent. In practice, comparing the corresponding components is usually the quickest way to verify whether two vectors are collinear.
Question 12: Direction Cosines of a Vector
12. Find the direction cosines of the vector \hat i+2\hat j+3\hat k.
Think First… Direction cosines are obtained by dividing each component of the vector by its magnitude.
Solution
Given vector, \vec a=\hat i+2\hat j+3\hat k
Its magnitude is
|\vec a|=\sqrt{1^2+2^2+3^2}=\sqrt{14}
Therefore, the direction cosines are
l=\frac{1}{\sqrt{14}},\qquad m=\frac{2}{\sqrt{14}},\qquad n=\frac{3}{\sqrt{14}}
Hence, the direction cosines of the given vector are \left(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\right).
Maths Better Tip! The direction cosines of a vector are the cosines of the angles it makes with the positive X-, Y– and Z-axes. They are usually denoted by l,\;m,\;n.
- If \vec a=a_1\hat i+a_2\hat j+a_3\hat k, then its direction cosines are \displaystyle l=\frac{a_1}{|\vec a|},\quad m=\frac{a_2}{|\vec a|},\quad n=\frac{a_3}{|\vec a|}.
- The direction cosines always satisfy the identity l^2+m^2+n^2=1.
A quick check is to verify that the values of l,\;m,\;n satisfy l^2+m^2+n^2=1. If they do not, recheck your calculations.
Question 13: Vector Algebra 10.2
13. Find the direction cosines of the vector joining the points A(1,2,-3) and B(-1,-2,1), directed from A to B.
Think First… First find the vector from A to B, then divide each of its components by its magnitude to obtain the direction cosines.
Solution
The vector joining A to B is
\overrightarrow{AB}=(-1-1)\hat i+(-2-2)\hat j+\left(1-(-3)\right)\hat k
\therefore\;\overrightarrow{AB}=-2\hat i-4\hat j+4\hat k
Its magnitude is
|\overrightarrow{AB}|=\sqrt{(-2)^2+(-4)^2+4^2}=\sqrt{36}=6
Hence, the direction cosines are
l=-\frac13,\qquad m=-\frac23,\qquad n=\frac23
Hence, the direction cosines of the vector joining A to B are \left(-\frac13,-\frac23,\frac23\right).
Question 14: Vector Algebra 10.2
14. Show that the vector \hat i+\hat j+\hat k is equally inclined to the axes OX,\ OY,\ OZ.
Think First… A vector is equally inclined to the coordinate axes if its three direction cosines are equal.
Solution
Given vector, \vec a=\hat i+\hat j+\hat k
Its magnitude is
|\vec a|=\sqrt{1^2+1^2+1^2}=\sqrt3
Therefore, the direction cosines are
l=\frac1{\sqrt3},\qquad m=\frac1{\sqrt3},\qquad n=\frac1{\sqrt3}
Since, l=m=n=\frac1{\sqrt3}
the vector makes equal angles with the positive x-, y- and z-axes.
Hence, the vector \hat i+\hat j+\hat k is equally inclined to the axes OX, OY and OZ.
Maths Better Tip! If a vector is equally inclined to the positive X-, Y– and Z-axes, then it makes the same angle \theta with each axis.
Since the direction cosines are equal, l=m=n=\cos\theta. Using l^2+m^2+n^2=1, we get 3\cos^2\theta=1, so \cos\theta=\frac{1}{\sqrt3} and \theta=\cos^{-1}\!\left(\frac1{\sqrt3}\right)\approx54.7^\circ, not 60^\circ (a common mistake!).
Question 15: Section Formula
15. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are \hat i+2\hat j-\hat k and -\hat i+\hat j+\hat k respectively, in the ratio 2:1 (i) internally (ii) externally.
Think First… Use the internal or external section formula according to how the point divides the line segment.
Solution
Given, \vec a=\hat i+2\hat j-\hat k,\qquad \vec b=-\hat i+\hat j+\hat k
(i) Internally
Using the internal section formula,
\displaystyle \vec r=\frac{m\vec b+n\vec a}{m+n}
We get, \displaystyle \vec r=\frac{2\vec b+\vec a}{2+1}
=\frac{2(-\hat i+\hat j+\hat k)+(\hat i+2\hat j-\hat k)}{3}
Simplifying,
=\frac{-\hat i+4\hat j+\hat k}{3}
Hence, the required position vector is
\displaystyle \vec r=-\frac13\hat i+\frac43\hat j+\frac13\hat k
(ii) Externally
Using the external section formula,
\displaystyle \vec r=\frac{m\vec b-n\vec a}{m-n},\qquad (m\ne n)
We get, \displaystyle \vec r=\frac{2\vec b-\vec a}{2-1}
=2(-\hat i+\hat j+\hat k)-(\hat i+2\hat j-\hat k)
Simplifying,
=-3\hat i+3\hat k
Hence, the required position vector is
\displaystyle \vec r=-3\hat i+3\hat k
Maths Better Tip! To remember the section formula, think of the ratio as acting crosswise. If a point R divides the line joining P and Q in the ratio m:n internally, multiply the position vector of P by n and that of Q by m, then divide by m+n. For external division, simply replace n by -n in the internal section formula.
Question 16: Mid-point Formula
16. Find the position vector of the midpoint of the vector joining the points P(2,3,4) and Q(4,1,-2).
Think First… The position vector of the midpoint is obtained by taking the average of the corresponding coordinates of the two points.
Solution
The position vectors of the given points are
\vec a=2\hat i+3\hat j+4\hat k,\qquad \vec b=4\hat i+\hat j-2\hat k
Using the midpoint formula,
We have, \displaystyle \vec r=\frac{\vec a+\vec b}{2}
=\frac{(2\hat i+3\hat j+4\hat k)+(4\hat i+\hat j-2\hat k)}{2}
=\frac{6\hat i+4\hat j+2\hat k}{2}=3\hat i+2\hat j+\hat k
Hence, the required position vector of the midpoint is 3\hat i+2\hat j+\hat k.
Question 17: Vector Algebra 10.2
17. Show that the points A,\ B,\ C with position vectors 3\hat i-4\hat j-4\hat k,\;2\hat i-\hat j+\hat k,\;\hat i-3\hat j-5\hat k, respectively, form the vertices of a right-angled triangle.
Think First… Find the lengths of the three sides. If the square of the longest side equals the sum of the squares of the other two sides, the triangle is right angled.
Solution
The coordinates of the given points are
A(3,-4,-4),\quad B(2,-1,1),\quad C(1,-3,-5)
Using the distance formula,
AB=\sqrt{(2-3)^2+(-1+4)^2+(1+4)^2}=\sqrt{35}
BC=\sqrt{(1-2)^2+(-3+1)^2+(-5-1)^2}=\sqrt{41}
CA=\sqrt{(3-1)^2+(-4+3)^2+(-4+5)^2}=\sqrt6
Now, we have
AB^2+CA^2=35+6=41=BC^2
Therefore, by Pythagoras’ theorem, the triangle is right angled at B.
Hence, the points A, B and C form the vertices of a right angled triangle.
Question 18: Vector Algebra 10.2 – MCQ
18. In the given triangle ABC, which of the following is not true?
- (A) \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec0
- (B) \overrightarrow{AB}+\overrightarrow{BC}-\overrightarrow{AC}=\vec0
- (C) \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{AC}=\vec0
- (D) \overrightarrow{AB}-\overrightarrow{CB}+\overrightarrow{CA}=\vec0
Solution
Think First… Use the triangle law of vector addition: \overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}. Rewrite each option using this identity.
From the triangle law,
\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}
- (A) becomes \overrightarrow{AC}+\overrightarrow{CA}=\overrightarrow{AC}-\overrightarrow{AC}=\vec0, which is true.
- (B) becomes \overrightarrow{AC}-\overrightarrow{AC}=\vec0, which is true.
- (C) becomes \overrightarrow{AC}+\overrightarrow{AC}=2\overrightarrow{AC}\neq\vec0 (in general), so it is not true.
- (D) Since \overrightarrow{CB}=-\overrightarrow{BC}, it becomes \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec0, which is true.
✅ Hence, the correct answer is (C).
Maths Better Note. In some editions of the NCERT textbook, option (C) appears to have a printing error. It should be \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{AC}=\vec0. With this corrected option, the answer is (C).
Question 19: Vector Algebra 10.2 – MCQ
19. If \vec a and \vec b are two collinear vectors, then which of the following are incorrect?
- (A) \vec b=\lambda\vec a, for some scalar \lambda
- (B) \vec a=\pm\vec b
- (C) The respective components of \vec a and \vec b are not proportional.
- (D) Both the vectors \vec a and \vec b have same direction, but different magnitudes.
Solution
Think First… Collinear vectors are scalar multiples of each other. They may have the same or opposite directions, and their corresponding components are proportional.
- (A) is correct because collinear vectors satisfy \vec b=\lambda\vec a for some scalar \lambda.
- (B) is incorrect because \vec a=\pm\vec b holds only when the magnitudes are equal. Collinear vectors need not have equal magnitudes.
- (C) is incorrect because the corresponding components of collinear vectors are proportional.
- (D) is incorrect because collinear vectors may have the same direction or opposite directions depending on the sign of the scalar multiple.
✅ Hence, the incorrect statements are (B), (C) and (D).
Common Mistakes to Avoid
- Confusing a vector with its magnitude: A vector has both magnitude and direction, whereas its magnitude is only a scalar. Do not interchange \vec a and |\vec a|.
- Making mistakes while finding the magnitude: Use |\vec a|=\sqrt{a_1^2+a_2^2+a_3^2} and remember to square each component correctly, especially when they are negative.
- Finding the unit vector incorrectly: A unit vector is obtained by dividing the vector by its magnitude, i.e., \hat a=\dfrac{\vec a}{|\vec a|}. Never divide the magnitude by the vector.
- Making errors while finding direction cosines: Divide each component of the vector by its magnitude. Also remember that the direction cosines satisfy l^2+m^2+n^2=1.
- Checking collinearity incorrectly: Two vectors are collinear only if one is a scalar multiple of the other, or equivalently, their corresponding components are proportional.
- Using the wrong ratio in the section formula: For internal division in the ratio m:n, use \dfrac{m\vec b+n\vec a}{m+n}. Do not interchange m and n.
- Confusing internal and external division: The formulas for internal and external division are different. Read the question carefully before applying the section formula.
- Making sign errors in vector subtraction: While finding vectors like \overrightarrow{AB}, subtract the coordinates of A from B carefully and pay special attention to negative signs.
- Not using vector identities: Apply identities such as \overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC} and \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec0 whenever required.
- Skipping the final verification: Whenever possible, verify your answer by checking the magnitude, direction, ratio, or vector identity used in the solution.
Continue Learning
In Exercise 10.2, we learned how to find the magnitude, unit vector, direction cosines and direction ratios of a vector. This exercise also covers collinear vectors, the section formula, basic vector identities and a few multiple-choice questions based on these concepts.
Before starting Exercise 10.3, revise the following concepts:
- How to represent a vector in component form as \vec a=a_1\hat i+a_2\hat j+a_3\hat k.
- How to find the magnitude of a vector using |\vec a|=\sqrt{a_1^2+a_2^2+a_3^2}.
- How to find the unit vector in the direction of a given vector using \hat a=\dfrac{\vec a}{|\vec a|}.
- How to find the direction cosines of a vector and the relation l^2+m^2+n^2=1.
- How to check whether two vectors are collinear by using scalar multiples or proportional corresponding components.
- How to apply the section formula for internal and external division of a line segment joining two points.
- Basic vector identities such as \overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC} and \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec0.
- The importance of simplifying the final answer and expressing it in the form expected in NCERT.
Explore More
A good understanding of the concepts covered in Exercise 10.2 forms the foundation for many topics in Vector Algebra and Three-Dimensional Geometry. Always begin by writing the vector in its component form, then apply the appropriate formula carefully to find its magnitude, unit vector or direction cosines.
While checking collinearity, verify that one vector is a scalar multiple of the other or that the corresponding components are proportional. For problems based on section formula, identify whether the division is internal or external before substituting the values. Finally, use basic vector identities whenever required and simplify the final answer to match the NCERT form. With regular practice, you will be able to solve vector problems accurately and confidently.
All the best and keep learning 👍
