Vector Algebra: Chapter 10 Links
Current page: Vector Algebra Miscellaneous NCERT Solutions. Go to other exercises and content from this chapter :
10.1 | 10.2 | 10.3 | 10.4 | MCQs
Vector Algebra Miscellaneous Exercise is a comprehensive revision of the entire chapter. The questions in this exercise combine all the important concepts studied throughout the chapter, including position vectors, direction ratios, unit vectors, section formula, scalar (dot) product, vector (cross) product, collinearity, direction cosines and applications of vectors in geometry.
\displaystyle \hat a=\frac{\vec a}{|\vec a|};\qquad \vec a\cdot\vec b=|\vec a||\vec b|\cos\theta;\qquad \vec a\times\vec b=|\vec a||\vec b|\sin\theta\,\hat n
This exercise helps you revise how different vector concepts are connected. Some questions require direct application of formulas, while others involve combining multiple ideas such as vector addition, section formula, dot product and cross product to arrive at the solution.
By solving these questions, you will strengthen your overall understanding of Vector Algebra and improve your confidence for the CBSE Class 12 Board Exams, CUET, and other competitive examinations where mixed-concept questions are frequently asked.
Key Concepts
Before solving the Miscellaneous Exercise, quickly revise the important formulas and concepts from the entire Vector Algebra chapter. Most questions in this exercise combine two or more of these concepts.
1. Position Vector and Magnitude
If a point is P(x,y,z), then its position vector is
\displaystyle \overrightarrow{OP}=x\hat i+y\hat j+z\hat k
and its magnitude is
\displaystyle |\vec r|=\sqrt{x^2+y^2+z^2}
2. Unit Vector and Direction Cosines
If \displaystyle \vec a=a_1\hat i+a_2\hat j+a_3\hat k, then
\displaystyle \hat a=\frac{\vec a}{|\vec a|},\qquad l=\frac{a_1}{|\vec a|},\qquad m=\frac{a_2}{|\vec a|},\qquad n=\frac{a_3}{|\vec a|}
where l, m and n are the direction cosines of the vector.
3. Section Formula
If a point divides the line joining vectors \vec a and \vec b in the ratio m:n, then its position vector is
Internally: \displaystyle \frac{m\vec b+n\vec a}{m+n}
Externally: \displaystyle \frac{m\vec b-n\vec a}{m-n}
4. Scalar (Dot) Product
If the angle between two vectors is \theta, then
\displaystyle \vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=a_1b_1+a_2b_2+a_3b_3
The angle between two non-zero vectors is given by
\displaystyle \cos\theta=\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}
5. Vector (Cross) Product
The vector product of two vectors is
\displaystyle \vec a\times\vec b=|\vec a||\vec b|\sin\theta\,\hat n
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\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}
6. Important Points
- The vector sum of the three sides of a triangle taken in order is \vec0.
- The diagonal of a parallelogram represents the sum of its adjacent side vectors.
- If \vec a\cdot\vec b=0, then the vectors are perpendicular (provided both are non-zero).
- If \vec a\times\vec b=\vec0, then the vectors are parallel or anti-parallel.
- Use the section formula carefully to distinguish between internal and external division.
- Always simplify vectors first before finding their magnitude, angle or unit vector.
Quick Reference Table
| Concept | Formula / Result |
|---|---|
| Position Vector | \displaystyle \overrightarrow{OP}=x\hat i+y\hat j+z\hat k |
| Magnitude of a Vector | \displaystyle |\vec a|=\sqrt{a_1^2+a_2^2+a_3^2} |
| Unit Vector | \displaystyle \hat a=\frac{\vec a}{|\vec a|} |
| Direction Cosines | \displaystyle l=\frac{a_1}{|\vec a|},\quad m=\frac{a_2}{|\vec a|},\quad n=\frac{a_3}{|\vec a|} |
| Relation Between Direction Cosines | \displaystyle l^2+m^2+n^2=1 |
| Vector Addition | \displaystyle \vec a+\vec b=(a_1+b_1)\hat i+(a_2+b_2)\hat j+(a_3+b_3)\hat k |
| Scalar Multiplication | \displaystyle \lambda\vec a=(\lambda a_1)\hat i+(\lambda a_2)\hat j+(\lambda a_3)\hat k |
| Internal Division | \displaystyle \frac{m\vec b+n\vec a}{m+n} |
| External Division | \displaystyle \frac{m\vec b-n\vec a}{m-n} |
| Scalar (Dot) Product | \displaystyle \vec a\cdot\vec b=|\vec a||\vec b|\cos\theta |
| Dot Product (Component Form) | \displaystyle \vec a\cdot\vec b=a_1b_1+a_2b_2+a_3b_3 |
| Angle Between Two Vectors | \displaystyle \cos\theta=\frac{\vec a\cdot\vec b}{|\vec a||\vec b|} |
| Projection of \vec a on \vec b | \displaystyle \frac{\vec a\cdot\vec b}{|\vec b|}=\vec a\cdot\hat b |
| Cross Product | \displaystyle \vec a\times\vec b=|\vec a||\vec b|\sin\theta\,\hat n |
| Cross Product (Component Form) | \displaystyle \vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix} |
| Area of Parallelogram | \displaystyle |\vec a\times\vec b| |
| Area of Triangle | \displaystyle \frac12|\vec a\times\vec b| |
| Perpendicular Vectors | \displaystyle \vec a\cdot\vec b=0\qquad(\vec a,\vec b\ne\vec0) |
| Parallel Vectors | \displaystyle \vec a\times\vec b=\vec0 |
| Dot Products of Unit Vectors | \displaystyle \hat i\cdot\hat i=\hat j\cdot\hat j=\hat k\cdot\hat k=1,\qquad \hat i\cdot\hat j=\hat j\cdot\hat k=\hat k\cdot\hat i=0 |
| Cross Products of Unit Vectors | \displaystyle \hat i\times\hat j=\hat k,\qquad \hat j\times\hat k=\hat i,\qquad \hat k\times\hat i=\hat j,\qquad \hat j\times\hat i=-\hat k,\qquad \hat k\times\hat j=-\hat i,\qquad \hat i\times\hat k=-\hat j,\qquad \hat i\times\hat i=\hat j\times\hat j=\hat k\times\hat k=\vec0 |
Before You Begin… Before solving a question, first identify the concept involved—position vector, section formula, unit vector, dot product, cross product, or direction cosines. Once you recognise the concept, choosing the correct formula and method becomes much easier.
Let us now solve all the NCERT questions of Vector Algebra Miscellaneous Exercise step by step.
Question 1: Vector Algebra Miscellaneous Exercise
1. Write down a unit vector in the XY-plane, making an angle of 30^\circ with the positive direction of the x -axis.
Solution:
Think First… A unit vector in the XY-plane making an angle \theta with the positive x-axis is given by \cos\theta\,\hat i+\sin\theta\,\hat j.
Here, \displaystyle \theta=30^\circ
Using the formula
\displaystyle \hat a=\cos\theta\,\hat i+\sin\theta\,\hat j
Substituting \theta=30^\circ, we get
\displaystyle \hat a=\cos30^\circ\,\hat i+\sin30^\circ\,\hat j
Using
\displaystyle \cos30^\circ=\frac{\sqrt3}{2},\qquad \sin30^\circ=\frac12
Therefore,
\displaystyle \hat a=\frac{\sqrt3}{2}\hat i+\frac12\hat j
Hence, the required unit vector is \displaystyle \frac{\sqrt3}{2}\hat i+\frac12\hat j.
Question 2: Vector Algebra Miscellaneous
2. Find the scalar components and magnitude of the vector joining the points P(x_1,y_1,z_1) and Q(x_2,y_2,z_2).
Solution:
Think First… The vector joining two points is obtained by subtracting the position vector of the initial point from that of the terminal point. Its scalar components are the coefficients of \hat i,\hat j,\hat k, and its magnitude is found using the distance formula.
The position vectors of the given points are
\displaystyle \overrightarrow{OP}=x_1\hat i+y_1\hat j+z_1\hat k,\qquad \overrightarrow{OQ}=x_2\hat i+y_2\hat j+z_2\hat k
Therefore, the vector joining P to Q is
\displaystyle \overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}
\displaystyle =(x_2-x_1)\hat i+(y_2-y_1)\hat j+(z_2-z_1)\hat k
Hence, the scalar components of the vector are
\displaystyle x_2-x_1,\qquad y_2-y_1,\qquad z_2-z_1
The magnitude of the vector is
\displaystyle |\overrightarrow{PQ}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}
Hence, the scalar components are x_2-x_1,\;y_2-y_1,\;z_2-z_1 and the magnitude is \displaystyle |\overrightarrow{PQ}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}.
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, this Miscellaneous Exercise brings together all the important concepts of Vector Algebra covered in this chapter. As you work through all the questions, you will apply ideas such as vector addition, section formula, direction cosines, dot product and cross product in a variety of geometrical and application-based problems. These questions provide excellent practice for the CBSE Board Examination and CUET, while also strengthening your foundation for Three-Dimensional Geometry. Now, let’s continue with the next question.
Question 3: Triangle law of Addition of Vectors
3. A girl walks 4 km towards west, then she walks 3 km in a direction 30^\circ east of north and stops. Determine the girl’s displacement from her initial point of departure.
Solution:
Think First… Take east as the positive x-direction and north as the positive y-direction. Resolve each displacement into its horizontal and vertical components, then add the vectors.
Take the positive x-axis towards the east and the positive y-axis towards the north. Refer to the diagram.
Let O be the girl’s initial position, A the position after walking 4 km west, and B her final position.
Thus, the required displacement is represented by \overrightarrow{OB}.
By the Triangle Law of Vector Addition,
\displaystyle \overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}\qquad ...(1)
Since OA is 4 km towards west,
\displaystyle \overrightarrow{OA}=-4\hat i
Now consider the second displacement AB. It is 3 km in a direction 30^\circ east of north. Therefore, resolve AB into its horizontal and vertical components as shown in the diagram.
In \triangle AMB, the horizontal (eastward) component, AM is
\displaystyle 3\sin30^\circ=\frac32
and, the vertical (northward) component, BM is
\displaystyle 3\cos30^\circ=\frac{3\sqrt3}{2}
Hence, \displaystyle \overrightarrow{AB}=\frac32\hat i+\frac{3\sqrt3}{2}\hat j
Substituting the values in equation (1), we get
\displaystyle \overrightarrow{OB}=-4\hat i+\frac32\hat i+\frac{3\sqrt3}{2}\hat j
\displaystyle =-\frac52\hat i+\frac{3\sqrt3}{2}\hat j
Hence, the girl’s displacement from her initial point of departure is \displaystyle -\frac52\hat i+\frac{3\sqrt3}{2}\hat j km.
Question 4: Vector Algebra Miscellaneous
4. If \vec a=\vec b+\vec c, then is it true that |\vec a|=|\vec b|+|\vec c|? Justify your answer.
Solution:
Think First… The magnitude of the sum of two vectors is not generally equal to the sum of their magnitudes. Equality holds only when the two vectors are parallel and act in the same direction.
We have, \displaystyle \vec a=\vec b+\vec c
Take P, Q and R to be the vertices of a triangle such that
\displaystyle \overrightarrow{PQ}=\vec b,\qquad \overrightarrow{QR}=\vec c,\qquad \overrightarrow{PR}=\vec a
Then, by the triangle law of vector addition,
\displaystyle \vec a=\vec b+\vec c
Also, by the triangle inequality,
\displaystyle |\vec a|<|\vec b|+|\vec c|
Hence,
\displaystyle |\vec a|\ne|\vec b|+|\vec c|
Hence, the given statement is not true in general.
Maths Better Tip… Whenever a question involves the position of points, displacement, or section formula, draw a rough sketch first. A simple diagram often makes it easier to identify the required vectors, apply the Triangle Law of Vector Addition, and avoid sign errors.
Question 5: Vector Algebra Miscellaneous
5. Find the value of x for which x(\hat i+\hat j+\hat k) is a unit vector.
Solution:
Think First… A unit vector has magnitude 1. So, first find the magnitude of the given vector and equate it to 1.
Let \displaystyle \vec a=x(\hat i+\hat j+\hat k)
Then,
\displaystyle |\vec a|=\sqrt{x^2+x^2+x^2}=\sqrt{3x^2}=\sqrt3\,|x|
Since \vec a is a unit vector,
i.e. \displaystyle |\vec a|=1
Therefore, \displaystyle \sqrt3\,|x|=1
\displaystyle |x|=\frac{1}{\sqrt3}
Hence, the required value of x is \displaystyle \pm\frac{1}{\sqrt3}.
Question 6: Vector Algebra Miscellaneous
6. Find a vector of magnitude 5 units, and parallel to the resultant of the vectors \vec a=2\hat i+3\hat j-\hat k and \vec b=\hat i-2\hat j+\hat k.
Solution:
Think First… First find the resultant of the given vectors. Then obtain its unit vector and multiply it by the required magnitude.
Given, \displaystyle \vec a=2\hat i+3\hat j-\hat k,\; \vec b=\hat i-2\hat j+\hat k
The resultant of the two vectors is
\displaystyle \vec a+\vec b=(2+1)\hat i+\bigl(3-2\bigr)\hat j+\bigl(-1+1\bigr)\hat k
\displaystyle =3\hat i+\hat j
Its magnitude is
\displaystyle |\vec a+\vec b|=\sqrt{3^2+1^2}=\sqrt{10}
Therefore, the unit vector parallel to the resultant is
\displaystyle \frac{3\hat i+\hat j}{\sqrt{10}}
Thus, a vector of magnitude 5 units parallel to the resultant is
\displaystyle 5\left(\frac{3\hat i+\hat j}{\sqrt{10}}\right)=\frac{15}{\sqrt{10}}\hat i+\frac{5}{\sqrt{10}}\hat j
\displaystyle =\frac{3\sqrt{10}}{2}\hat i+\frac{\sqrt{10}}{2}\hat j
Hence, the required vector is \displaystyle \frac{3\sqrt{10}}{2}\hat i+\frac{\sqrt{10}}{2}\hat j.
Question 7: Unit Vector
7. If \vec a=\hat i+\hat j+\hat k,\;\vec b=2\hat i-\hat j+3\hat k and \vec c=\hat i-2\hat j+\hat k, find a unit vector parallel to the vector 2\vec a-\vec b+3\vec c.
Solution:
Think First… First evaluate the given vector. Then divide it by its magnitude to obtain the required unit vector.
Given, \displaystyle \vec a=\hat i+\hat j+\hat k,\quad \vec b=2\hat i-\hat j+3\hat k,\quad \vec c=\hat i-2\hat j+\hat k
Now, \displaystyle 2\vec a-\vec b+3\vec c
\displaystyle =2(\hat i+\hat j+\hat k)-(2\hat i-\hat j+3\hat k)+3(\hat i-2\hat j+\hat k)
\displaystyle =(2\hat i+2\hat j+2\hat k)-2\hat i+\hat j-3\hat k+3\hat i-6\hat j+3\hat k=3\hat i-3\hat j+2\hat k
Its magnitude is
\displaystyle |2\vec a-\vec b+3\vec c|=\sqrt{3^2+(-3)^2+2^2}=\sqrt{22}
Therefore, the required unit vector is
\displaystyle \frac{3\hat i-3\hat j+2\hat k}{\sqrt{22}}=\frac{3}{\sqrt{22}}\hat i-\frac{3}{\sqrt{22}}\hat j+\frac{2}{\sqrt{22}}\hat k
Hence, a unit vector parallel to 2\vec a-\vec b+3\vec c is \displaystyle \frac{3}{\sqrt{22}}\hat i-\frac{3}{\sqrt{22}}\hat j+\frac{2}{\sqrt{22}}\hat k.
Maths Better Tip… Before moving to the next question, quickly verify your final answer. If you have found a unit vector, its magnitude must be 1. If you have found an angle, it must lie between 0 and \pi. A quick check often catches sign and calculation errors.
Question 8: Collinear Points Using Vectors
8. Show that the points A(1,-2,-8),\;B(5,0,-2) and C(11,3,7) are collinear, and find the ratio in which B divides AC.
Solution:
Think First… If the vectors joining two pairs of points are scalar multiples of each other, then the points are collinear. The ratio of the corresponding vectors gives the ratio in which the middle point divides the line segment.
We have
\displaystyle \overrightarrow{AB}=(5-1)\hat i+\bigl(0-(-2)\bigr)\hat j+\bigl(-2-(-8)\bigr)\hat k
\displaystyle =4\hat i+2\hat j+6\hat k
Also,
\displaystyle \overrightarrow{BC}=(11-5)\hat i+(3-0)\hat j+\bigl(7-(-2)\bigr)\hat k
\displaystyle =6\hat i+3\hat j+9\hat k
Clearly, \displaystyle \overrightarrow{BC}=\frac32\,\overrightarrow{AB}
Hence, the vectors \overrightarrow{AB} and \overrightarrow{BC} are parallel. Therefore, the points A, B and C are collinear.
Also, since \displaystyle \overrightarrow{BC}=\frac32\,\overrightarrow{AB}, their magnitudes are in the ratio
\displaystyle AB:BC=2:3
Hence, the points A, B and C are collinear and B divides AC internally in the ratio 2:3.
Question 9: Vector Algebra – Section Formula
9. Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are (2\vec a+\vec b) and (\vec a-3\vec b) externally in the ratio 1:2. Also, show that P is the midpoint of the line segment RQ.
Solution:
Think First… Since the division is external, use the external section formula. Then compare the midpoint of RQ with the given position vector of P.
We have, \displaystyle \vec{OP}=2\vec a+\vec b,\qquad \vec{OQ}=\vec a-3\vec b
R────────P───────Q 1 2
Since R divides PQ externally in the ratio 1:2, using the external section formula,
We get, \displaystyle \vec{OR}=\frac{1(\vec a-3\vec b)-2(2\vec a+\vec b)}{1-2}
\displaystyle =\frac{\vec a-3\vec b-4\vec a-2\vec b}{-1}
\displaystyle =3\vec a+5\vec b
Now, the position vector of the midpoint of RQ is
\displaystyle \frac{\vec{OR}+\vec{OQ}}{2}=\frac{(3\vec a+5\vec b)+(\vec a-3\vec b)}{2}
\displaystyle =\frac{4\vec a+2\vec b}{2}=2\vec a+\vec b
But \displaystyle \vec{OP}=2\vec a+\vec b
Therefore, P is the midpoint of the line segment RQ.
Hence, the position vector of R is \displaystyle 3\vec a+5\vec b, and P is the midpoint of RQ.
Question 10: Vector Algebra Miscellaneous
10. The two adjacent sides of a parallelogram are 2\hat i-4\hat j+5\hat k and \hat i-2\hat j-3\hat k. Find the unit vector parallel to its diagonal. Also, find its area.
Solution:
Think First… The diagonal of a parallelogram is the sum of its adjacent side vectors. The area of the parallelogram is the magnitude of their cross product.
Let
\displaystyle \vec a=2\hat i-4\hat j+5\hat k,\qquad \vec b=\hat i-2\hat j-3\hat k
The diagonal of the parallelogram is
\displaystyle \vec a+\vec b=(2+1)\hat i+(-4-2)\hat j+(5-3)\hat k
\displaystyle =3\hat i-6\hat j+2\hat k
Its magnitude is
\displaystyle |\vec a+\vec b|=\sqrt{3^2+(-6)^2+2^2}=\sqrt{49}=7
Therefore, the required unit vector is
\displaystyle \frac{1}{7}(3\hat i-6\hat j+2\hat k)
Now,
\displaystyle \vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-4&5\\1&-2&-3\end{vmatrix}
\displaystyle =22\hat i+11\hat j+0\hat k
Hence, the area of the parallelogram is
\displaystyle |\vec a\times\vec b|=\sqrt{22^2+11^2}=\sqrt{605}=11\sqrt5
Hence, the required unit vector is \displaystyle \frac{1}{7}(3\hat i-6\hat j+2\hat k) and the area of the parallelogram is \displaystyle 11\sqrt5.
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 are doing 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 11: Misc. Exercise Chapter 10
11. Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are \displaystyle \pm\left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right).
Solution:
Think First… If a vector is equally inclined to the three coordinate axes, then it makes equal angles with them. Hence, all its direction cosines are equal. Use the relation between the direction cosines to find their value.
It is given that the vector is equally inclined to the axes, so let the direction cosines of the given vector be
\displaystyle l=m=n
Using the relation
\displaystyle l^2+m^2+n^2=1
Substituting l=m=n, we get
\displaystyle 3l^2=1
\displaystyle l=\pm\frac1{\sqrt3}
Since l=m=n,
\displaystyle l=m=n=\pm\frac1{\sqrt3}
Hence, the direction cosines are \displaystyle \pm\left(\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right).
Question 12: Vector Algebra Miscellaneous
12. Let \vec a=\hat i+4\hat j+2\hat k,\;\vec b=3\hat i-2\hat j+7\hat k and \vec c=2\hat i-\hat j+4\hat k. Find a vector \vec d which is perpendicular to both \vec a and \vec b and satisfies \vec c\cdot\vec d=15.
Solution:
Think First… A vector perpendicular to two given vectors is parallel to their cross product. First find \vec a\times\vec b, then use the given condition to determine the required vector.
Let the required vector be \vec d.
Since \vec d is perpendicular to both \vec a and \vec b, it must be parallel to their vector (cross) product. Therefore,
\displaystyle \vec d=\lambda(\vec a\times\vec b)
Now, find \vec a\times\vec b.
\displaystyle \vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\1&4&2\\3&-2&7\end{vmatrix}
Expanding the determinant,
We get, \displaystyle \vec a\times\vec b=\hat i\bigl(4\cdot7-2(-2)\bigr)-\hat j\bigl(1\cdot7-2\cdot3\bigr)+\hat k\bigl(1(-2)-4\cdot3\bigr)
\displaystyle =\hat i(28+4)-\hat j(7-6)+\hat k(-2-12)
\displaystyle =32\hat i-\hat j-14\hat k
Therefore, \displaystyle \vec d=\lambda(32\hat i-\hat j-14\hat k)
Using the given condition
\displaystyle \vec c\cdot\vec d=15
Substituting \vec c=2\hat i-\hat j+4\hat k and \vec d=\lambda(32\hat i-\hat j-14\hat k),
We get, \displaystyle (2\hat i-\hat j+4\hat k)\cdot\lambda(32\hat i-\hat j-14\hat k)=15
\displaystyle \lambda(64+1-56)=15
\displaystyle 9\lambda=15 \Rightarrow \lambda=\frac53[
Hence,
\displaystyle \vec d=\frac53(32\hat i-\hat j-14\hat k)=\frac13(160\hat i-5\hat j-70\hat k)
Hence, the required vector is \displaystyle \frac13(160\hat i-5\hat j-70\hat k).
NCERT Correction! The printed NCERT answer appears to have a sign error in the k-component. The vector obtained above satisfies all the given conditions exactly.
Question 13: Vector Algebra Miscellaneous
13. The scalar product of the vector \hat i+\hat j+\hat k with a unit vector along the sum of the vectors 2\hat i+4\hat j-5\hat k and \lambda\hat i+2\hat j+3\hat k is equal to 1. Find the value of \lambda.
Solution:
Think First… First find the resultant vector, then its unit vector. Finally, use the given dot product to determine the value of \lambda.
The sum of the given vectors is
\displaystyle (2\hat i+4\hat j-5\hat k)+(\lambda\hat i+2\hat j+3\hat k)=(\lambda+2)\hat i+6\hat j-2\hat k
Let \displaystyle \vec r=(\lambda+2)\hat i+6\hat j-2\hat k
The required unit vector along \vec r is obtained using the formula
\displaystyle \hat r=\frac{\vec r}{|\vec r|}
So, \displaystyle \hat r=\frac{(\lambda+2)\hat i+6\hat j-2\hat k}{\sqrt{(\lambda+2)^2+6^2+(-2)^2}}
\displaystyle =\frac{(\lambda+2)\hat i+6\hat j-2\hat k}{\sqrt{(\lambda+2)^2+40}}
Given that the scalar product of the vector \hat i+\hat j+\hat k with this unit vector is equal to 1, we have
\displaystyle (\hat i+\hat j+\hat k)\cdot\hat r=1
Substituting the value of \hat r and using dot product, we get
\displaystyle \frac{(\lambda+2)+6-2}{\sqrt{(\lambda+2)^2+40}}=1
or \displaystyle \frac{\lambda+6}{\sqrt{(\lambda+2)^2+40}}=1
Squaring both sides,
We get, \displaystyle (\lambda+6)^2=(\lambda+2)^2+40
\displaystyle \lambda^2+12\lambda+36=\lambda^2+4\lambda+44
\displaystyle 8\lambda=8 \Rightarrow \lambda=1
Hence, the required value of \lambda is 1.
Question 14: Vector Algebra
14. If \vec a,\vec b,\vec c are mutually perpendicular vectors of equal magnitudes, show that the vector \vec a+\vec b+\vec c is equally inclined to \vec a,\vec b and \vec c.
Solution:
Think First… To show that a vector is equally inclined to three vectors, show that it makes the same angle with each of them. Use the dot product formula for the angle between two vectors.
Let \displaystyle \vec r=\vec a+\vec b+\vec c
Since the given vectors are mutually perpendicular and have equal magnitudes, let
\displaystyle |\vec a|=|\vec b|=|\vec c|=m
Also,
\displaystyle \vec a\cdot\vec b=\vec b\cdot\vec c=\vec c\cdot\vec a=0
Now, \displaystyle \vec r\cdot\vec a=(\vec a+\vec b+\vec c)\cdot\vec a
\displaystyle =\vec a\cdot\vec a+\vec b\cdot\vec a+\vec c\cdot\vec a
\displaystyle =m^2+0+0=m^2
Similarly,
\displaystyle \vec r\cdot\vec b=m^2,\quad \vec r\cdot\vec c=m^2
Also, the magnitude of \vec r is
\displaystyle |\vec r|=\sqrt{\vec r\cdot\vec r}=\sqrt{(\vec a+\vec b+\vec c)\cdot(\vec a+\vec b+\vec c)}
Expanding the dot product,
\displaystyle |\vec r|=\sqrt{\vec a\cdot\vec a+\vec b\cdot\vec b+\vec c\cdot\vec c+2\vec a\cdot\vec b+2\vec b\cdot\vec c+2\vec c\cdot\vec a}
Since \vec a,\vec b and \vec c are mutually perpendicular,
\displaystyle \vec a\cdot\vec b=\vec b\cdot\vec c=\vec c\cdot\vec a=0
Also, \displaystyle \vec a\cdot\vec a=\vec b\cdot\vec b=\vec c\cdot\vec c=m^2. Therefore,
\displaystyle |\vec r|=\sqrt{m^2+m^2+m^2}=\sqrt{3m^2}=\sqrt3\,m
Therefore,
\displaystyle \cos\alpha=\frac{\vec r\cdot\vec a}{|\vec r|\,|\vec a|}=\frac{m^2}{(\sqrt3\,m)(m)}=\frac1{\sqrt3}
Similarly,
\displaystyle \cos\beta=\frac1{\sqrt3},\qquad \cos\gamma=\frac1{\sqrt3}
Thus, \displaystyle \alpha=\beta=\gamma
Hence, the vector \vec a+\vec b+\vec c is equally inclined to \vec a,\vec b and \vec c.
NCERT Correction! The printed NCERT question appears to contain a typographical error. The correct vector is \vec a+\vec b+\vec c, not \vec c\cdot\vec d=15, since the latter is a scalar.
Maths Better Tip! Remember, \vec a\cdot\vec b=0 does not always mean that one of the vectors is the zero vector. It can also happen when two non-zero vectors are perpendicular. Always check the given information before drawing a conclusion.
Question 15: Perpendicular Vectors
15. Prove that (\vec a+\vec b)\cdot(\vec a+\vec b)=|\vec a|^2+|\vec b|^2 if and only if \vec a,\vec b are perpendicular, given \vec a\ne\vec0,\;\vec b\ne\vec0.
Solution
Think First… Since this is an “if and only if” statement, we prove both implications separately.
(⇒) Assume that (\vec a+\vec b)\cdot(\vec a+\vec b)=|\vec a|^2+|\vec b|^2
Using the distributive property of the dot product,
\displaystyle (\vec a+\vec b)\cdot(\vec a+\vec b)=\vec a\cdot\vec a+\vec a\cdot\vec b+\vec b\cdot\vec a+\vec b\cdot\vec b
Since the dot product is commutative and \vec a\cdot\vec a=|\vec a|^2,\;\vec b\cdot\vec b=|\vec b|^2,
\displaystyle (\vec a+\vec b)\cdot(\vec a+\vec b)=|\vec a|^2+2\vec a\cdot\vec b+|\vec b|^2
Using the given condition,
\displaystyle |\vec a|^2+2\vec a\cdot\vec b+|\vec b|^2=|\vec a|^2+|\vec b|^2
Subtracting |\vec a|^2+|\vec b|^2 from both sides,
\displaystyle 2\vec a\cdot\vec b=0
\displaystyle \vec a\cdot\vec b=0
Since \vec a\ne\vec0 and \vec b\ne\vec0, the vectors are perpendicular.
(⇐) Conversely, assume that \vec a\perp\vec b
Since the vectors are perpendicular,
\displaystyle \vec a\cdot\vec b=0
Again, using the distributive property of the dot product,
\displaystyle (\vec a+\vec b)\cdot(\vec a+\vec b)=\vec a\cdot\vec a+\vec a\cdot\vec b+\vec b\cdot\vec a+\vec b\cdot\vec b
Since \vec a\cdot\vec b=\vec b\cdot\vec a=0,
\displaystyle (\vec a+\vec b)\cdot(\vec a+\vec b)=|\vec a|^2+|\vec b|^2
Hence proved.
Question 16: Vector Algebra Miscellaneous – MCQ
16. If \theta is the angle between two vectors \vec a and \vec b, then \vec a\cdot\vec b\ge0 only when
- (A) 0<\theta<\dfrac{\pi}{2}
- (B) 0\le\theta\le\dfrac{\pi}{2}
- (C) 0<\theta<\pi
- (D) 0\le\theta\le\pi
Solution
Think First… The sign of the dot product depends on the value of \cos\theta, since \vec a\cdot\vec b=|\vec a||\vec b|\cos\theta.
We know that
\displaystyle \vec a\cdot\vec b=|\vec a||\vec b|\cos\theta
Since |\vec a| and |\vec b| are always positive for non-zero vectors,
\displaystyle \vec a\cdot\vec b\ge0\iff\cos\theta\ge0
Now, \cos\theta\ge0 for
\displaystyle 0\le\theta\le\frac{\pi}{2}
✅️ Hence, the correct answer is (B).
Question 17: Vector Algebra Miscellaneous – MCQ
17. Let \vec a and \vec b be two unit vectors and \theta is the angle between them. Then \vec a+\vec b is a unit vector if
- (A) \theta=\dfrac{\pi}{4}
- (B) \theta=\dfrac{\pi}{3}
- (C) \theta=\dfrac{\pi}{2}
- (D) \theta=\dfrac{2\pi}{3}
Solution
Think First… Use the identity |\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b. Since both vectors are unit vectors, substitute |\vec a|=|\vec b|=1.
Since \vec a+\vec b is a unit vector,
\displaystyle |\vec a+\vec b|=1
Squaring both sides,
\displaystyle |\vec a+\vec b|^2=1
Now,
\displaystyle |\vec a+\vec b|^2=|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b
Since |\vec a|=|\vec b|=1,
We get, \displaystyle 1=1+1+2\cos\theta
\displaystyle 2\cos\theta=-1
or \displaystyle \cos\theta=-\frac12
Therefore,
\displaystyle \theta=\frac{2\pi}{3}
✅️ Hence, the correct answer is (D).
Question 18: Vector Algebra Miscellaneous – MCQ
18. The value of \hat i\cdot(\hat j\times\hat k)+\hat j\cdot(\hat i\times\hat k)+\hat k\cdot(\hat i\times\hat j) is
- (A) 0
- (B) -1
- (C) 1
- (D) 3
Solution
Think First… Use the standard cross products of the unit vectors and then evaluate each dot product separately.
Using the identities
\displaystyle \hat j\times\hat k=\hat i,\qquad \hat i\times\hat k=-\hat j,\qquad \hat i\times\hat j=\hat k
Therefore,
\displaystyle \hat i\cdot(\hat j\times\hat k)=\hat i\cdot\hat i=1
\displaystyle \hat j\cdot(\hat i\times\hat k)=\hat j\cdot(-\hat j)=-1
and \displaystyle \hat k\cdot(\hat i\times\hat j)=\hat k\cdot\hat k=1
So, the given expression becomes
\displaystyle 1-1+1=1
✅️ Hence, the correct answer is (C).
Question 19: Vector Algebra Miscellaneous – MCQ
19. If \theta is the angle between any two vectors \vec a and \vec b, then |\vec a\cdot\vec b|=|\vec a\times\vec b| when \theta is equal to
- (A) 0
- (B) \dfrac{\pi}{4}
- (C) \dfrac{\pi}{2}
- (D) \pi
Solution
Think First… Use the formulas for the magnitudes of the dot product and the cross product, then equate them.
Given that, \displaystyle |\vec a\cdot\vec b|=|\vec a\times\vec b|
Also, we know that
\displaystyle |\vec a\cdot\vec b|=|\vec a|\,|\vec b|\,|\cos\theta| (taking modulus on both sides of \vec a\cdot\vec b=|\vec a|\,|\vec b|\cos\theta).
and,
\displaystyle |\vec a\times\vec b|=|\vec a|\,|\vec b|\,\sin\theta (since 0\le\theta\le\pi and hence \sin\theta\ge0).
Therefore,
\displaystyle |\cos\theta|=\sin\theta
For 0\le\theta\le\pi, this relation is satisfied when
\displaystyle \theta=\frac{\pi}{4}\quad\text{or}\quad\theta=\frac{3\pi}{4}
Among the given options, only \displaystyle \frac{\pi}{4} is listed.
✅️ Hence, the correct answer is (B).
Common Mistakes to Avoid
- Using the wrong vector operation: First identify whether the question requires vector addition, scalar multiplication, dot product or cross product. Applying the wrong operation leads to an incorrect solution.
- Making sign errors while working with components: Add, subtract and multiply the corresponding components carefully. Most mistakes occur due to incorrect handling of negative signs.
- Using an incorrect formula: Before solving, choose the appropriate formula for magnitude, unit vector, section formula, dot product, cross product or angle between vectors.
- Ignoring the magnitude while finding a unit vector: Always divide the vector by its magnitude. A unit vector must have magnitude 1.
- Confusing dot product and cross product: The dot product gives a scalar, whereas the cross product gives a vector. Use the correct operation according to the question.
- Making mistakes while expanding the determinant: While finding a cross product, follow the signs of the cofactors carefully. A single sign error changes the final vector completely.
- Using the wrong section formula: Read the question carefully to determine whether the point divides the line segment internally or externally, and use the corresponding formula.
- Forgetting standard properties of vectors: Use identities such as \vec a\cdot\vec a=|\vec a|^2, \vec a\cdot\vec b=\vec b\cdot\vec a and \vec a\times\vec a=\vec0 whenever applicable.
- Not verifying the required condition: After obtaining the answer, check that it satisfies the given condition, such as being a unit vector, perpendicular to another vector, or having the required magnitude or direction.
- Skipping the final answer in the required form: Express the final answer exactly as asked—in vector form, component form, ratio, magnitude, unit vector or angle.
Continue Learning
With the completion of the Miscellaneous Exercise, you have now covered all the important concepts of Vector Algebra. In this chapter, you learned how to represent vectors, perform vector operations, use the scalar and vector products, find angles and projections, determine areas, and solve problems involving section formula, direction cosines and geometric applications.
Before moving on to Three Dimensional Geometry, make sure you are comfortable with the following concepts, as they will be used frequently throughout the next chapter:
- Representing vectors in component form and finding their magnitudes and unit vectors.
- Adding, subtracting and multiplying vectors by scalars.
- Using the section formula to find the position vector of a point dividing a line segment internally or externally.
- Finding the scalar (dot) product to determine angles, projections and perpendicularity of vectors.
- Using the vector (cross) product to determine a vector perpendicular to two given vectors and to find the area of a parallelogram or triangle.
- Applying direction cosines and direction ratios to determine the direction of a vector.
- Interpreting vector equations and solving geometrical problems involving points, lines and planes.
- Choosing the correct vector formula and presenting the final answer in the required NCERT form.
These concepts form the foundation of Three Dimensional Geometry, where vectors will be used extensively to study the equations of lines and planes, angles, distances and their geometric applications.
Explore More
A strong understanding of the concepts covered in Vector Algebra forms the foundation for studying Three-Dimensional Geometry. While solving problems, first identify whether the question involves vector operations, the section formula, the scalar (dot) product, the vector (cross) product, direction cosines, or the equation of a vector. Choosing the correct approach at the beginning makes the solution much simpler.
While solving questions, perform vector calculations carefully, paying special attention to signs and component-wise operations. Use the appropriate vector identities and properties wherever required, verify that the final answer satisfies the given conditions, and present it in the form expected in NCERT. With regular practice, you will be able to solve both algebraic and geometrical problems involving vectors accurately and confidently, making the transition to Three-Dimensional Geometry smooth and enjoyable.
All the best and keep learning 👍
