Vector Algebra: Chapter 10 Links
Current page: Vector Algebra 10.4 NCERT Solutions. Go to other exercises and content from this chapter :
10.1 | 10.2 | 10.3 | Misc. | MCQs
Vector Algebra 10.4 introduces the vector (cross) product of two vectors and its geometric applications. In this exercise, you will learn how to find the cross product of two vectors, determine a vector perpendicular to given vectors, calculate the area of a triangle or parallelogram, and evaluate cross products using the determinant method.
\displaystyle \vec a\times\vec b=|\vec a||\vec b|\sin\theta\,\hat n;\;\left|\vec a\times\vec b\right|=|\vec a||\vec b|\sin\theta;\;\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}
The main idea is to understand that the cross product produces a vector perpendicular to both given vectors, with its direction determined by the right-hand thumb rule. You will also learn how to use the determinant method to evaluate cross products and apply them to find areas and unit vectors.
By practising these questions, you will develop a clear understanding of the properties and applications of the vector product. These concepts form an important part of Vector Algebra and are frequently tested in the CBSE board examinations as well as CUET.
Key Concepts
Before solving Exercise 10.4, it is important to understand the vector (cross) product of two vectors, its direction, properties, determinant method, and its applications in finding perpendicular vectors and areas.
1. Vector (Cross) Product of Two Vectors
If θ is the angle between two non-zero vectors \vec a and \vec b, then their vector (cross) product is defined as
\displaystyle \vec a\times\vec b=|\vec a||\vec b|\sin\theta\,\hat n
where \hat n is a unit vector perpendicular to both \vec a and \vec b, whose direction is determined by the right-hand thumb rule.
2. Magnitude of the Cross Product
The magnitude of the cross product is
\displaystyle |\vec a\times\vec b|=|\vec a||\vec b|\sin\theta
This formula is frequently used to find the area of a triangle or a parallelogram.
3. Cross Product Using Components
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}
This determinant method is used in most numerical problems.
4. Special Cases of Cross Product
- If two vectors are parallel or anti-parallel, then \vec a\times\vec b=\vec0.
- If two vectors are perpendicular, then \displaystyle |\vec a\times\vec b|=|\vec a||\vec b|.
- \vec a\times\vec a=\vec0.
- \vec a\times(-\vec a)=\vec0.
- \hat i\times\hat j=\hat k,\qquad \hat j\times\hat k=\hat i,\qquad \hat k\times\hat i=\hat j.
- \hat j\times\hat i=-\hat k,\qquad \hat k\times\hat j=-\hat i,\qquad \hat i\times\hat k=-\hat j.
5. Properties of Vector Product
- Not commutative: \displaystyle \vec a\times\vec b=-\,\vec b\times\vec a
- Distributive over addition: \displaystyle \vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c
- Scalar multiplication: \displaystyle (\lambda\vec a)\times\vec b=\lambda(\vec a\times\vec b)=\vec a\times(\lambda\vec b)
6. Area Using Cross Product
If \vec a and \vec b are adjacent sides of a parallelogram, then
\displaystyle \text{Area of parallelogram}=|\vec a\times\vec b|
and
\displaystyle \text{Area of triangle}=\frac12|\vec a\times\vec b|
7. Unit Vector Perpendicular to Two Vectors
A unit vector perpendicular to both \vec a and \vec b is
\displaystyle \pm\frac{\vec a\times\vec b}{|\vec a\times\vec b|}
The two signs represent the two opposite perpendicular directions.
8. Important Points
- The cross product of two vectors is always a vector.
- Its direction is determined by the right-hand thumb rule.
- Use the determinant method whenever vectors are given in i, j, k form.
- Remember that reversing the order of vectors changes the sign of the cross product.
- Use \frac12|\vec a\times\vec b| for the area of a triangle and |\vec a\times\vec b| for the area of a parallelogram.
- To find a unit vector perpendicular to two vectors, first find their cross product and then divide by its magnitude.
Quick Reference Table
| Vector (Cross) Product: \displaystyle \vec a\times\vec b=|\vec a||\vec b|\sin\theta\,\hat n |
| Magnitude of Cross Product: \displaystyle |\vec a\times\vec b|=|\vec a||\vec b|\sin\theta |
| Cross Product in 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} |
| Angle Between Two Vectors: \displaystyle \sin\theta=\frac{|\vec a\times\vec b|}{|\vec a||\vec b|},\qquad \theta=\sin^{-1}\!\left(\frac{|\vec a\times\vec b|}{|\vec a||\vec b|}\right) |
| 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 |
| Reverse Order: \displaystyle \hat j\times\hat i=-\hat k,\qquad \hat k\times\hat j=-\hat i,\qquad \hat i\times\hat k=-\hat j |
| Cross Product with Itself: \displaystyle \vec a\times\vec a=\vec0,\qquad \hat i\times\hat i=\hat j\times\hat j=\hat k\times\hat k=\vec0 |
| Vector Product Properties: \displaystyle \vec a\times\vec b=-\vec b\times\vec a,\quad \vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c,\quad (\lambda\vec a)\times\vec b=\lambda(\vec a\times\vec b) |
| Area of a Triangle: \displaystyle \frac12|\vec a\times\vec b| |
| Area of a Parallelogram: \displaystyle |\vec a\times\vec b| |
| Unit Vector Perpendicular to \displaystyle \vec a and \displaystyle \vec b: \displaystyle \pm\frac{\vec a\times\vec b}{|\vec a\times\vec b|} |
Before You Begin… Before starting a question, identify whether you need to find the cross product, a unit vector perpendicular to two vectors, or the area of a triangle or parallelogram. Also, pay attention to the order of the vectors, as reversing the order changes the direction of the cross product.
Let us now solve all the NCERT questions of Vector Algebra 10.4 step by step.
Question 1: Vector (Cross) Product of Two Vectors
1. Find |\vec a\times\vec b|, if \vec a=\hat i-7\hat j+7\hat k and \vec b=3\hat i-2\hat j+2\hat k.
Solution
Think First… When vectors are given in component form, first find their cross product using the determinant method. Then calculate its magnitude.
We have,
\displaystyle \vec a=\hat i-7\hat j+7\hat k,\qquad \vec b=3\hat i-2\hat j+2\hat k
Using the determinant formula,
\displaystyle \vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-7&7\\3&-2&2\end{vmatrix}
Expanding the determinant along first row,
We get, \displaystyle \vec a\times\vec b=\hat i\bigl((-7)(2)-7(-2)\bigr)-\hat j\bigl((1)(2)-7(3)\bigr)+\hat k\bigl((1)(-2)-(-7)(3)\bigr)
\displaystyle =\hat i(-14+14)-\hat j(2-21)+\hat k(-2+21)
\displaystyle =19\hat j+19\hat k
Therefore,
\displaystyle |\vec a\times\vec b|=\sqrt{0^2+19^2+19^2}=\sqrt{722}=19\sqrt2
Hence, \displaystyle |\vec a\times\vec b|=\mathbf{19\sqrt2}.
Question 2: Vector Algebra 10.4
2. Find a unit vector perpendicular to each of the vectors \vec a+\vec b and \vec a-\vec b, where \vec a=3\hat i+2\hat j+2\hat k and \vec b=\hat i+2\hat j-2\hat k
Solution
Think First… A vector perpendicular to two given vectors is obtained by taking their cross product. Divide this vector by its magnitude to get the required unit vector.
We have,
\displaystyle \vec a=3\hat i+2\hat j+2\hat k,\qquad \vec b=\hat i+2\hat j-2\hat k
First, find the vectors
\displaystyle \vec a+\vec b=4\hat i+4\hat j
and \displaystyle \vec a-\vec b=2\hat i+4\hat k
A vector perpendicular to both, is
\displaystyle (\vec a+\vec b)\times(\vec a-\vec b)=\begin{vmatrix}\hat i&\hat j&\hat k\\4&4&0\\2&0&4\end{vmatrix}
Expanding the determinant along first row,
\displaystyle =\hat i(16)-\hat j(16)+\hat k(-8)=16\hat i-16\hat j-8\hat k
Its magnitude is
\displaystyle |16\hat i-16\hat j-8\hat k|=\sqrt{16^2+(-16)^2+(-8)^2}=\sqrt{576}=24
Therefore, the required unit vectors are
\displaystyle \pm\frac{16\hat i-16\hat j-8\hat k}{24}
Dividing each component by 24, we get
\displaystyle \pm\left(\frac23\hat i-\frac23\hat j-\frac13\hat k\right)
Hence, the required unit vectors are \displaystyle \pm\left(\frac23\hat i-\frac23\hat j-\frac13\hat k\right).
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 10.4 focuses on the vector (cross) product and its geometric applications. So far, you have learned how to find the cross product of two vectors and determine a unit vector perpendicular to two given vectors. In the remaining questions, you will apply these concepts to prove vector identities, solve problems involving parallel and perpendicular vectors, and calculate the areas of triangles and parallelograms. These ideas are frequently tested in the CBSE Board Examination, CUET, and form an important foundation for Three-Dimensional Geometry. Now, let’s proceed to the next question.
Question 3: Vector Algebra 10.4
3. If a unit vector \vec a makes angles \frac{\pi}{3} with \hat i, \frac{\pi}{4} with \hat j and an acute angle \theta with \hat k. Find \theta and hence, the components of \vec a.
Solution
Think First… The components of a unit vector are its direction cosines. Use the identity \displaystyle \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1 to find the unknown angle.
Let the given angles be,
\displaystyle \alpha=\frac{\pi}{3},\qquad \beta=\frac{\pi}{4},\qquad \gamma=\theta
Since \vec a is a unit vector, its direction cosines satisfy
\displaystyle \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1
Substituting the given values,
\displaystyle \cos^2\frac{\pi}{3}+\cos^2\frac{\pi}{4}+\cos^2\theta=1
We get, \displaystyle \left(\frac12\right)^2+\left(\frac{1}{\sqrt2}\right)^2+\cos^2\theta=1
\displaystyle \frac14+\frac12+\cos^2\theta=1
\displaystyle \cos^2\theta=\frac14
Since \theta is acute,
\displaystyle \therefore \cos\theta=\frac12 \Rightarrow \theta=\frac{\pi}{3}
Also, the components of the unit vector are
\displaystyle \vec a=\cos\alpha\,\hat i+\cos\beta\,\hat j+\cos\gamma\,\hat k
\displaystyle =\frac12\hat i+\frac{1}{\sqrt2}\hat j+\frac12\hat k
Hence, \displaystyle \theta=\frac{\pi}{3} and the components of \vec a are \displaystyle \frac12,\;\frac{1}{\sqrt2},\;\frac12.
Question 4: Vector Algebra 10.4
4. Show that
\displaystyle (\vec a-\vec b)\times(\vec a+\vec b)=2(\vec a\times\vec b)
Solution
Think First… Expand the left-hand side using the distributive property of the vector product. Then use \displaystyle \vec a\times\vec a=\vec0,\;\vec b\times\vec b=\vec0,\;\vec b\times\vec a=-\,\vec a\times\vec b.
Consider the left-hand side,
\displaystyle (\vec a-\vec b)\times(\vec a+\vec b)
Using the distributive property of vector product, we get
\displaystyle L.H.S.=\vec a\times\vec a+\vec a\times\vec b-\vec b\times\vec a-\vec b\times\vec b\qquad ...(1)
Now, we know
- \displaystyle \vec a\times\vec a=\vec0
- \displaystyle \vec b\times\vec b=\vec0
- \displaystyle \vec b\times\vec a=-\,\vec a\times\vec b
Therefore, equation (1) becomes
\displaystyle (\vec a-\vec b)\times(\vec a+\vec b)=\vec0+\vec a\times\vec b+\vec a\times\vec b-\vec0
\displaystyle =2(\vec a\times\vec b)
Hence,
\displaystyle (\vec a-\vec b)\times(\vec a+\vec b)=2(\vec a\times\vec b)
Maths Better Tip… The vector product is distributive over addition but not commutative. Whenever you expand expressions involving cross products, remember that \vec b\times\vec a=-\,\vec a\times\vec b and \vec a\times\vec a=\vec0. These two properties simplify most vector identities in just a few steps.
Question 5: Vector Algebra 10.4
5. Find \lambda and \mu if (2\hat i+6\hat j+27\hat k)\times(\hat i+\lambda\hat j+\mu\hat k)=\vec0.
Solution
Think First… If the cross product of two vectors is zero, then the vectors are parallel. Equate the corresponding components to find the unknowns.
We have,
\displaystyle (2\hat i+6\hat j+27\hat k)\times(\hat i+\lambda\hat j+\mu\hat k)=\vec0
Since the cross product is zero, the two vectors are parallel.
Therefore, the corresponding components are proportional
\displaystyle \frac{2}{1}=\frac{6}{\lambda}=\frac{27}{\mu}
Thus, \displaystyle \frac{6}{\lambda}=2 \Rightarrow \lambda=3
Also, \displaystyle \frac{27}{\mu}=2 \Rightarrow \mu=\frac{27}{2}
Hence, \displaystyle \lambda=3,\qquad \mu=\frac{27}{2}
Alternate Method
Using the determinant formula,
\displaystyle (2\hat i+6\hat j+27\hat k)\times(\hat i+\lambda\hat j+\mu\hat k)=\begin{vmatrix}\hat i&\hat j&\hat k\\2&6&27\\1&\lambda&\mu\end{vmatrix}=\vec0
Expanding the determinant along first row,
\displaystyle \hat i(6\mu-27\lambda)-\hat j(2\mu-27)+\hat k(2\lambda-6)=\vec0
Equating the corresponding components,
\displaystyle 6\mu-27\lambda=0,\qquad 2\mu-27=0,\qquad 2\lambda-6=0
From the last two equations,
\displaystyle \lambda=3,\qquad \mu=\frac{27}{2}
These values also satisfy the first equation since
\displaystyle 6\left(\frac{27}{2}\right)-27(3)=81-81=0
Hence, \displaystyle \lambda=3,\qquad \mu=\frac{27}{2}
Question 6: Vector Algebra
6. Given that \vec a\cdot\vec b=0 and \vec a\times\vec b=\vec0. What can you conclude about the vectors \vec a and \vec b?
Solution
Think First… The dot product of two non-zero vectors is zero only if they are perpendicular, whereas their cross product is zero only if they are parallel. A pair of non-zero vectors cannot be both perpendicular and parallel at the same time.
Given,
\displaystyle \vec a\cdot\vec b=0\qquad\text{and}\qquad \vec a\times\vec b=\vec0
Since
- \displaystyle \vec a\cdot\vec b=0 implies that \vec a and \vec b are perpendicular (if both are non-zero).
- \displaystyle \vec a\times\vec b=\vec0 implies that \vec a and \vec b are parallel (or one of them is the zero vector).
A pair of non-zero vectors cannot be both perpendicular and parallel simultaneously.
Therefore, at least one of the vectors must be the zero vector.
Hence, either \displaystyle |\vec a|=0 or \displaystyle |\vec b|=0.
Maths Better Tip… Remember the two important conditions: \vec a\cdot\vec b=0 indicates that the vectors are perpendicular (provided both are non-zero), whereas \vec a\times\vec b=\vec0 indicates that the vectors are parallel (or one of them is the zero vector). This distinction is frequently tested in CBSE and CUET MCQs.
Question 7: Distributive Property
7. Let the vectors \vec a,\vec b,\vec c be given as \vec a=a_1\hat i+a_2\hat j+a_3\hat k,\;\vec b=b_1\hat i+b_2\hat j+b_3\hat k,\;\vec c=c_1\hat i+c_2\hat j+c_3\hat k. Show that \vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c.
Solution
Think First… Write \vec b+\vec c in component form and evaluate the cross product using the determinant method. Then compare the result with \vec a\times\vec b+\vec a\times\vec c.
We have
\displaystyle \vec b+\vec c=(b_1+c_1)\hat i+(b_2+c_2)\hat j+(b_3+c_3)\hat k
Therefore,
\displaystyle \vec a\times(\vec b+\vec c)=\begin{vmatrix}\hat i&\hat j&\hat k\\a_1&a_2&a_3\\b_1+c_1&b_2+c_2&b_3+c_3\end{vmatrix}
Expanding the determinant,
\displaystyle \vec a\times(\vec b+\vec c)=\hat i\bigl(a_2(b_3+c_3)-a_3(b_2+c_2)\bigr)-\hat j\bigl(a_1(b_3+c_3)-a_3(b_1+c_1)\bigr)+\hat k\bigl(a_1(b_2+c_2)-a_2(b_1+c_1)\bigr)
Removing the brackets,
\displaystyle =\hat i(a_2b_3-a_3b_2+a_2c_3-a_3c_2)-\hat j(a_1b_3-a_3b_1+a_1c_3-a_3c_1)+\hat k(a_1b_2-a_2b_1+a_1c_2-a_2c_1)
Grouping the terms,
\displaystyle =(\hat i(a_2b_3-a_3b_2)-\hat j(a_1b_3-a_3b_1)+\hat k(a_1b_2-a_2b_1))+(\hat i(a_2c_3-a_3c_2)-\hat j(a_1c_3-a_3c_1)+\hat k(a_1c_2-a_2c_1))
\displaystyle =\vec a\times\vec b+\vec a\times\vec c
Hence,
\displaystyle \vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c
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 8: Vector Algebra 10.4
8. If either \vec a=\vec0 or \vec b=\vec0 , then \vec a\times\vec b=\vec0. Is the converse true? Justify your answer with an example.
Solution
Think First… Remember that the cross product of two vectors is zero not only when one of the vectors is the zero vector, but also when the two non-zero vectors are parallel (collinear).
The given statement is true because if either \vec a=\vec0 or \vec b=\vec0, then
\displaystyle \vec a\times\vec b=\vec0
However, the converse is not true.
Consider the non-zero vectors
\displaystyle \vec a=\hat i+\hat j,\qquad \vec b=2\hat i+2\hat j
Here, \vec b=2\vec a, so the vectors are parallel (collinear).
Therefore,
\displaystyle \vec a\times\vec b=\vec0
even though neither \vec a nor \vec b is the zero vector.
Hence, the converse is not true. In fact, there are infinitely many such examples, since the cross product of any two non-zero parallel (collinear) vectors is always zero.
Question 9: Area of Triangle
9. Find the area of the triangle with vertices A(1,1,2),\;B(2,3,5)\text{ and }C(1,5,5).
Solution
Think First… To find the area of a triangle, first form two vectors from the same vertex. Then use the formula \displaystyle \text{Area}=\frac12\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|.
Using the coordinates of the given points, and taking A as the common vertex, we have
\displaystyle \overrightarrow{AB}=(2-1)\hat i+(3-1)\hat j+(5-2)\hat k=\hat i+2\hat j+3\hat k
and \displaystyle \overrightarrow{AC}=(1-1)\hat i+(5-1)\hat j+(5-2)\hat k=4\hat j+3\hat k
Now, \displaystyle \overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&3\\0&4&3\end{vmatrix}
\displaystyle =\hat i(6-12)-\hat j(3-0)+\hat k(4-0)
\displaystyle =-6\hat i-3\hat j+4\hat k
Hence,
\displaystyle \left|\overrightarrow{AB}\times\overrightarrow{AC}\right|=\sqrt{(-6)^2+(-3)^2+4^2}=\sqrt{61}
Therefore, the area of the triangle is
\displaystyle \frac12\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|=\frac{\sqrt{61}}{2}
Hence, the required area of the triangle is \displaystyle \frac{\sqrt{61}}{2}
Question 10: Area of Parallelogram
10. Find the area of the parallelogram whose adjacent sides are determined by the vectors \vec a=\hat i-\hat j+3\hat k and \vec b=2\hat i-7\hat j+\hat k.
Solution
Think First… If the adjacent sides of a parallelogram are represented by vectors \vec a and \vec b, then its area is given by \displaystyle |\vec a\times\vec b|.
Given,
\displaystyle \vec a=\hat i-\hat j+3\hat k,\qquad \vec b=2\hat i-7\hat j+\hat k
The area of the parallelogram is
\displaystyle |\vec a\times\vec b|
Now,
\displaystyle \vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&3\\2&-7&1\end{vmatrix}
Expanding the determinant,
\displaystyle \vec a\times\vec b=\hat i\bigl((-1)(1)-3(-7)\bigr)-\hat j\bigl((1)(1)-3(2)\bigr)+\hat k\bigl((1)(-7)-(-1)(2)\bigr)
\displaystyle =20\hat i+5\hat j-5\hat k
Therefore,
\displaystyle |\vec a\times\vec b|=\sqrt{20^2+5^2+(-5)^2}=\sqrt{450}=15\sqrt2
Hence, the required area of the parallelogram is \displaystyle 15\sqrt2.
Maths Better Tip! Whenever a question asks for the area of a triangle or a parallelogram, first form two vectors from the same vertex and then use the cross product. Remember that the magnitude of the cross product gives the area of the parallelogram, while half of it gives the area of the triangle.
- The area of a parallelogram with adjacent sides \vec a and \vec b is \displaystyle |\vec a\times\vec b|.
- The area of a triangle formed by the same vectors is \displaystyle \frac12|\vec a\times\vec b|.
- When the vertices of a triangle are given, always form two vectors from the same vertex before applying the cross product formula.
- If the question asks only for the area, use the magnitude of the cross product and ignore its direction.
Remember, area is always a positive scalar quantity, so take the magnitude of the cross product before writing the final answer.
Question 11: Vector Algebra 10.4 – MCQ
11. Let the vectors \vec a and \vec b be such that |\vec a|=3 and |\vec b|=\dfrac{\sqrt2}{3}, then \vec a\times\vec b is a unit vector if the angle between \vec a and \vec b is
- (A) \dfrac{\pi}{6}
- (B) \dfrac{\pi}{4}
- (C) \dfrac{\pi}{3}
- (D) \dfrac{\pi}{2}
Solution
Think First… The magnitude of the cross product is given by |\vec a\times\vec b|=|\vec a||\vec b|\sin\theta. Since \vec a\times\vec b is a unit vector, its magnitude must be 1.
Given that, \displaystyle |\vec a|=3,\quad |\vec b|=\frac{\sqrt2}{3}
Since \vec a\times\vec b is a unit vector,
\displaystyle \therefore |\vec a\times\vec b|=1
Using the formula
\displaystyle |\vec a\times\vec b|=|\vec a||\vec b|\sin\theta
We get, \displaystyle 1=3\left(\frac{\sqrt2}{3}\right)\sin\theta
\displaystyle 1=\sqrt2\sin\theta
\displaystyle \Rightarrow \sin\theta=\frac{1}{\sqrt2}
Thus, \displaystyle \theta=\frac{\pi}{4}
✅️ Hence, the correct answer is (B).
Question 12: Vector Algebra 10.4 – MCQ
12. The area of a rectangle having vertices A,\;B,\;C\text{ and }D with position vectors -\hat i+\frac12\hat j+4\hat k,\;\hat i+\frac12\hat j+4\hat k,\;\hat i-\frac12\hat j+4\hat k,\;-\hat i-\frac12\hat j+4\hat k respectively is
- (A) \dfrac12
- (B) 1
- (C) 2
- (D) 4
Solution
Think First… Find two adjacent sides of the rectangle as vectors. The area of the rectangle is the magnitude of their cross product, i.e., |\overrightarrow{AB}\times\overrightarrow{BC}|.
Taking AB and BC as the adjacent sides of the rectangle, using the coordinates of the given vertices, we have
\displaystyle \overrightarrow{AB}=(1-(-1))\hat i+\left(\frac12-\frac12\right)\hat j+(4-4)\hat k=2\hat i
and
\displaystyle \overrightarrow{BC}=(1-1)\hat i+\left(-\frac12-\frac12\right)\hat j+(4-4)\hat k=-\hat j
Now,
\displaystyle \overrightarrow{AB}\times\overrightarrow{BC}=(2\hat i)\times(-\hat j)=-2\hat k
Hence,
\displaystyle |\overrightarrow{AB}\times\overrightarrow{BC}|=|-2\hat k|=2
Therefore, the area of the rectangle is \displaystyle 2
✅️ Hence, the correct answer is (C).
Common Mistakes to Avoid
- Using the wrong cross product formula: Remember that \vec a\times\vec b=|\vec a||\vec b|\sin\theta\,\hat n. Use sin θ, not cos θ.
- Making mistakes while expanding the determinant: Always expand the determinant carefully and remember that the middle term has a negative sign.
- Ignoring the order of vectors: The vector product is not commutative. Reversing the order changes the sign, i.e., \vec a\times\vec b=-\,\vec b\times\vec a.
- Using the wrong area formula: Use \dfrac12|\vec a\times\vec b| for a triangle and |\vec a\times\vec b| for a parallelogram.
- Forgetting to divide while finding a unit vector: After finding \vec a\times\vec b, divide it by its magnitude to obtain a unit vector perpendicular to both vectors.
- Confusing parallel and perpendicular vectors: If \vec a\times\vec b=\vec0, the vectors are parallel (or one of them is the zero vector). If \vec a\cdot\vec b=0, the vectors are perpendicular (provided both are non-zero).
- Making sign errors while forming vectors between points: To obtain \overrightarrow{PQ}, subtract the coordinates of P from those of Q. A wrong vector leads to an incorrect cross product and hence a wrong area.
- Forgetting the basic cross products of unit vectors: Remember \hat i\times\hat j=\hat k,\;\hat j\times\hat k=\hat i,\;\hat k\times\hat i=\hat j and their reverse orders give negative vectors.
- Using the wrong trigonometric function while finding the angle: For the cross product, use \sin\theta=\dfrac{|\vec a\times\vec b|}{|\vec a||\vec b|}, not the dot product formula.
- Skipping the final verification: Check whether the required quantity is the vector itself, its magnitude, a unit vector, or the area before writing the final answer.
Continue Learning
In Exercise 10.4, you learned how to use the vector (cross) product to find a vector perpendicular to two given vectors, calculate the area of triangles and parallelograms, apply the determinant method, and solve problems involving the properties of the cross product. These concepts have important applications in geometry, physics, engineering, and three-dimensional mathematics.
Before starting the Miscellaneous Exercise, revise the important concepts from both Exercises 10.3 and 10.4. The Miscellaneous Exercise combines questions on the dot product and the cross product, so a clear understanding of the following topics will help you solve the mixed problems more confidently:
- How to find the vector (cross) product using \vec a\times\vec b=|\vec a||\vec b|\sin\theta\,\hat n and the determinant method.
- How to calculate the magnitude of the cross product and use \displaystyle \sin\theta=\frac{|\vec a\times\vec b|}{|\vec a||\vec b|} whenever required.
- The properties of the vector product, especially \vec a\times\vec b=-\vec b\times\vec a, \vec a\times\vec a=\vec0, and the distributive law.
- How to find a unit vector perpendicular to two given vectors using \displaystyle \pm\frac{\vec a\times\vec b}{|\vec a\times\vec b|}.
- How to find the area of a triangle using \displaystyle \frac12|\vec a\times\vec b| and the area of a parallelogram using \displaystyle |\vec a\times\vec b|.
- How to form vectors joining two points before applying the cross product in geometry problems.
- When to use the dot product and when to use the cross product in vector problems.
- The importance of simplifying the final answer and expressing it in the form expected in NCERT.
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A good understanding of the concepts covered in Exercise 10.4 is essential for solving problems involving the vector (cross) product, areas of triangles and parallelograms, and vectors perpendicular to two given vectors. Always begin by identifying what the question asks—whether it requires finding the cross product, its magnitude, a unit vector, or the area of a geometric figure.
While solving questions, expand the determinant carefully, pay special attention to signs, and remember that reversing the order of the vectors changes the direction of the cross product. Use the properties of the vector product wherever required, simplify the final answer, and express it in the form expected in NCERT. With regular practice, you will be be able to solve problems involving vector products, perpendicular vectors and areas accurately and confidently.
All the best and keep learning 👍
