Skip to content
Related Articles
Get the best out of our app
GeeksforGeeks App
Open App
geeksforgeeks
Browser
Continue

Related Articles

Class 12 RD Sharma Solutions – Chapter 25 Vector or Cross Product – Exercise 25.1 | Set 2

Improve Article
Save Article
Like Article
author
isha412k
contributor
11 published articles
Improve Article
Save Article
Like Article

Question 13. If |\vec{a}|=13 , |\vec{b}|=5 and \vec{a}.\vec{b}=60 , find |\vec{a}\times\vec{b}|

Solution:

We know that,

=> \vec{a}.\vec{b} = |\vec{a}||\vec{b}|\cos\theta

=>|\vec{a}.\vec{b}| = |\vec{a}||\vec{b}||\sin\theta|

=> 60 = 13\times5\times\cos\theta

=> 65\cos\theta = 60

=> \cos\theta = \dfrac{12}{13}

Also,

=>| \vec{a}\times\vec{b}| = |\vec{a}||\vec{b}||\cos\theta||\hat{n}|

And \sin^2\theta + \cos^2\theta = 1

=> \sin\theta = \sqrt{1-\cos^2\theta}

=> \sin\theta = \sqrt{1-(\dfrac{12}{13})^2}

=> \sin\theta = \sqrt{\dfrac{25}{169}}

=> \sin\theta = \dfrac{5}{13}

=> |\vec{a}\times\vec{b}| = 13\times5\times\dfrac{5}{13}

=>| \vec{a}\times\vec{b}| = 25

Question 14. Find the angle between 2 vectors \vec{a} and  \vec{b} , if  |\vec{a}\times\vec{b}| = \vec{a}.\vec{b}

Solution:

Given |\vec{a}\times\vec{b}| = \vec{a}.\vec{b}

=>|\vec{a}||\vec{b}|\sin\theta|\hat{n}| = |\vec{a}||\vec{b}|\cos\theta

=> |\vec{a}||\vec{b}|\sin\theta = |\vec{a}||\vec{b}|\cos\theta , as \hat{n}  is a unit vector.

=> \sin\theta = \cos\theta

=> \tan\theta = 1

=> \theta = \dfrac{\pi}{4}

Question 15. If \vec{a}\times\vec{b} = \vec{b}\times\vec{c} \neq \vec{0} , then show that \vec{a}+\vec{c}=m\vec{b} , where m is any scalar.

Solution:

Given that \vec{a}\times\vec{b} = \vec{b}\times\vec{c} \neq \vec{0}

=> \vec{a}\times\vec{b} - \vec{b}\times\vec{c} = \vec{0}

=> \vec{a}\times\vec{b} -[-(\vec{c}\times\vec{a})] = \vec{0}

=> \vec{a}\times\vec{b} + \vec{c}\times\vec{b} = \vec{0}

Using distributive property,

=> (\vec{a}+\vec{c})\times\vec{b}=\vec{0}

If two vectors are parallel, then their cross-product is 0 vector.

=> (\vec{a}+\vec{c})  and \vec{b}  are parallel vectors.

=> (\vec{a}+\vec{c}) = m\vec{b}

Hence proved.

 Question 16. If |\vec{a}|=2 , |\vec{b}|=7 and \vec{a}\times\vec{b} = 3\hat{i}+2\hat{j}+6\hat{k} , find the angle between \vec{a}  and \vec{b}

Solution:

Given that, |\vec{a}|=2 , |\vec{b}|=7 and \vec{a}\times\vec{b} = 3\hat{i}+2\hat{j}+6\hat{k}

We know that,

=> \vec{a}\times\vec{b} =|\vec{a}||\vec{b}|\sin\theta\hat{n}

=> |\vec{a}\times\vec{b}| =|\vec{a}||\vec{b}|\sin\theta\|\hat{n}|

=> |\vec{a}\times\vec{b}| =|\vec{a}||\vec{b}|\sin\theta

=> \sqrt{3^2+2^2+6^2} = 2\times7\times\sin\theta

=> \sqrt{9+4+36} = 14\sin\theta

=> 7 = 14\sin\theta

=> \sin\theta = \dfrac{1}{2}

=> \theta = \dfrac{\pi}{6}

Question 17. What inference can you draw if \vec{a}\times\vec{b}=\vec{0} and \vec{a}.\vec{b}=0

Solution:

Given, \vec{a}\times\vec{b}=\vec{0} and \vec{a}.\vec{b}=0

=>\vec{a}\times\vec{b} = \vec{0}

=> |\vec{a}|\vec{b}|\sin\theta\hat{n} = \vec{0}

Either of the following conditions is true,

1. |\vec{a}| = 0

2. |\vec{b}| =0

3. |\vec{a}| = |\vec{b}| = 0

4. \vec{a} is parallel to \vec{b}

=> \vec{a}.\vec{b}=0

=> |\vec{a}||\vec{b}|\cos\theta = 0

Either of the following conditions is true,

1. |\vec{a}| = 0

2. |\vec{b}| =0

3. |\vec{a}| = |\vec{b}| = 0

4. \vec{a} is perpendicular to \vec{b}

Since both these conditions are true, that implies atleast one of the following conditions is true,

1. |\vec{a}| = 0

2. |\vec{b}| =0

3. |\vec{a}| = |\vec{b}| = 0

Question 18. If \vec{a} , \vec{b} and \vec{c} are 3 unit vectors such that \vec{a}\times\vec{b} = \vec{c} , \vec{b}\times\vec{c}=\vec{a} and \vec{c}\times\vec{a} = \vec{b} . Show that \vec{a} , \vec{b} and \vec{c}  form an orthogonal right handed triad of unit vectors.

Solution:

Given, \vec{a}\times\vec{b} = \vec{c} , \vec{b}\times\vec{c}=\vec{a} and \vec{c}\times\vec{a} = \vec{b}

As,

=> \vec{c} = \vec{a}\times\vec{b}

=> \vec{c}  is perpendicular to both \vec{a} and \vec{b} .

Similarly,

=> \vec{a}  is perpendicular to both \vec{b} and \vec{c}

=> \vec{b}  is perpendicular to both \vec{a} and \vec{c}

=> \vec{a} , \vec{b} and \vec{c}  are mutually perpendicular.

As, \vec{a} , \vec{b} and \vec{c}  are also unit vectors,

=> \vec{a} , \vec{b} and \vec{c}  form an orthogonal right-handed triad of unit vectors

Hence proved.

Question 19. Find a unit vector perpendicular to the plane ABC, where the coordinates of A, B, and C are A(3, -1, 2), B(1, -1, 3), and C(4, -3, 1).

Solution:

Given A(3, -1, 2), B(1, -1, 3) and C(4, -3, 1).

Let,

=> \vec{a} = A = 3\hat{i}- \hat{j} +2\hat{k}

=> \vec{b} = B = \hat{i} -\hat{j} + 3\hat{k}

=> \vec{c} = C = 4\hat{i}-3\hat{j}+\hat{k}

Plane ABC has two vectors \vec{AB}  and \vec{AC}

=> \vec{AB} = \vec{b} - \vec{a}

=> \vec{AB} = (\hat{i}-\hat{j}-3\hat{k})-(3\hat{i}-\hat{j}+2\hat{k})

=> \vec{AB}= (1-3)\hat{i}+ (-1+1)\hat{j} +(-3-2)\hat{k}

=> \vec{AB} = -2\hat{i}-5\hat{k}

=> \vec{AC} = \vec{c} - \vec{a}

=> \vec{AC} = (4\hat{i}-3\hat{j}+\hat{k})-(3\hat{i}-\hat{j}+2\hat{k})

=> \vec{AC} = (4-3)\hat{i}+ (-3+1)\hat{j} +(1-2)\hat{k}

=> \vec{AC} = \hat{i}-2\hat{i}-\hat{k}

A vector perpendicular to both \vec{AB} and \vec{AC} is given by,

=> \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}

=>  \vec{AB}\times\vec{AC}= \dfrac{1}{7}\times\dfrac{1}{7}\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-2 & 0 & -5\\1 & -2& -1\end{vmatrix}

=> \vec{AB}\times\vec{AC} = \hat{i}[(0)(-1)-(-2)(-5)] -\hat{j}[(-2)(-1)-(1)(-5)] +\hat{k}[(-2)(-2)-(1)(0)]

=> \vec{AB}\times\vec{AC} = \hat{i}[0-10]-\hat{j}[2+5]+\hat{k}[4-0]

=> \vec{AB}\times\vec{AC} = -10\hat{j}-7\hat{j}+4\hat{k}

To find the unit vector,

=> \hat{p} = \dfrac{\vec{AB}\times\vec{AC}}{|\vec{AB}\times\vec{AC}|}

=> \hat{p} = \dfrac{1}{\sqrt{(-10^2)+(-7)^2+4^2}}(-10\hat{j}-7\hat{j}+4\hat{k})

=> \hat{p} = \dfrac{1}{\sqrt{100+49+16}}(-10\hat{j}-7\hat{j}+4\hat{k})

=> \hat{p} = \dfrac{1}{\sqrt{165}}(-10\hat{j}-7\hat{j}+4\hat{k})

Question 20. If a, b and c are the lengths of sides BC, CA and AB of a triangle ABC, prove that \vec{BC} +\vec{CA} +\vec{AB} = \vec{0} and deduce that \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}

Solution:

Given that |\vec{BC}|=a , |\vec{CA}|=b and |\vec{AB}| = c

From triangle law of vector addition, we have

=> \vec{AB} + \vec{BC} = \vec{AC}

=> \vec{AB} + \vec{BC} = -\vec{CA}

=> \vec{AB} + \vec{BC} + \vec{CA} = \vec{0}

=> \vec{BC} +\vec{CA} +\vec{AB} = \vec{0}

=> \vec{a} + \vec{b} + \vec{c} = \vec{0}

=> \vec{a}\times(\vec{a}+\vec{b}+\vec{c}) = \vec{a}\times\vec{0}

=> \vec{a}\times\vec{a} + \vec{a}\times\vec{b} + \vec{a}\times\vec{c} = \vec{0}

=> \vec{0} + \vec{a}\times\vec{b} + \vec{a}\times\vec{c} = \vec{0}

=> \vec{a}\times\vec{b} + \vec{a}\times\vec{c} = \vec{0}

=> \vec{a}\times\vec{b} = -\vec{a}\times\vec{c}

=> \vec{a}\times\vec{b} = \vec{c}\times\vec{a}

=> |\vec{a}||\vec{b}|\sin C = |\vec{c}||\vec{a}|\sin B

=> |\vec{b}|\sin C = |\vec{c}|\sin B

=> b\sin C = c\sin B

=> \dfrac{b}{\sin B} = \dfrac{c}{\sin C}

Similarly,

=> \vec{b}\times(\vec{a}+\vec{b}+\vec{c}) = \vec{b}\times\vec{0}

=> \vec{b}\times\vec{a} + \vec{b}\times\vec{b} + \vec{b}\times\vec{c} = \vec{0}

=> \vec{0} + \vec{b}\times\vec{a} + \vec{b}\times\vec{c} = \vec{0}

=> \vec{b}\times\vec{a} + \vec{b}\times\vec{c} = \vec{0}

=> \vec{b}\times\vec{a} = -\vec{b}\times\vec{c}

=> \vec{b}\times\vec{a} = \vec{c}\times\vec{b}

=> |\vec{b}||\vec{a}|\sin C = |\vec{c}||\vec{b}|\sin A

=> |\vec{a}|\sin C = |\vec{c}|\sin A

=> a\sin C = c\sin A

=> \dfrac{a}{\sin A} = \dfrac{c}{\sin C}

=> \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{sin C}

Hence proved.

Question 21. If \vec{a} = \hat{i}-2\hat{j}+3\hat{k} and \vec{b}=2\hat{i}+3\hat{j}-5\hat{k} , then find \vec{a}\times\vec{b} . Verify that \vec{a} and \vec{a}\times\vec{b}  are perpendicular to each other.

Solution:

Given, \vec{a} = \hat{i}-2\hat{j}+3\hat{k} and \vec{b}=2\hat{i}+3\hat{j}-5\hat{k}

=> \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}

=> \vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & -2 & 3\\2 & 3 & -5\end{vmatrix}

=> \vec{a}\times\vec{b} = \hat{i}[(-2)(-5)-(3)(3)]-\hat{j}[(1)(-5)-(2)(3)]+\hat{k}[(1)(3)-(2)(-2)]

=> \vec{a}\times\vec{b} = \hat{i}[10-9]-\hat{j}[-5-6]+\hat{k}[3+4]

=> \vec{a}\times\vec{b} = \hat{i}+11\hat{j}+7\hat{k}

Two vectors are perpendicular if their dot product is zero.

=> (\vec{a}\times\vec{b}).\vec{a} = (\hat{i}-2\hat{j}+3\hat{k}).(\hat{i}+11\hat{j}+7\hat{k})

=> (\vec{a}\times\vec{b}).\vec{a} = \hat{i}.\hat{i}-2\hat{j}.11\hat{j}+3\hat{k}.7\hat{k}

=> (\vec{a}\times\vec{b}).\vec{a} = 1-22+21

=> (\vec{a}\times\vec{b}).\vec{a} =0

Hence proved.

Question 22. If \vec{p}  and \vec{q}  are unit vectors forming an angle of 30\degree , find the area of the parallelogram having \vec{a}=\vec{p}+2\vec{q} and \vec{b}=2\vec{p}+\vec{q}  as its diagonals.

Solution:

Given \vec{p} and \vec{q} forming an angle of 30\degree .

Area of a parallelogram having diagonals \vec{a}  and \vec{b}  is \dfrac{1}{2}|\vec{a}\times\vec{b}|

=> \vec{p}\times\vec{q} = |\vec{p}||\vec{q}|\sin 30\degree \hat{n}

=> \vec{p}\times\vec{q} = 1\times1\times\dfrac{1}{2}\times \hat{n}

=> \vec{p}\times\vec{q} = \dfrac{1}{2} \hat{n}

Thus area is,

=> Area = \dfrac{1}{2}|(\vec{p}+2\vec{q})\times( 2\vec{p}+\vec{q})|

=> Area = \dfrac{1}{2}|\vec{p}\times( 2\vec{p}+\vec{q})+2\vec{q}\times( 2\vec{p}+\vec{q})|

=> Area = \dfrac{1}{2}|\vec{p}\times\vec{q}+4(\vec{q}\times\vec{p})|

=> Area = \dfrac{1}{2}|\vec{p}\times\vec{q}+4(-\vec{p}\times\vec{q})|

=> Area = \dfrac{1}{2}|-3(\vec{p}\times\vec{q})|

=> Area = \dfrac{3}{2}|(\vec{p}\times\vec{q})|

=> Area = \dfrac{3}{2}|\dfrac{1}{2} \hat{n}|

=> Area = \dfrac{3}{2}\times\dfrac{3}{2}\times1

=> Area = \dfrac{3}{4}  square units

Question 23. For any two vectors \vec{a}  and \vec{b}  , prove that |\vec{a}\times\vec{b}|^2 = \begin{vmatrix} \vec{a}.\vec{a}& \vec{a}.\vec{b}\\\vec{b}.\vec{a} & \vec{b}.\vec{b}\end{vmatrix}  

Solution:

We know that,

=> \vec{a}\times\vec{b}= |\vec{a}||\vec{b}|\sin\theta\hat{n}

=> |\vec{a}\times\vec{b}|= |\vec{a}||\vec{b}|\sin\theta|\hat{n}|

=> |\vec{a}\times\vec{b}|= |\vec{a}||\vec{b}|\sin\theta

=> |\vec{a}\times\vec{b}|^2= |\vec{a}|^2|\vec{b}|^2\sin^2\theta

=> |\vec{a}\times\vec{b}|^2= |\vec{a}|^2|\vec{b}|^2(1-\cos^2\theta)

=> |\vec{a}\times\vec{b}|^2= |\vec{a}|^2|\vec{b}|^2 -(|\vec{a}|^2|\vec{b}|^2\cos^2\theta)

=> |\vec{a}\times\vec{b}|^2= |\vec{a}|^2|\vec{b}|^2 -(|\vec{a}|.|\vec{b}|)^2

=> |\vec{a}\times\vec{b}|^2= (\vec{a}.\vec{a})(\vec{b}.\vec{b})-(\vec{a}.\vec{b})(\vec{b}.\vec{a})

=> |\vec{a}\times\vec{b}|^2 = \begin{vmatrix} \vec{a}.\vec{a}& \vec{a}.\vec{b}\\\vec{b}.\vec{a} & \vec{b}.\vec{b}\end{vmatrix}

Hence proved.

Question 24. Define \vec{a}\times\vec{b} and prove that |\vec{a}\times\vec{b}|=(\vec{a}.\vec{b})\tan \theta , where \theta  is the angle between \vec{a} and \vec{b}

Solution:

Definition of \vec{a}\times\vec{b} : Let \vec{a}  and \vec{b} be 2 non-zero, non-parallel vectors. Then \vec{a}\times\vec{b} , is defined as a vector with the magnitude of |\vec{a}||\vec{b}|\sin\theta , and which is perpendicular to both the vectors \vec{a}  and \vec{b} .

We know that,

=> \vec{a}\times\vec{b} = |\vec{a}||\vec{b}|\sin\theta\hat{n}

=> |\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta|\hat{n}|

=> |\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta  ……………..(eq.1)

And as,

=> \vec{a}.\vec{b} = |\vec{a}||\vec{b}|\cos\theta

=> |\vec{a}||\vec{b}| = \dfrac{\vec{a}.\vec{b}}{\cos\theta}

Substituting in (eq.1),

=> |\vec{a}\times\vec{b}| = \dfrac{\vec{a}.\vec{b}}{\cos\theta} \sin\theta

=> |\vec{a}\times\vec{b}| = (\vec{a}.\vec{b})\tan\theta


My Personal Notes arrow_drop_up
Last Updated : 28 Mar, 2021
Like Article
Save Article
Similar Reads
1. Class 12 RD Sharma Solutions - Chapter 25 Vector or Cross Product - Exercise 25.1 | Set 3
2. Class 12 RD Sharma Solutions - Chapter 25 Vector or Cross Product - Exercise 25.1 | Set 1
3. Class 12 RD Sharma Solutions - Chapter 24 Scalar or Dot Product - Exercise 24.1 | Set 1
4. Class 12 RD Sharma Solutions - Chapter 24 Scalar or Dot Product - Exercise 24.1 | Set 2
5. Class 12 RD Sharma Solutions - Chapter 24 Scalar or Dot Product - Exercise 24.1 | Set 3
6. Class 12 RD Sharma Solutions - Chapter 26 Scalar Triple Product - Exercise 26.1
7. Class 12 RD Sharma Solutions - Chapter 24 Scalar or Dot Product - Exercise 24.2
8. Class 12 RD Sharma Solutions - Chapter 9 Continuity - Exercise 9.1 | Set 2
9. Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.3 | Set 1
10. Class 12 RD Sharma Solutions - Chapter 33 Binomial Distribution - Exercise 33.1 | Set 1
Related Tutorials
1. Linear Algebra
2. Maths
3. Physics
4. Chemistry
5. CBSE Class 12 Syllabus 2023-2024
Previous
Class 12 RD Sharma Solutions - Chapter 25 Vector or Cross Product - Exercise 25.1 | Set 1
Next
Class 12 RD Sharma Solutions - Chapter 25 Vector or Cross Product - Exercise 25.1 | Set 3
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
isha412k
isha412k
Vote for difficulty
Article Tags :
Report Issue
We use cookies to ensure you have the best browsing experience on our website. By using our site, you acknowledge that you have read and understood our Cookie Policy & Privacy Policy
Lightbox