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
Summary
This set focuses on applying cross product properties and calculations in various scenarios.
Key points include:
1. Cross product properties (anticommutative, distributive, etc.)
2. Geometric applications (area, volume, perpendicularity)
3. Vector triple product and its expansion
4. Using cross products to solve geometric problems
Practice Problems
1. If a = 3i - 2j + k, b = i + 2j - 3k, and c = 2i + j - k, find a · (b × c).
2. Prove that a × (b × c) + b × (c × a) + c × (a × b) = 0.
3. If a = i + 2j - k, b = 2i - j + 3k, and c = i + j + k, find the volume of the parallelepiped formed by these vectors.
4. Show that (a × b) × (c × d) = [a b d]c - [a b c]d.
5. Find a vector perpendicular to both a = 2i - j + 3k and b = i + 2j - k, and has a magnitude of 5 units.
6. Prove that (a × b) · (c × d) = (a · c)(b · d) - (a · d)(b · c).
7. If a, b, and c are unit vectors such that a + b + c = 0, prove that a × b = b × c = c × a.
8. Find the area of the triangle formed by the points A(1, -1, 2), B(3, 2, -1), and C(4, -3, 5).
9. Prove that |a × (b × c)| = |b||c|sin θ, where θ is the angle between b and c.
10. If a · b = 5, b · c = -2, c · a = 3, |a| = 2, |b| = 3, and |c| = 4, find |a × b + b × c + c × a|.
Explore
Basic Arithmetic
Algebra
Geometry
Trigonometry & Vector Algebra
Calculus
Probability and Statistics
Practice