Class 12 RD Sharma Solutions Chapter 25 Vector or Cross Product Exercise 25.1 | Set 2 (original) (raw)
Last Updated : 27 Aug, 2024
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,
- |\vec{a}| = 0
2. |\vec{b}| =0
3. |\vec{a}| = |\vec{b}| = 0
- \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
- |\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|.