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Class 12 RD Sharma Solutions – Chapter 25 Vector or Cross Product – Exercise 25.1 | Set 2
  • Last Updated : 28 Mar, 2021

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

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