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Class 12 RD Sharma Solutions – Chapter 25 Vector or Cross Product – Exercise 25.1 | Set 3

  • Last Updated : 28 Mar, 2021

Question 25. If |\vec{a}|=\sqrt{26}  |\vec{b}|= 7   and |\vec{a}\times\vec{b}|=35  , find \vec{a}.\vec{b}

Solution:

We know that,

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=> \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}|

=> 35 = \sqrt{26}\times7|\sin\theta|\times1

=> \sin\theta = \dfrac{35}{7\sqrt{26}}

=> \sin\theta = \dfrac{5}{\sqrt{26}}

As \cos^2\theta + \sin^2\theta =1  ,

=> \cos\theta = \sqrt{1-\sin^2\theta}

=> \cos\theta = \sqrt{1-(\dfrac{5}{\sqrt{26}})^2}



=> \cos\theta = \sqrt{1-\dfrac{25}{26}}

=> \cos\theta = \dfrac{1}{\sqrt{26}}

Thus,

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

=> \vec{a}.\vec{b} = 7\sqrt{26}\times\dfrac{1}{\sqrt{26}}

=> \vec{a}.\vec{b} = 7

Question 26. Find the area of the triangle formed by O, A, B when \vec{OA} = \hat{i}+2\hat{j}+3\hat{k}  \vec{OB} = -3\hat{i}-2\hat{j}+\hat{k}

Solution:

The area of a triangle whose adjacent sides are given by \vec{a}   and \vec{b}   is \dfrac{1}{2}|\vec{a}\times\vec{b}|

=> \vec{OA}\times\vec{OB} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\b_1 & b_2 & b_3\end{vmatrix}

=> \vec{OA}\times\vec{OB} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2 & 3\\-3 & -2 & 1\end{vmatrix}



=> \vec{OA}\times\vec{OB} = \hat{i}[2+6]-\hat{j}[1+9]+\hat{k}[-2+6]

=> \vec{OA}\times\vec{OB} = 8\hat{i}-10\hat{j}+4\hat{k}

=> Area = \dfrac{1}{2} |\vec{OA}\times\vec{OB}|

=> Area = \dfrac{1}{2}\sqrt{8^2+(-10^2)+4^2}

=> Area = \dfrac{1}{2}\sqrt{64+100+16}

=> Area = \dfrac{1}{2}\times6\sqrt{45}

=> Area = 3\sqrt{5}   square units.

Question 27. Let \vec{a}=\hat{i}+4\hat{j}+2\hat{k}  \vec{b}=3\hat{i}-2\hat{j}+7\hat{k}   and \vec{c} = 2\hat{i}-\hat{j}+4\hat{k}  . Find a vector \vec{d}   which is perpendicular to both \vec{a}   and \vec{b}  and \vec{c}.\vec{d} = 15

Solution:

Given that \vec{d}   is perpendicular to both \vec{a}   and \vec{b}  .

=> \vec{d}.\vec{a} =0   ……….(1)



=> \vec{d}.\vec{b} =0   ……….(2)

Also,

=> \vec{c}.\vec{d} = 15   …….(3)

Let \vec{d} = d_1\hat{i}+d_2\hat{j}+d_3\hat{k}

From eq(1),

=> d1 + 4d2 + 2d3 = 0 

From eq(2),

=> 3d1 – 2d2 + 7d3 = 0

From eq(3),

=> 2d1 – d2 + 4d3 = 15 

On solving the 3 equations we get,

d1 = 160/3, d2 = -5/3, and d3 = -70/3, 

=> \vec{d} = \dfrac {1}{3}(160\hat{i}-5\hat{j}-70\hat{k})

Question 28. 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:

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

Let \vec{c} = \vec{a}+\vec{b}

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

=> \vec{c} = \hat{i}[3+1] +\hat{j}[2+2] +\hat{k}[2-2]

=> \vec{c} = 4\hat{i} +4\hat{j}

Let \vec{d} = \vec{a}-\vec{b}



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

=> \vec{d} = \hat{i}[3-1] +\hat{j}[2-2] +\hat{k}[2+2]

=> \vec{d} = 2\hat{i} +4\hat{k}

A vector perpendicular to both \vec{c}   and \vec{d}   is,

=> \vec{c}\times\vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\c_1 & c_2 & c_3\\d_1 & d_2 & d_3\end{vmatrix}

=> \vec{c}\times\vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\4 & 4 & 0\\2 & 0 & 4\end{vmatrix}

=> \vec{c}\times\vec{d} = \hat{i}[16-0]-\hat{j}[16-0]+\hat{k}[0-8]

=> \vec{c}\times\vec{d} = 16\hat{i}-16\hat{j}-8\hat{k}

To find the unit vector,

=> \hat{p} = \dfrac{\vec{c}\times\vec{d}}{|\vec{c}\times\vec{d}|}



=> \hat{p} = \dfrac{1}{\sqrt{16^2+(-16)^2+(-8)^2}}(16\hat{i}-16\hat{j}-8\hat{k})

=> \hat{p} = \dfrac{1}{\sqrt{256+256+64}}(16\hat{i}-16\hat{j}-8\hat{k})

=> \hat{p} = \dfrac{1}{24}(16\hat{i}-16\hat{j}-8\hat{k})

=> \hat{p} = \dfrac{1}{3}(2\hat{i}-2\hat{j}-\hat{k})

Question 29. Using vectors, find the area of the triangle with the vertices A(2, 3, 5), B(3, 5, 8), and C(2, 7, 8).

Solution:

Given, A(2, 3, 5), B(3, 5, 8), and C(2, 7, 8)

Let,

=> \vec{a} = A = 2\hat{i}+3\hat{j}+5\hat{k}

=> \vec{b} = B = 3\hat{i}+5\hat{j}+8\hat{k}

=> \vec{c} = C = 2\hat{2}+7\hat{j}+8\hat{k}



Then,

=> \vec{AB} = \vec{b}-\vec{a}

=> \vec{AB} = (3\hat{i}+5\hat{j}+8\hat{k})-(2\hat{i}+3\hat{j}+5\hat{k})

=> \vec{AB} = \hat{i}[3-2]+\hat{j}[5-3]+\hat{k}[8-5]

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

=> \vec{AC} = \vec{c}-\vec{a}

=> \vec{AC} = (2\hat{2}+7\hat{j}+8\hat{k})-(2\hat{i}+3\hat{j}+5\hat{k})

=> \vec{AC} = \hat{i}[2-2]+\hat{j}[7-3]+\hat{k}[8-5]

=> \vec{AC} = 4\hat{j}+3\hat{k}

The area of a triangle whose adjacent sides are given by \vec{a}   and \vec{b}   is \dfrac{1}{2}|\vec{a}\times\vec{b}|



=> \vec{AB}\times\vec{AC}= \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2& 3\\0 & 4 & 3\end{vmatrix}

=> \vec{AB}\times\vec{AC} = \hat{i}[6-12]-\hat{j}[3-0]+\hat{k}[4-0]

=> \vec{AB}\times\vec{AC} = -6\hat{i}-3\hat{j}+4\hat{k}

=> Area = \dfrac{1}{2}|\vec{AB}\times\vec{AC}|

=> Area = \dfrac{1}{2}\sqrt{(-6)^2+(-3)^2+4^2}

=> Area = √61/2

Question 30. If \vec{a}=2\hat{i}-3\hat{j}+\hat{k}  \vec{b}=-\hat{i}+\hat{k}  \vec{c}=2\hat{j}-\hat{k}   are three vectors, find the area of the parallelogram having diagonals (\vec{a}+\vec{b})   and (\vec{b}+\vec{c})  .

Solution:

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

Let,

=> \vec{c} = (\vec{a}+\vec{b})



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

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

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

=> \vec{d} = (\vec{b}+\vec{c})

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

=> \vec{d} = -\hat{i}+2\hat{j}

The area of the parallelogram having diagonals \vec{c}   and \vec{d}   is \dfrac{1}{2}|\vec{c}\times\vec{d}|

=> \vec{c}\times\vec{d} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & -3& 2\\-1 & 2 & 0\end{vmatrix}

=> \vec{c}\times\vec{d} = \hat{i}[0-4] -\hat{j}[0+2]+\hat{k}[2-3]

=> \vec{c}\times\vec{d} = -4\hat{i}-2\hat{j}-\hat{k}



=> Area = \dfrac{1}{2}|\vec{c}\times\vec{d}|

=> Area = \dfrac{1}{2}\sqrt{(-4)^2+(-2)^2+(-1)^2}

=> Area = \dfrac{1}{2}\sqrt{21}

=> Area = √21/2

Question 31. The two adjacent sides of a parallelogram are 2\hat{i}-4\hat{j}+5\hat{k}  and \hat{i}-2\hat{j}-3\hat{k}  . Find the unit vector parallel to one of its diagonals. Also, find its area.

Solution:

Given a parallelogram ABCD and its 2 sides AB and BC.

By triangle law of addition,

=> \vec{AC} = \vec{AB}+\vec{BC}

=> \vec{AC} = (2\hat{i}-4\hat{j}+5\hat{k})+(\hat{i}-2\hat{j}-3\hat{k})

=> \vec{AC} = \hat{i}[2+1] +\hat{j}[-4-2]+\hat{k}[5-3]

=> \vec{AC} = 3\hat{i}-6\hat{j}+2\hat{k}

Unit vector is,

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

=> \hat{p} = \dfrac{1}{\sqrt{3^2+(-6)^2+2^2}}(3\hat{i}-6\hat{j}+2\hat{k})

=> \hat{p} = \dfrac{1}{\sqrt{49}}(3\hat{i}-6\hat{j}+2\hat{k})

=> \hat{p} = \dfrac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})

Area of a parallelogram whose adjacent sides are given is |\vec{a}\times\vec{b}|

=> |\vec{AB}\times\vec{BC}| = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2 & -4 & 5\\1 & -2 & -3\end{vmatrix}

=> |\vec{AB}\times\vec{BC}| = \hat{i}[12+10]-\hat{j}[-6-5]+\hat{k}[-4+4]

=> |\vec{AB}\times\vec{BC}| = 22\hat{i}+11\hat{j}



Thus area is,

=> Area = |22\hat{i}+11\hat{j}|

=> Area = \sqrt{22^2+11^2}

=> Area = \sqrt{605}

=> Area = 11 √5 square units

Question 32. If either \vec{a}=0  or \vec{b}=0  , then \vec{a}\times\vec{b}=\vec{0}  . Is the converse true? Justify with example.

Solution:

Let us take two parallel non-zero vectors \vec{a}   and \vec{b}

=> \vec{a}\times\vec{b} = \vec{0}

For example,

\vec{a} = \hat{i}   and \vec{b}=2\hat{i}

=> \vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 0 & 0\\2 & 0 & 0\end{vmatrix}

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

But,

=> |\vec{a}| = \sqrt{1^2} =1

=> |\vec{b}| = \sqrt{2^2} =2

Hence the converse may not be true.

Question 33. If \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}  and \vec{c} = c_1\hat{i}+c_2\hat{j}+c_3\hat{k}  , then verify that \vec{a}\times(\vec{b}\times\vec{c})=\vec{a}\times\vec{b}+\vec{a}\times\vec{c}  .

Solution:

Given, \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}  and \vec{c} = c_1\hat{i}+c_2\hat{j}+c_3\hat{k}

=> (\vec{b}+\vec{c}) = (b_1+c_1)\hat{i}+(b_2+c_2)\hat{j}+(b_3+c_3)\hat{k}

=> \vec{a}\times(\vec{b}\times\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}



=> \vec{a}\times(\vec{b}\times\vec{c}) = \hat{i}[a_2(b_3+c_3)-a_3(b_2+c_2)]-\hat{j}[a_1(b_3+c_3)-a_3(b_1+c_1)]+\hat{k}[a_1(b_2+c_2)-a_2(b_1+c_1)]

=> \vec{a}\times(\vec{b}\times\vec{c}) = \hat{i}[a_2b_3+a_2c_3-a_3b_2-a_3c_2]+\hat{j}[-a_1b_3-a_1c_3+a_3b_1+a_3c_1]+\hat{k}[a_1b_2+a_1c_2-a_2b_1-a_2c_1]   …..eq(1)

Now,

=> \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} = \hat{i}[a_2b_3-b_2a_3]-\hat{j}[a_1b_3-b_1a_3]+\hat{k}[a_1b_2-b_1a_2]

And,

=> \vec{a}\times\vec{c} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\a_1 & a_2 & a_3\\c_1 & c_2 & c_3\end{vmatrix}

=> \vec{a}\times\vec{c} = \hat{i}[a_2b_3-c_2a_3]-\hat{j}[a_1c_3-c_1a_3]+\hat{k}[a_1c_2-c_1a_2]

Thus, 

=> \vec{a}\times\vec{b}+\vec{a}\times\vec{c} = (\hat{i}[a_2b_3-b_2a_3]-\hat{j}[a_1b_3-b_1a_3]+\hat{k}[a_1b_2-b_1a_2]) + (\hat{i}[a_2b_3-c_2a_3]-\hat{j}[a_1c_3-c_1a_3]+\hat{k}[a_1c_2-c_1a_2])



=> \vec{a}\times\vec{b}+\vec{a}\times\vec{c} = \hat{i}[a_2b_3+a_2c_3-a_3b_2-a_3c_2]+\hat{j}[-a_1b_3-a_1c_3+a_3b_1+a_3c_1]+\hat{k}[a_1b_2+a_1c_2-a_2b_1-a_2c_1]   …eq(2)

Thus eq(1) = eq(2)

Hence proved.

Question 34(i). Using vectors find the area of the triangle with the vertices A(1, 1, 2), B(2, 3, 5), and C(1, 5, 5).

Solution:

Given, A(1, 1, 2), B(2, 3, 5), and C(1, 5, 5)

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

=> \vec{b} = B = 2\hat{i}+3\hat{j}+5\hat{k}

=> \vec{c} = C = \hat{i}+5\hat{j}+5\hat{k}

Now 2 sides of the triangle are given by,

=> \vec{AB} = \vec{b}-\vec{a}

=> \vec{AB} = (2\hat{i}+3\hat{j}+5\hat{k})-(\hat{i}+\hat{j}+2\hat{k})

=> \vec{AB} = \hat{i}[2-1] +\hat{j}[3-1]+\hat{j}[5-2]

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

=> \vec{AC} = \vec{c}-\vec{a}

=> \vec{AC} = (\hat{i}+5\hat{j}+5\hat{k})-(\hat{i}+\hat{j}+2\hat{k})

=> \vec{AC} = \hat{i}[1-1] +\hat{j}[5-1]+\hat{j}[5-2]

=> \vec{AC} = 4\hat{j}+3\hat{k}

Area of the triangle whose adjacent sides are given is \dfrac{1}{2}|\vec{a}\times\vec{b}|

=> \vec{AB}\times\vec{AC} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2 & 3\\0 & 4 & 3\end{vmatrix}

=> \vec{AB}\times\vec{AC} = \hat{i}[6-12]-\hat{j}[3-0]+\hat{k}[4-0]



=> \vec{AB}\times\vec{AC} = -6\hat{i}-3\hat{j}+4\hat{k}

Thus area of the triangle is,

=> Area = \dfrac{1}{2}\sqrt{(-6)^2+(-3)^2+4^2}

=> Area = \dfrac{1}{2}\sqrt{36+9+16}

=> Area = √61/2

Question 34(ii). Using vectors find the area of the triangle with the vertices A(1, 2, 3), B(2, -1, 4), and C(4, 5, -1).

Solution:

Given, A(1, 2, 3), B(2, -1, 4), and C(4, 5, -1)

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

=> \vec{b} = B = 2\hat{i}-1\hat{j}+4\hat{k}

=> \vec{c} = C = 4\hat{i}+5\hat{j}-1\hat{k}

Now 2 sides of the triangle are given by,

=> \vec{AB} = \vec{b}-\vec{a}

=> \vec{AB} = (2\hat{i}-1\hat{j}+4\hat{k})-(4\hat{i}+5\hat{j}-1\hat{k})

=> \vec{AB} = \hat{i}[2-1] +\hat{j}[-1-2]+\hat{j}[4-3]

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

=> \vec{AC} = \vec{c}-\vec{a}

=> \vec{AC} = (4\hat{i}+5\hat{j}-1\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})

=> \vec{AC} = \hat{i}[4-1] +\hat{j}[5-2]+\hat{j}[-1-3]

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

Area of the triangle whose adjacent sides are given is \dfrac{1}{2}|\vec{a}\times\vec{b}|



=> \vec{AB}\times\vec{AC} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & -3 & 1\\3 & 3 & -4\end{vmatrix}

=> \vec{AB}\times\vec{AC} = \hat{i}[12-3]-\hat{j}[-4-3]+\hat{k}[3+9]

=> \vec{AB}\times\vec{AC} = 9\hat{i}+7\hat{j}+12\hat{k}

Thus area of the triangle is,

=> Area = \dfrac{1}{2}\sqrt{(9)^2+(7)^2+12^2}

=> Area = \dfrac{1}{2}\sqrt{81+49+144}

=> Area = √274/2

Question 35. Find all the vectors of magnitude 10\sqrt{3}   that are perpendicular to the plane of \hat{i}+2\hat{j}+\hat{k}  and -\hat{i}+3\hat{j}+4\hat{k}  .

Solution:

Given, \vec{a} = \hat{i}+2\hat{j}+\hat{k}  and \vec{b}=\hat{i}+3\hat{j}+4\hat{k}

A vector perpendicular to both \vec{a}   and \vec{b}   is,

=> \vec{a}\times\vec{b} =  \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1 & 2 & 1\\-1 & 3 & 4\end{vmatrix}

=> \vec{a}\times\vec{b} = \hat{i}[8-3]-\hat{j}[4+1]+\hat{k}[3+2]

=> \vec{a}\times\vec{b} = 5\hat{i}-5\hat{j}+5\hat{k}

Unit vector is,

=> \hat{p} = \dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}

=> \hat{p} = \dfrac{1}{\sqrt{5^2+(-5)^2+5^2}}(5\hat{i}-5\hat{j}+5\hat{k})

=> \hat{p} = \dfrac{1}{5\sqrt{3}}(5\hat{i}-5\hat{j}+5\hat{k})

=> \hat{p} = \dfrac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})

Now vectors of magnitude 10\sqrt{3}   are given by,

=> 10\sqrt{3}\hat{p} = \pm10\sqrt{3}\times \dfrac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})



=> Required vectors, \pm10(\hat{i}-\hat{j}+\hat{k})

Question 36. The adjacent sides of a parallelogram are 2\hat{i}-4\hat{j}-5\hat{k}  and 2\hat{i}+2\hat{j}+3\hat{k}  . Find the 2 unit vectors parallel to its diagonals. Also, find its area of the parallelogram.

Solution:

Given, \vec{AB}=2\hat{i}-4\hat{j}-5\hat{k}   and \vec{BC} = 2\hat{i}+2\hat{j}+3\hat{k}

=> \vec{AC} = \vec{AB} + \vec{BC}

=> \vec{AC} = (2\hat{i}-4\hat{j}-5\hat{k})+(2\hat{i}+2\hat{j}+3\hat{k})

=> \vec{AC} = 4\hat{i}-\hat{j}-2\hat{k}

Unit vector is,

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

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

=> \hat{p} = \dfrac{1}{\sqrt{21}}(4\hat{i}-\hat{j}-2\hat{k})

Area is given by |\vec{AB}\times\vec{BC}|  ,




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