Math, asked by supraja2003, 10 months ago

If a=2i+2j+k, a.b=14, axb=3i+j-8k,
then b =
1) 5i-j+2k 2) 5i+j-k
3) 5i+j+2k 4) 5i+j-2k​

Answers

Answered by Anonymous
7

Answer:

\large\boxed{\sf{(3)\;5\hat{ i   } + \hat{j}+2\hat{k}}}

Step-by-step explanation:

From the question, given that,

 \vec{ a  } = 2\hat{   i} + 2\hat{j   } + \hat{ k  } \\  \\ \vec{  a }.\vec{  b} = 14 \\  \\ \vec{  a } \times \vec{   b} = 3\hat{  i } + \hat{  j } - 8\hat{  k }

To find the \vec{b}, let's assume that,

\vec{  b } = x\hat{  i } + y\hat{ j  } +z \hat{  k }

Therefore, we will get the dot product as,

 =  > \vec{  a }.\vec{ b  } = 2x + 2y + z \\  \\  =  > 2x + 2y + z = 14 \:  \:  \:  \:  \:  \:  \:............ (1)

Also, finding the cross product, we get,

 =  > \vec{ a  } \times \vec{   b} = (2\hat{  i } + 2\hat{  j } + \hat{  k }) \times (x\hat{ i  } + y\hat{   j} + z\hat{   k})

But, we know that,

  • \hat{ i  } \times \hat{   i} = \hat{  j } \times \hat{ j  } = \hat{  k } \times \hat{  k } = 0
  • \hat{ i  } \times \hat{   j} = \hat{ k  }
  • \hat{  j } \times \hat{ k  } = \hat{  i }
  • \hat{k   } \times \hat{ i  } = \hat{ j  }

Therefore, we will get,

 =  > \vec{   a}  \times \vec{b   } = (2z - y)\hat{  i } + (x - 2z)\hat{  j } + (2 - 2x)\hat{k   } \\  \\  =  > (2z - y)\hat{  i } + (x - 2z)\hat{  j } + (2y - 2x)\hat{k   } = 3\hat{  i } + \hat{  j } - 8\hat{ k  }

Comparing the coefficient, we have,

2z - y = 3 \:  \:  \:  \:  \: .......(2) \\  \\ x - 2z = 1  \:  \:  \:  \:  \:  \:  \:  \:  \: .............(3)\\  \\ 2y - 2x =  - 8 \:  \:  \:  \:  \:  \: ..........(4)

From eqn (2)and (3), we have,

 =  > x = 1 + 2z   \\   \\  =  > y = 2z - 3

Now, substituting these values in eqn (1), we get,

 =  > 2(1 + 2z) + 2(2z - 3) + z = 14 \\  \\  =  > 2 + 4z + 4z - 6 + z = 14 \\  \\  =  > 9z - 4 = 14 \\  \\  =  > 9z = 14 + 4 = 18 \\  \\  =  > z =  \frac{18}{9}   \\  \\  =  > z= 2

Therefore, we will get,

 =  > x = 1 +( 2 \times 2) = 1 + 4 \\  \\  =  > x = 5

and,

 =  > y =( 2 \times 2) - 3  = 4 - 3\\  \\  =  > y = 1

Hence, we get the required vector,

 \bold{(3) \: \vec{  b }  = 5\hat{  i } + \hat{ j  } + 2\hat{ k  }}

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