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torisob [31]
4 years ago
8

Given, Q =x^b/x^c If Q=1 then proof that b=c

Mathematics
1 answer:
LuckyWell [14K]4 years ago
4 0

Answer:

  1. Q = x^ b / x^c
  2. Q = x^(b-c)
  3. 1 = x^ (b-c)
  4. x^0 = x^ (b-c)
  5. 0 = b - c
  6. b = c

Step-by-step explanation:

  1. if the same base are in division then the powers are subtracted
  2. x^0 = 1
  3. if there is same base in an equation then the power of the base can be the equation
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Given:

The two vectors are:

\overrightarrow{a}=2\hat{i}-\hat{j}+\hat{k}

\overrightarrow{b}=\hat{i}-3\hat{j}+5\hat{k}

To find:

The value of |\overrightarrow{a}\times \overrightarrow{b}|.

Solution:

We have,

\overrightarrow{a}=2\hat{i}-\hat{j}+\hat{k}

\overrightarrow{b}=\hat{i}-3\hat{j}+5\hat{k}

The cross product of these two vectors is:

\overrightarrow{a}\times \overrightarrow{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&-1&1\\1&-3&5\end{vmatrix}

\overrightarrow{a}\times \overrightarrow{b}=\hat{i}[(-1)(5)-(1)(-3)]-\hat{j}[(2)(5)-(1)(1)]+\hat{k}[(2)(-3)-(-1)(1)]

\overrightarrow{a}\times \overrightarrow{b}=\hat{i}[-5+3]-\hat{j}[10-1]+\hat{k}[-6+1]

\overrightarrow{a}\times \overrightarrow{b}=-2\hat{i}-9\hat{j}-5\hat{k}

Now the magnitude of the cross product is:

|\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{(-2)^2+(-9)^2+(-5)^2}

|\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{4+81+25}

|\overrightarrow{a}\times \overrightarrow{b}|=\sqrt{110}

Therefore, the value of |\overrightarrow{a}\times \overrightarrow{b}| is \sqrt{110}.

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Answer:

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Step-by-step explanation:

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Answer:

1, 000, 000

Step-by-step explanation:
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Read 2 more answers
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