Question

If $A=2i-3j+4k$, its components in $yz$ plane and $zx$ plane are respectively:
A. $\sqrt{13}$ and 5
B. 5 and $2\sqrt{5}$
C. $2\sqrt{5}$ and $\sqrt{13}$
D. $\sqrt{13}$ and $\sqrt{29}$

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Answer

To solve the problem, we need to find the components of the vector $ \mathbf{A} = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k} $ in the $ yz $ and $ zx $ planes.

In the $ yz $ Plane:

  • Consider only the $ y $ and $ z $ components of vector $ \mathbf{A} $.
  • The $ y $ component is $ -3 $ and the $ z $ component is $ 4 $.
  • The magnitude of the component of $ \mathbf{A} $ in the $ yz $ plane can be calculated using the formula: $$ \sqrt{y^2 + z^2} $$ Here: $$ y = -3 \quad \text{and} \quad z = 4 $$ Thus: $$ \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$

In the $ zx $ Plane:

  • Consider only the $ z $ and $ x $ components of vector $ \mathbf{A} $.
  • The $ z $ component is $ 4 $ and the $ x $ component is $ 2 $.
  • The magnitude of the component of $ \mathbf{A} $ in the $ zx $ plane can be calculated using the formula: $$ \sqrt{z^2 + x^2} $$ Here: $$ z = 4 \quad \text{and} \quad x = 2 $$ Thus: $$ \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} $$

Therefore, the components of vector $ \mathbf{A} $ in the $ yz $ and $ zx $ planes respectively are:

  • In the $ yz $ plane: $ \mathbf{5} $
  • In the $ zx $ plane: $ 2\sqrt{5} $

Thus, the correct answer is:

B. 5 and $ 2\sqrt{5} $


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