Question

please Answer these Questions step by step .......... ​

Answer 1

Solution:-

2) The projection length of the vector 3i + 1j + 2k on y-z Plane is

$\to \rm\: vector \: = 3i \: + 1j \: + 2k \:$

we know that

=> y is called j and k is called z

To find projection length of Y - Z , use this formula

$\rm \: length \: = \sqrt{ {y}^{2} + {z}^{2} } = \sqrt{1 {}^{2} + {2}^{2} } \\ = \sqrt{1 + 4} = \sqrt{5}$

Answer

$\rm \: = \sqrt{5}$

2) The projection length of the vector 3i + 1j + 2k on X - Z Plane is

$\to \rm\: vector \: = 3i \: + 1j \: + 2k \:$

=> x is called i and k is called z

$\rm \: length \: = \sqrt{ {x}^{2} + {z}^{2} } = \sqrt{3 {}^{2} + {2}^{2} } \\ = \sqrt{9 + 4} = \sqrt{13}$

Answer

$= \sqrt{13}$

4) if A = 2i - 3j + 4k, its components along xy plane is

$\rm \:vector \to \: 2i - 3j + 4k$

Component along xy plane is

$\rm \: components \: along \: xy \: plane \: is \\ = \rm \sqrt{ {x}^{2} + {y}^{2} } = \sqrt{ {2}^{2} + ( - 3) {}^{2} } = \sqrt{4 + 9} = \sqrt{13}$

5. if A = 2i - 3j + 4k, its components along yz plane is

$\rm \:vector \to \: 2i - 3j + 4k$

Component along yz plane is

$\rm \: components \: along \: yz \: plane \: is \\ = \rm \sqrt{ {y}^{2} + {z}^{2} } = \sqrt{ {( - 3)}^{2} + ( 4) {}^{2} } = \sqrt{9 + 16} = \sqrt{25} = 5$

6.if A = 2i - 3j + 4k, its components along xz plane is

$\rm \:vector \to \: 2i - 3j + 4k$

Component along xzplane is

$\rm \: components \: along \: xz \: plane \: is \\ = \rm \sqrt{ {x}^{2} + {z}^{2} } = \sqrt{ {( 2)}^{2} + ( 4) {}^{2} } = \sqrt{4+ 16} = \sqrt{20} = 2 \sqrt{5}$

7. The projection length of the vector 2i - 4j - 6k on xy Plane is

$\rm \to \: vector = 2i - 4j - 6k$

The projection length of xy is

$\rm \: \sqrt{ {x}^{2} + {y}^{2} } = \sqrt{ {2}^{2} + (- 4) {}^{2} } \\ = \sqrt{4 + 16} = \sqrt{20} = 2 \sqrt{5}$

Answer 2

Answer:

ask the answer above is correct akka......