Question

A vector of magnitude √7 units makes equal angles 53° each with X and Y axes. its component along
z axis :-
(A) 1.2 units
(B) 1.4 units
(C) 1.6 units
(D) 1.8 units

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Answer 1

Answer:

1.4

Explanation:

look kido we have two ways to do that

1. (cos alpha)^2+(cos beta)^2+(cos gamma)^2=1

well i can do the derivation but it will just stretch the answer

cos 53 *cos53+cos53*cos53+(cos gama)^2=1

9/25+9/25+(cos gamma)^2=1

(cos gamma)^2=1-18/25

(cos gamma)^2=7/25

(cos gamma)=root7/5 eq 1

now

Az=|A|*cos theta

proof:cos theta=base/hyptenius

cos theta=Az/|A|

Az=|A|*cos theta

so Az=root 7* root 7/5(value of cos theta from eq 1)

Az=7/5

Az=1.4 Units

now the second method

its simple

|A|^2=Ax^2+Ay^2+Az^2

7=(cos 53* root 7)^2+(cos 53* root 7)^2+Az^2

7=(root 7*3/5)^2+(root7*3/5)^2+Az^2

7=7*9/25+7*9/25+Az^2

7-126/25=Az^2

175-126/25=Az^2

49/25=Az^2

7/5=Az

Az=1.6

im bad at explanation but hope it helps

Answer 2

Answer:

The correct answer is option (B) 1.4 units.

Explanation:

Given:

A vector of magnitude √7 units makes equal angles 53° each with X and Y axes.

To find:

the component along z axis

Step 1

If the angle made by a vector with x-axis exists a, with y-axis exists b and with z-axis exists c, then we have,

$(\cos a)^{2} +(\cos b)^{2} +(\cos c)^{2} =1$

According to the question, a = b = 53 degree

We have,

$\cos 53=3 / 5$

$(3 / 5)^{2} +(3 / 5)^{2} +(\cos c)^{2} =1$

$18 / 25+(\cos \mathrm{c})^{2} =1$

$(\cos \mathrm{c})^{2} =1-18 / 25$

$(\cos \mathrm{c})^{2} =7 / 25$

$\cos \mathrm{c}=\sqrt{7} / 25$

Step 2

Component of the vector along z-axis = magnitude of the vector $*$ cos c

$&=\sqrt{7 } * \sqrt{7 / 25}$

$&=7 / \sqrt{ 25}$

$&=7 / 5$

7 / 5 = 1.4 units

So, the component of the vector along the z-axis exists 7 / 5.

The component along the z-axis is 1.4 units.

Therefore, the correct answer is option (B) 1.4 units.

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