Question

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Give proper answer

Answer 1

Final Answer : 1: A)
${ \cos }^{ - 1} ( \frac{1}{ \sqrt{3} })$
2 : A)
${ \cos}^{ - 1 } ( \frac{ \sqrt{2} }{ \sqrt{3} } )$

Solution : 1:
Let the side of cube be of length 'a' units .
Then,
A = (0,0,0)
B = (0,0,a)
C= ( a, a, 0)
D = (a, a, a)
Here, we need to find angle between AB and AD.
Vector AB = (0,0,a) - (0,0,0) = (0,0,a)
Vector AD = (a, a, a) -(0,0,0) = ( a, a, a)

Now, Angle between AB and AD is given by :

cos (α) = AB. AD/ (|AB| | AD| )

$= \frac{(0.a + 0.a + a.a)}{( \sqrt{ {0}^{2} + {0}^{2} + {a}^{2} } ) \times ( \sqrt{ {a}^{2} + {a}^{2} + {a}^{2} } )} \\ = \frac{ {a}^{2} }{a.a \sqrt{3} } \\ = \frac{1}{ \sqrt{3} }$

Hence,
$\alpha = { \cos }^{ - 1} ( \frac{1}{ \sqrt{3} } )$

Solution : 2 :
Angle between AC and AD.
AC = (a, a, 0)-(0,0,0) = (a, a, 0)

Now ,
cos(β) = AC. AD/(|AC| |AD|)

$= \frac{a.a + a.a + 0.a}{ \sqrt{ {a}^{2} + {a}^{2} + {0}^{2} } \sqrt{ {a}^{2} + {a}^{2} + {a}^{2} } } \\ = \frac{2 {a}^{2} }{(a \sqrt{2})(a \sqrt{3} ) } = \frac{ \sqrt{2} }{ \sqrt{3} }$
$= > \cos( \beta ) = \frac{ \sqrt{2} }{ \sqrt{3} } \\ = > \beta = { \cos }^{ - 1} ( \frac{ \sqrt{2} }{ \sqrt{3} } )$

Hence, Required angle is β .

Answer 2

option a is the correct answer buddy

Answer 3

option a is the correct answer buddy

Answer 4

option a is the correct answer buddy