Question

solve all the questions ..☺☺

Answer 1

hello dear  brother !!


question 40 )

⇒ given : A = 2i - 4j + 6k  and  B = 3i + j + 2k 

we know that , to find angle between two vectors the formula that is used is ,

cosФ  = $\frac{A.B}{|A| |B|}$

implies ,

cosФ =  $\frac{6-4+12}{ \sqrt{4+16+36}. \sqrt{9 + 1 + 4} }$

implies,

cosФ =  $\frac{14}{ \sqrt{56} \sqrt{14} }$  =  1/2 

therefore cosФ = 1/2 implies Ф = 60° 
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question 41 )

given that : A = i + j + k and B = -i -j +2k 

implies ,

cosФ = $\frac{A.B}{|A| |B|}$  

A.B = -1 -1 + 2 = 0 , since A.B = 0 

cosФ = 0  implies , Ф  = 90° 
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question 42 )

we know that if two vectors are perpendicular then their dot product is equals to zero therefore , 

A.B = 0 

implies ,  (2i + 3j - 6k ) ( 3i - mj + 6k ) = 0 

therefore ,  6 - 3m - 36 = 0   

m = - 10 
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question 43 )

i )  A = i + j + k and B = -2i -2j -2k

cosФ  = $\frac{A.B}{|A|.|B|}$  

ccos Ф = $\frac{-6}{ \sqrt{3} \sqrt{12} }$  = - 1

therefore , Ф = 180°

ii )  A = -2i + 2j - k and B = 3i + 6j +2k 

cosФ = $\frac{-6 + 12 -2}{ \sqrt{9} \sqrt{49} }$

implies , cosФ = 4 / 21 = 0.19 

we know cos60° = 1/2 = 0.5 , while comparing the answer that we arrived it is clearly noticed that the value lies between 0 - 0.5 therefore the value of Ф lies between 60° and 90° , appoximately  79° 

iii ) 

A = 4i + 6j - 3k and B = -2i -5j + 7k

cosФ = $\frac{-8-35-21}{ \sqrt{16 + 36 + 9} \sqrt{4 + 25 + 49} }$

cosФ  =$\frac{-64}{ \sqrt{61} \sqrt{78} }$

cosФ  = -64 / 68.9 

cosФ = - 0.412

we know cos(148.9) ≈ -0.319

therefore by equating we get Ф = 148 

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HOPE IT HELPS !!