Answer
Verified
386.7k+ views
Hint: We are given magnitude of a and b as 1 because they are unit vectors and there angle with \[c=x\overrightarrow{a}+y\overrightarrow{b}+(\overrightarrow{a}\times \overrightarrow{b})\] is \[\alpha \] , so we will first calculate dot product of a and b with c, then we will square the c vector because we know the magnitude of c vector also , then after solving we will get the results .
Complete step-by-step answer:
We are given \[\overrightarrow{a}\] and \[\overrightarrow{b}\] be two unit vector such that \[\overrightarrow{a}.\overrightarrow{b}=0\], it means magnitude of vector a and b is 1 and angle between them is 90, also given a vector \[c=x\overrightarrow{a}+y\overrightarrow{b}+(\overrightarrow{a}\times \overrightarrow{b})\] and \[|c|=2\]
vector c is inclined at some angle \[\alpha \] to both a and b, for this let’s take dot product of c and a
then c and b
\[\overrightarrow{c}.\overrightarrow{a}=(x\overrightarrow{a}+y\overrightarrow{b}+(\overrightarrow{a}\times \overrightarrow{b})).\overrightarrow{a}\] , solving LHS and RHS differently and applying formula \[\overrightarrow{a}.\overrightarrow{b}=|a||b|cos\alpha \], we get
\[\overrightarrow{c}.\overrightarrow{a}=|c||a|cos\alpha \], now on putting \[|\overrightarrow{c}|=2\],\[|\overrightarrow{a}|=1\] and \[\alpha \] angle between them
Which on solving both side we get \[2\times 1\times \cos \alpha =x\]
Similarly applying this for c and b we get \[2\times 1\times \cos \alpha =y\]
Now on putting values of x and y in vector c we get equation like \[c=2\cos \alpha \overrightarrow{a}+2\cos \alpha \overrightarrow{b}+(\overrightarrow{a}\times \overrightarrow{b})\]
Further solving gives \[c=2\cos \alpha (\overrightarrow{a}+\overrightarrow{b})+(\overrightarrow{a}\times \overrightarrow{b})\]
Now on squaring both side equation will look like, applying formula \[{{(\overrightarrow{a}+\overrightarrow{b})}^{2}}=|\overrightarrow{a}{{|}^{2}}+|\overrightarrow{b}{{|}^{2}}+2\overrightarrow{a}.\overrightarrow{b}\]\[|c{{|}^{2}}=4{{\cos }^{2}}\alpha {{(\overrightarrow{a}+\overrightarrow{b})}^{2}}+{{(\overrightarrow{a}\times \overrightarrow{b})}^{2}}+2\cos \alpha (\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}\times \overrightarrow{b})\]
Now here we know that \[|\overrightarrow{c}|=2\],\[{{(\overrightarrow{a}+\overrightarrow{b})}^{2}}=1+1=2\] , \[{{(\overrightarrow{a}\times \overrightarrow{b})}^{2}}=1\times 1\times \sin 90=1\] and \[(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}\times \overrightarrow{b})=0\] because a and b are perpendicular
So, on putting values of these in equation \[|c{{|}^{2}}=4{{\cos }^{2}}\alpha {{(\overrightarrow{a}+\overrightarrow{b})}^{2}}+{{(\overrightarrow{a}\times \overrightarrow{b})}^{2}}+2\cos \alpha (\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}\times \overrightarrow{b})\]
We get \[4=8{{\cos }^{2}}\alpha +1\], which on solving gives
\[3=8{{\cos }^{2}}\alpha \], hence answer is 3
Note: Some of the students might have a doubt that how can we write this \[(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}\times \overrightarrow{b})=0\]
It is because cross product of two vectors is always perpendicular those vectors so on taking dot product with the corresponding vectors it results into 0
Complete step-by-step answer:
We are given \[\overrightarrow{a}\] and \[\overrightarrow{b}\] be two unit vector such that \[\overrightarrow{a}.\overrightarrow{b}=0\], it means magnitude of vector a and b is 1 and angle between them is 90, also given a vector \[c=x\overrightarrow{a}+y\overrightarrow{b}+(\overrightarrow{a}\times \overrightarrow{b})\] and \[|c|=2\]
vector c is inclined at some angle \[\alpha \] to both a and b, for this let’s take dot product of c and a
then c and b
\[\overrightarrow{c}.\overrightarrow{a}=(x\overrightarrow{a}+y\overrightarrow{b}+(\overrightarrow{a}\times \overrightarrow{b})).\overrightarrow{a}\] , solving LHS and RHS differently and applying formula \[\overrightarrow{a}.\overrightarrow{b}=|a||b|cos\alpha \], we get
\[\overrightarrow{c}.\overrightarrow{a}=|c||a|cos\alpha \], now on putting \[|\overrightarrow{c}|=2\],\[|\overrightarrow{a}|=1\] and \[\alpha \] angle between them
Which on solving both side we get \[2\times 1\times \cos \alpha =x\]
Similarly applying this for c and b we get \[2\times 1\times \cos \alpha =y\]
Now on putting values of x and y in vector c we get equation like \[c=2\cos \alpha \overrightarrow{a}+2\cos \alpha \overrightarrow{b}+(\overrightarrow{a}\times \overrightarrow{b})\]
Further solving gives \[c=2\cos \alpha (\overrightarrow{a}+\overrightarrow{b})+(\overrightarrow{a}\times \overrightarrow{b})\]
Now on squaring both side equation will look like, applying formula \[{{(\overrightarrow{a}+\overrightarrow{b})}^{2}}=|\overrightarrow{a}{{|}^{2}}+|\overrightarrow{b}{{|}^{2}}+2\overrightarrow{a}.\overrightarrow{b}\]\[|c{{|}^{2}}=4{{\cos }^{2}}\alpha {{(\overrightarrow{a}+\overrightarrow{b})}^{2}}+{{(\overrightarrow{a}\times \overrightarrow{b})}^{2}}+2\cos \alpha (\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}\times \overrightarrow{b})\]
Now here we know that \[|\overrightarrow{c}|=2\],\[{{(\overrightarrow{a}+\overrightarrow{b})}^{2}}=1+1=2\] , \[{{(\overrightarrow{a}\times \overrightarrow{b})}^{2}}=1\times 1\times \sin 90=1\] and \[(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}\times \overrightarrow{b})=0\] because a and b are perpendicular
So, on putting values of these in equation \[|c{{|}^{2}}=4{{\cos }^{2}}\alpha {{(\overrightarrow{a}+\overrightarrow{b})}^{2}}+{{(\overrightarrow{a}\times \overrightarrow{b})}^{2}}+2\cos \alpha (\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}\times \overrightarrow{b})\]
We get \[4=8{{\cos }^{2}}\alpha +1\], which on solving gives
\[3=8{{\cos }^{2}}\alpha \], hence answer is 3
Note: Some of the students might have a doubt that how can we write this \[(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}\times \overrightarrow{b})=0\]
It is because cross product of two vectors is always perpendicular those vectors so on taking dot product with the corresponding vectors it results into 0
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Which type of bond is stronger ionic or covalent class 12 chemistry CBSE
What organs are located on the left side of your body class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
When people say No pun intended what does that mea class 8 english CBSE