Question

Let \[\overrightarrow{a}\] and \[\overrightarrow{b}\] be two unit vector such that\[\overrightarrow{a}.\overrightarrow{b}=0\]. For some\[x,y\in R\], let \[c=x\overrightarrow{a}+y\overrightarrow{b}+(\overrightarrow{a}\times \overrightarrow{b})\]. If \[|c|=2\] and the vector c is inclined at some angle \[\alpha \] to both a and b then the value of \[8{{\cos }^{2}}\alpha \] is …. .

Answer 1

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