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

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