Question

If a=i+3j-k and b=2i-j+k determine unit vector parallel to a×b also find sine of angle between a and b

Answer 1

Given,

• $\vec{\sf{a}}=\sf{\hat i+3\hat j-\hat k}$

• $\vec{\sf{b}}=\sf{2\hat i-\hat j+\hat k}$

The cross product of these two vectors is given by,

$\longrightarrow\vec{\sf{a}}\times\vec{\sf{b}}=\left|\begin{array}{ccc}\sf{\hat i}&\sf{\hat j}&\sf{\hat k}\\\sf{1}&\sf{3}&\sf{-1}\\\sf{2}&\sf{-1}&\sf{1}\end{array}\right|$

$\longrightarrow\vec{\sf{a}}\times\vec{\sf{b}}=\sf{(3-1)\,\hat i-(1+2)\,\hat j+(-1-6)\,\hat k}$

$\longrightarrow\vec{\sf{a}}\times\vec{\sf{b}}=\sf{2\,\hat i-3\,\hat j-7\,\hat k}$

The magnitude of this vector is,

$\longrightarrow\left|\vec{\sf{a}}\times\vec{\sf{b}}\right|=\left|\sf{2\,\hat i-3\,\hat j-7\,\hat k}\right|$

$\longrightarrow\left|\vec{\sf{a}}\times\vec{\sf{b}}\right|=\sf{\sqrt{2^2+(-3)^2+(-7)^2}}$

$\longrightarrow\left|\vec{\sf{a}}\times\vec{\sf{b}}\right|=\sf{\sqrt{4+9+49}}$

$\longrightarrow\left|\vec{\sf{a}}\times\vec{\sf{b}}\right|=\sf{\sqrt{62}}$

Hence the unit vector parallel to $\vec{\sf{a}}\times\vec{\sf{b}}$ is,

$\longrightarrow\mathsf{\hat a}\times\mathsf{\hat b}=\dfrac{\vec{\sf{a}}\times\vec{\sf{b}}}{\left|\vec{\sf{a}}\times\vec{\sf{b}}\right|}$

$\longrightarrow\underline{\underline{\mathsf{\hat a}\times\mathsf{\hat b}=\mathsf{\dfrac{1}{\sqrt{62}}}\left(\sf{2\,\hat i-3\,\hat j-7\,\hat k\right)}}}$

Magnitude of each vector $\vec{\sf{a}}$ and $\vec{\sf{b}}$ is,

$\longrightarrow\sf{a}=\left|\hat i+3\hat j-\hat k\right|$

$\longrightarrow\sf{a}=\sqrt{1^2+3^2+(-1)^2}$

$\longrightarrow\sf{a}=\sqrt{1+9+1}$

$\longrightarrow\sf{a}=\sqrt{11}$

And,

$\longrightarrow\sf{b}=\left|2\hat i-\hat j+\hat k\right|$

$\longrightarrow\sf{b}=\sqrt{2^2+(-1)^2+1^2}$

$\longrightarrow\sf{b}=\sqrt{4+1+1}$

$\longrightarrow\sf{b}=\sqrt{6}$

Let $\theta$ be the angle between $\vec{\sf{a}}$ and $\vec{\sf{b}}.$

Sine of this angle is given by,

$\longrightarrow\left|\vec{\sf{a}}\times\vec{\sf{b}}\right|=\sf{ab\sin\theta}$

$\longrightarrow\sin\theta=\dfrac{\left|\vec{\sf{a}}\times\vec{\sf{b}}\right|}{\sf{ab}}$

$\longrightarrow\sin\theta=\dfrac{\sf{\sqrt{62}}}{\sf{\sqrt{11}\times\sqrt{6}}}$

$\longrightarrow\sin\theta=\dfrac{\sf{\sqrt{62}}}{\sf{\sqrt{66}}}$

$\longrightarrow\underline{\underline{\sin\theta=\sf{\sqrt{\dfrac{31}{33}}}}}$

Answer 2

Hope it help you guys...!!!?✌️✌️