Question

With reference to a right handed system of mutually perpendicular unit vectors i, j, k, $\alpha =3i-j$ and $\beta =2i+j-3k$. If $\beta ={{\beta }_{1}}+{{\beta }_{2}}$, where ${{\beta }_{1}}$ is parallel to $\alpha$ and ${{\beta }_{2}}$ is perpendicular to $\alpha$, then
(a) ${{\beta }_{1}}=\dfrac{3}{2}i+\dfrac{1}{2}j$ ${{\beta }_{1}}=\dfrac{3}{2}i+\dfrac{1}{2}j$
(b) ${{\beta }_{1}}=\dfrac{3}{2}i-\dfrac{1}{2}j$
(c) ${{\beta }_{2}}=\dfrac{1}{2}i+\dfrac{3}{2}j-3k$
(d) ${{\beta }_{2}}=\dfrac{1}{2}i-\dfrac{3}{2}j-3k$

Answer 1

Hint: We know that when two vectors are parallel to each other, say if, a and b are parallel to each other, then we can represent a in terms of b in the form of the relation a=kb, where k is a real constant. Here, we will find ${{\beta }_{1}}$ in the form of $\lambda \alpha$, where $\lambda$ is a real constant and then we will use the relation$\beta ={{\beta }_{1}}+{{\beta }_{2}}$ to find ${{\beta }_{2}}$.

Complete step-by-step answer:
Here, we can see that ${{\beta }_{1}}$ and $\alpha$ are parallel to each other, so we may take ${{\beta }_{1}}=\lambda \alpha$.
Or,${{\beta }_{1}}=\lambda (3i-j)$
${{\beta }_{1}}=3\lambda i-\lambda j...........(1)$

Now, we may check option (a), so on comparing equation (1) with option (a) we get:
$\begin{aligned} & 3\lambda =\dfrac{3}{2} \\\ & \lambda =\dfrac{1}{2} \\\ \end{aligned}$

And also:
$\begin{aligned} & -\lambda =\dfrac{1}{2} \\\ & \lambda =-\dfrac{1}{2} \\\ \end{aligned}$

Here, we can see that we are getting two different values for $\lambda$ , so in such case there will be no real value of $\lambda$.

Now, we may check for option (b) and again comparing equation (1) with option (b), we get:
$\begin{aligned} & 3\lambda =\dfrac{3}{2} \\\ & \lambda =\dfrac{1}{2} \\\ \end{aligned}$

And also:
$\begin{aligned} & -\lambda =-\dfrac{1}{2} \\\ & \lambda =\dfrac{1}{2} \\\ \end{aligned}$

Here, we can see that both the conditions give the value of $\lambda$ as $\dfrac{1}{2}$.
Hence, we have got a real value of $\lambda$ as $\dfrac{1}{2}$.

So, now we have ${{\beta }_{1}}=\dfrac{3}{2}i-\dfrac{1}{2}j$

And, we also have an equation If $\beta ={{\beta }_{1}}+{{\beta }_{2}}$……….. (2)

So, we may substitute here the values of $\beta$ and ${{\beta }_{1}}$ in equation (2) and so we have:
$2i+j-3k=\dfrac{3}{2}i-\dfrac{1}{2}j+{{\beta }_{2}}$
Or, $(2-\dfrac{3}{2})i+(1+\dfrac{1}{2})j-3k={{\beta }_{2}}$
Or, $\dfrac{1}{2}i+\dfrac{3}{2}j-3k={{\beta }_{2}}$
Hence, the correct options are (b) and (c).

Note: Here, it should be noted that since$\beta$ is parallel to $\alpha$, so $\beta$ can be represented as a scalar $\lambda$ multiplied by the vector $\alpha$ i.e. $\beta =\lambda \alpha$. To avoid mistakes calculations and comparisons must be done properly.