Question

If $\overline{a}=\widehat{i}+\widehat{j}+\widehat{k}$ , $\overline{b}=4\widehat{i}+3\widehat{j}+4\widehat{k}$ , $\overline{c}=\widehat{i}+\alpha \widehat{j}+\beta \widehat{k}$ , linearly dependent vectors and $|\overline{c}|=\sqrt{3}$ then
(a)$\alpha =1,\beta =-1$
(b)$\alpha =1,\beta =\pm 1$
(c)$\alpha =-1,\beta =\pm 1$
(d)$\alpha =\pm 1,\beta =1$

Answer 1

Hint: As it is given in the question that $\overline{a}\text{, }\overline{b}\text{ and }\overline{c}$ are linearly dependent vectors, so we can say that $l\overline{a}+m\overline{b}=\overline{c}$ . Also, it is given that $|\overline{c}|=\sqrt{3}$ , we can say that $\sqrt{1+{{\alpha }^{2}}+{{\beta }^{2}}}=\sqrt{3}$ .

Complete step-by-step answer:
It is given that $\overline{a}\text{, }\overline{b}\text{ and }\overline{c}$ are linearly dependent vectors, so this can be mathematically represented as
$l\overline{a}+m\overline{b}=\overline{c}$
Provided l, m are not zero at the same time. So, if we substitute the values of $\overline{a}\text{, }\overline{b}\text{ and }\overline{c}$ with the values given in the question, we get
$l\left( \widehat{i}+\widehat{j}+\widehat{k} \right)+m\left( 4\widehat{i}+3\widehat{j}+4\widehat{k} \right)=\left( \widehat{i}+\alpha \widehat{j}+\beta \widehat{k} \right)$
$\Rightarrow l\widehat{i}+l\widehat{j}+l\widehat{k}+4m\widehat{i}+3m\widehat{j}+4m\widehat{k}=\widehat{i}+\alpha \widehat{j}+\beta \widehat{k}$
$\Rightarrow \left( l+4m \right)\widehat{i}+\left( l+3m \right)\widehat{j}+\left( l+4m \right)\widehat{k}=\widehat{i}+\alpha \widehat{j}+\beta \widehat{k}$
We know, if two vectors are equal, there coefficients of all $\widehat{i},\text{ }\widehat{j}\text{ and }\widehat{k}$ must be equal. So, applying this for our above equation, we get
$\begin{align}
  & l+4m=1............(i) \\
 & l+3m=\alpha ...........(ii) \\
 & l+4m=\beta ..........(iii) \\
\end{align}$
So, if we see equation (i) and equation (iii), we can say that the value of $\beta =1$ .
Also, it is given that $|\overline{c}|=\sqrt{3}$ , we can say that $\sqrt{1+{{\alpha }^{2}}+{{\beta }^{2}}}=\sqrt{3}$ . And if we substitute the value $\beta =1$ and solve the equation, we get
 $\sqrt{1+{{\alpha }^{2}}+1}=\sqrt{3}$
On, squaring both sides, we get
$2+{{\alpha }^{2}}=3$
$\Rightarrow {{\alpha }^{2}}=1$
Now we know that ${{a}^{2}}=b$ implies $a=\pm \sqrt{b}$ . So, our equation becomes:
$\alpha =\pm 1$
So, we can conclude that the answer to the above question is option (d).

Note: It is important to remember the properties of the vector product and the scalar product for solving most of the problems related to vectors. Also, be careful about the calculations and the signs you are using while solving the calculation. It is important to know the relations between dependent and independent vectors as well.