Question

If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$, $\overrightarrow{b}=4\hat{i}+3\hat{j}+4\hat{k}$ and $\overrightarrow{c}=\hat{i}+\alpha \hat{j}+\beta \hat{k}$ are linearly dependent vectors and $|\overrightarrow{c}|=\sqrt{3}$ then the value of $\alpha$ may be.

• A. $1$
• B. $2$
• C. $-3$
• D. $4$

Answer 1

Answer: (A)

$1$

According to the question:

$\begin{array}{c}\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}\\ \overrightarrow{b}=4\hat{i}+3\hat{j}+4\hat{k}\\ \overrightarrow{c}=\hat{i}+\alpha \hat{j}+\beta \hat{k}\\ Sincea,b,carelinearlydependent,\\ \Rightarrow \left|\begin{array}{c}111\\ 434\\ 1\alpha \beta \end{array}\right|=0\\ \Rightarrow \left|\begin{array}{c}100\\ 4-10\\ 1(\alpha -1)(\beta -1)\end{array}\right|=0(by{c}_2\to c_2-c_{1}and{c}_3\to c_3-c_1)\\ \Rightarrow 1(-1)(\beta -1)\\ \therefore \beta =1\\ Now,|\overrightarrow{c}|=\sqrt{3}\\ \overrightarrow{c}=\hat{i}+\alpha \hat{j}+\beta \hat{k}\\ \Rightarrow |\overrightarrow{c}|=\sqrt{1+{\alpha }^2+{\beta }^2}\\ \Rightarrow \sqrt{3}=\sqrt{1+{\alpha }^2+{\beta }^2}\\ \Rightarrow 3=1+{\alpha }^2+{\beta }^2\\ \Rightarrow {\alpha }^2+{\beta }^2=2\\ \Rightarrow {\alpha }^2+1=2\\ \Rightarrow {\alpha }^2=\sqrt{1}\\ \therefore \alpha =1\\ sothatthevalueof\alpha is1,andthecoeerctoptionisA.\end{array}$