Question

Let $u$ be a vector coplanar with the vectors $\overrightarrow{a}=2\hat{i}+3\hat{j}-\hat{k}$ and $\overrightarrow{b}=\hat{j}+\hat{k}$. If vector $u$ is perpendicular to vector $a$ and $\overrightarrow{u}.\overrightarrow{b}=24$, then $u^2$ is equal to:

• A. $256$
• B. $84$
• C. $336$
• D. $315$

Answer 1

The correct option is C

$336$

Explanation for the correct option:

Finding the value of $u^2$

Given $\overrightarrow{a}=2\hat{i}+3\hat{j}-\hat{k}$ and $\overrightarrow{b}=\hat{j}+\hat{k}$, $\overrightarrow{u}.\overrightarrow{b}=24$

$\overrightarrow{u}=u_1\hat{i}+u_2\hat{j}+u_3\hat{k}$

Since vectors $\overrightarrow{u},\overrightarrow{a},\overrightarrow{b}$ are coplanar so

$\left|\begin{array}{ccc}{u}_1& u_2& u_3\\ 2& 3& -1\\ 0& 1& 1\end{array}\right|=0$

Solving the above matrix, we get :

$$\begin{gathered} 4u_1-2u_2+2u_3=0 \\ 2u_1-u_2+u_3=0...\left(1\mathbf{\right)}\mathbf{[}\mathbf{\text{Dividing throughout by}}\mathbf{}\mathbf{2}\mathbf{]} \\ \stackrel{\rightharpoonup }{u}·\stackrel{\rightharpoonup }{a}=0\mathbf{[}\mathbf{\text{Since}}\mathbf{}\mathit{u}\mathbf{\perp }\mathit{a}\mathbf{}\mathbf{\text{then dot product}}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{]} \\ 2u_1+3u_2-u_3=0...\left(2\right) \\ \stackrel{\rightharpoonup }{u}·\stackrel{\rightharpoonup }{b}=24\mathbf{[}\mathbf{\text{Since}}\mathbf{}\mathit{u}\mathbf{\perp }\mathit{b}\mathbf{}\mathbf{\text{then dot product}}\mathbf{}\mathbf{=}\mathbf{0}] \\ {u}_2+u_3=24...\left(3\right) \end{gathered}$$

Solving $\left(1\right),\left(2\right),and\left(3\right)$ using the simultaneous equation method,we get the value of ;

$u_1=-4,u_2=8,u_3=16$

$$\begin{gathered} \left|u^2\right|=\left|{u_1}^2+{u_2}^2+{u_3}^2\right| \\ \Rightarrow \left|u^2\right|=|{(-4)}^2+{\left(8\right)}^2+{\left(16\right)}^2| \\ \Rightarrow \left|u^2\right|=16+64+256 \\ \Rightarrow \left|u^2\right|=336 \end{gathered}$$

Hence, the correct option is (C)