Question

Let $x$ be a vector in the plane containing vectors $\overrightarrow{a}=2\hat{i}–\hat{j}+\hat{k}$ and $\overrightarrow{b}=\hat{i}+2\hat{j}–\hat{k}$. If the vector $x$ is perpendicular to $(3\hat{i}+2\hat{j}–\hat{k})$ and its projection on $a$ is $\frac{17\sqrt{6}}{2}$ , then the value of$|x{|}^2$ is equal to

Answer 1

Determine the value of $|x{|}^2$.

Step 1: Find the value of $\text{'}\overrightarrow{x}\text{'}$

Let $\overrightarrow{x}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}(\lambda \text{and}\mu \text{are scalar})$ [Since $\overrightarrow{x},\overrightarrow{a},\overrightarrow{b}$ containing in same plane]

Substitute $\overrightarrow{a}=2\hat{i}–\hat{j}+\hat{k}$ and $\overrightarrow{b}=\hat{i}+2\hat{j}–\hat{k}$

Now,

$$\begin{gathered} \overrightarrow{x}=\lambda \left(2\hat{i}–\hat{j}+\hat{k}\right)+\mu \left(\hat{i}+2\hat{j}+\hat{k}\right)......\left(1\right) \\ \overrightarrow{x}=\hat{i}(2\lambda +\mu )+\hat{j}(2\mu -\lambda )+\hat{k}(\lambda -\mu ) \end{gathered}$$

Since $x$ is perpendicular to $(3\hat{i}+2\hat{j}–\hat{k})$ [Given]

$\overrightarrow{x}.(3\hat{i}+2\hat{\hat{j}}-\hat{k})=0$

$$\begin{gathered} \Rightarrow 3\left(2\lambda +\mu \right)+2\left(2\mu -\lambda \right)-\left(\lambda -\mu \right)=0 \\ \Rightarrow 6\lambda +3\mu +4\mu -2\lambda -\lambda +\mu =0 \\ \Rightarrow 3\lambda +8\mu =0.....\left(2\right) \end{gathered}$$

Also projection of $\overrightarrow{x}\text{on}\overrightarrow{a}\text{is}\frac{17\sqrt{6}}{2}$ [Given]

$\frac{\overrightarrow{x}.\overrightarrow{a}}{\left|\overrightarrow{a}\right|}=\frac{17\sqrt{6}}{2}......\left(3\right)$

Step 2: Find the value of $\overrightarrow{x}.\overrightarrow{a}$ and $\left|\overrightarrow{a}\right|$

Find the value of $\overrightarrow{x}.\overrightarrow{a}$

$$\begin{gathered} \Rightarrow \overrightarrow{x}.\overrightarrow{a}=2\left(2\lambda +\mu \right)+\left(2\mu -\lambda \right)+\left(\lambda -\mu \right) \\ \Rightarrow \overrightarrow{x}.\overrightarrow{a}=6\lambda -\mu \end{gathered}$$

Find the value of $\left|\overrightarrow{a}\right|$

$$\begin{gathered} \left|\overrightarrow{a}\right|=\sqrt{{\left(2\right)}^2–{\left(1\right)}^2+{\left(1\right)}^2} \\ \left|\overrightarrow{a}\right|=\sqrt{4+1+1} \\ \left|\overrightarrow{a}\right|=\sqrt{6} \end{gathered}$$

Step 3: Find the value of ${\left|x\right|}^2$

Then, the Equation (3) becomes

$$\begin{gathered} \frac{6\lambda -\mu }{\sqrt{6}}=\frac{17\sqrt{6}}{2} \\ 6\lambda -\mu =\frac{17{\left(\sqrt{6}\right)}^2}{2} \\ 6\lambda -\mu =\frac{17\times 6}{2} \\ 6\lambda -\mu =17\times 2 \\ 6\lambda -\mu =51......\left(4\right) \end{gathered}$$

Consider the Equation (2) and multiply by $2$

$$\begin{gathered} 2\left(3\lambda +8\mu \right)=0 \\ 6\lambda +16\mu =0......\left(5\right) \end{gathered}$$

Substract Equation (4) by Equation (5)

$$\begin{gathered} \Rightarrow \left(6\lambda -\mu \right)-\left(6\lambda +16\mu \right)=51 \\ \Rightarrow 6\lambda -\mu -6\lambda -16\mu =51 \\ \Rightarrow -\mu -16\mu =51 \\ \Rightarrow -17\mu =51 \\ \Rightarrow 17\mu =-\frac{51}{17} \\ \Rightarrow \mu =-3 \end{gathered}$$

Substitute the vale $\mu =-3$ in the Equation (4)

$$\begin{gathered} \Rightarrow 6\lambda -\mu =51 \\ \Rightarrow 6\lambda -\left(-3\right)=51 \\ \Rightarrow 6\lambda +3=51 \\ \Rightarrow 6\lambda =51-3 \\ \Rightarrow 6\lambda =48 \\ \Rightarrow \lambda =\frac{48}{6} \\ \Rightarrow \lambda =8 \end{gathered}$$

Substitute the vale $\lambda =8$ and $\mu =-3$ in the Equation (1)

$$\begin{gathered} \Rightarrow \overrightarrow{x}=8\left(2\hat{i}–\hat{j}+\hat{k}\right)+\left(-3\right)\left(\hat{i}+2\hat{j}-\hat{k}\right) \\ \Rightarrow \overrightarrow{x}=16\hat{i}–8\hat{j}+8\hat{k}-3\hat{i}-6\hat{j}-3\hat{k} \\ \Rightarrow \overrightarrow{x}=13\hat{i}-14\hat{j}+11\hat{k} \\ \Rightarrow {\left|\overrightarrow{x}\right|}^2={\left(\sqrt{{13}^2-{14}^2+{11}^2}\right)}^2 \\ \Rightarrow {\left|\overrightarrow{x}\right|}^2=169+196+121 \\ \Rightarrow {\left|\overrightarrow{x}\right|}^2=486 \end{gathered}$$

Hence, the value of ${\left|x\right|}^2$ is $486$