Let $\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=-5 \hat{i}+7 \hat{j} \quad$ and $\quad \vec{c}=3 \hat{i}+y…

Let $\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=-5 \hat{i}+7 \hat{j} \quad$ and $\quad \vec{c}=3 \hat{i}+y \hat{j} \quad$ be three vectors such that $|\vec{a}-\vec{b}+\vec{c}|=\sqrt{141}$. If $y_1$ and $y_2$ are the values of $y$ satisfying the given condition, then $\left|\mathrm{y}_1-\mathrm{y}_2\right|=$
  1. 12
  2. 11
  3. 9
  4. 8

Solution

$\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=-5 \hat{i}+7 \hat{j}, \vec{c}=3 \hat{i}+y \hat{j}$ $\begin{aligned} & \vec{a}-\vec{b}+\vec{c}=(3 \hat{i}+\hat{j}-2 \hat{k})-(-5 \hat{i}+7 \hat{j})+(3 \hat{i}+y \hat{j}) \\ & \Rightarrow \vec{a}-\vec{b}+\vec{c}=11 \hat{i}+(y-6) \hat{j}-2 \hat{k} \\ & \Rightarrow|\vec{a}-\vec{b}+\vec{c}|=\sqrt{121+(y-6)^2+4} \\ & \Rightarrow \sqrt{141}=\sqrt{121+(y-6)^2+4} \\ & \Rightarrow 16=(y-6)^2 \\ & \Rightarrow y-6= \pm 4\end{aligned}$ Taking positive sign $y_1-6=4 \Rightarrow y_1=10$ Taking negative sign $y_2-6=-4 \Rightarrow y_2=2$ Now $\left|y_2-y_1\right|=|2-10|=8$

Asked in: AP EAMCET 2023 (17 May Shift 2)

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