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Let $\pi_1$ be the plane determined by the vectors $\hat{i}+\hat{j}$ and $\hat{j}+\hat{k}, \pi_2$ be the plane determined by the vectors $\hat{i}-\hat{j}$ and $\hat{i}+\hat{j}-\hat{k}$. Let $\vec{a}$ be a vector parallel to the line of intersection of $\pi_1$ and $\pi_2$. If $|\vec{a}|=\sqrt{14}$, then $|\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})|=$
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The correct answer is:
$2$
For $\pi_1: \vec{n}_1=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right|=\hat{i}-\hat{j}+\hat{k}$
For $\pi_2: \overrightarrow{\mathrm{n}}_2=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 0 \\ 1 & 1 & -1\end{array}\right|=\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
For line $: \vec{b}=\vec{n}_1 \times \vec{n}_2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 1 & 2\end{array}\right|=-3 \hat{i}-\hat{j}+2 \hat{k}$
$\therefore \vec{a}=\lambda(3 \hat{i}+\hat{j}-2 \hat{k})$
$|\vec{a}|=\sqrt{9 \lambda^2+\lambda^2+4 \lambda^2}=\sqrt{14} \Rightarrow \lambda^2=1 \Rightarrow \lambda= \pm 1$
Now, $|\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})|=| \pm 1(3 \hat{i}+\hat{j}-2 \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})|=2$
For $\pi_2: \overrightarrow{\mathrm{n}}_2=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 0 \\ 1 & 1 & -1\end{array}\right|=\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
For line $: \vec{b}=\vec{n}_1 \times \vec{n}_2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 1 & 2\end{array}\right|=-3 \hat{i}-\hat{j}+2 \hat{k}$
$\therefore \vec{a}=\lambda(3 \hat{i}+\hat{j}-2 \hat{k})$
$|\vec{a}|=\sqrt{9 \lambda^2+\lambda^2+4 \lambda^2}=\sqrt{14} \Rightarrow \lambda^2=1 \Rightarrow \lambda= \pm 1$
Now, $|\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})|=| \pm 1(3 \hat{i}+\hat{j}-2 \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})|=2$
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