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If $\triangle \mathrm{ABC}$ is right angled at $\mathrm{A}$, where $\mathrm{A} \equiv(4,2, x), \mathrm{B} \equiv(3,1,8)$ and $\mathrm{C} \equiv(2,-1,2)$, then the value of $x$ is
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The correct answer is:
$3$
Since $\triangle \mathrm{ABC}$ is right angled at $\mathrm{A}$,
$$
\begin{aligned}
& \overline{\mathrm{AB}} \cdot \overline{\mathrm{AC}}=0 \\
& \Rightarrow[-\hat{\mathrm{i}}-\hat{\mathrm{j}}+(8-x) \hat{\mathrm{k}}] \cdot(-2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+(2-x) \hat{\mathrm{k}})=0 \\
& \Rightarrow 2+3+(8-x)(2-x)=0 \\
& \Rightarrow x^2-10 x+21=0 \\
& \Rightarrow(x-3)(x-7)=0 \\
& \Rightarrow x=3 \text { or } x=7
\end{aligned}
$$
$$
\begin{aligned}
& \overline{\mathrm{AB}} \cdot \overline{\mathrm{AC}}=0 \\
& \Rightarrow[-\hat{\mathrm{i}}-\hat{\mathrm{j}}+(8-x) \hat{\mathrm{k}}] \cdot(-2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+(2-x) \hat{\mathrm{k}})=0 \\
& \Rightarrow 2+3+(8-x)(2-x)=0 \\
& \Rightarrow x^2-10 x+21=0 \\
& \Rightarrow(x-3)(x-7)=0 \\
& \Rightarrow x=3 \text { or } x=7
\end{aligned}
$$
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