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What are the values of $x$ for which the two vectors
$\left(\mathrm{x}^{2}-1\right) \hat{\mathrm{i}}+(\mathrm{x}+2) \hat{\mathrm{j}}+\mathrm{x}^{2} \hat{\mathrm{k}}$ and $2 \hat{\mathrm{i}}-\mathrm{x} \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ are orthogonal?
Options:
$\left(\mathrm{x}^{2}-1\right) \hat{\mathrm{i}}+(\mathrm{x}+2) \hat{\mathrm{j}}+\mathrm{x}^{2} \hat{\mathrm{k}}$ and $2 \hat{\mathrm{i}}-\mathrm{x} \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ are orthogonal?
Solution:
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Verified Answer
The correct answer is:
$\mathrm{x}=-\frac{1}{2}$ and $\mathrm{x}=1$
Ifvectors $\left(x^{2}-1\right) \hat{i}+(x+2) \hat{j}+x^{2} \hat{k}$ and $2 \hat{i}-x \hat{j}+3 \hat{k}$ are
orthogonal, then $2\left(x^{2}-1\right)-x(x+2)+3 x^{2}=0$
$\Rightarrow 2 x^{2}-2-x^{2}-2 x+3 x^{2}=0$
$\Rightarrow 4 \mathrm{x}^{2}-2 \mathrm{x}-2=0$
$\Rightarrow\left(2 \mathrm{x}^{2}-\mathrm{x}-1\right)=0$
$\Rightarrow(2 x+1)(x-1)=0$
$\Rightarrow \mathrm{x}=-\frac{1}{2}$ and $\mathrm{x}=1$
orthogonal, then $2\left(x^{2}-1\right)-x(x+2)+3 x^{2}=0$
$\Rightarrow 2 x^{2}-2-x^{2}-2 x+3 x^{2}=0$
$\Rightarrow 4 \mathrm{x}^{2}-2 \mathrm{x}-2=0$
$\Rightarrow\left(2 \mathrm{x}^{2}-\mathrm{x}-1\right)=0$
$\Rightarrow(2 x+1)(x-1)=0$
$\Rightarrow \mathrm{x}=-\frac{1}{2}$ and $\mathrm{x}=1$
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