Let vec{a}=x hat{i}+y hat{j}+z hat{k} and x=2 | vedclass

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(C) Given $\vec{a}=x \hat{i}+y \hat{j}+z \hat{k}$ and $x=2 y$. $|\vec{a}| = \sqrt{x^2+y^2+z^2} = \sqrt{(2y)^2+y^2+z^2} = \sqrt{5y^2+z^2}$. Given $|\ve

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$2 \sqrt{5} \hat{i}+\sqrt{5} \hat{j}-5 \hat{k}$

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(C) Given $\vec{a}=x \hat{i}+y \hat{j}+z \hat{k}$ and $x=2 y$. $|\vec{a}| = \sqrt{x^2+y^2+z^2} = \sqrt{(2y)^2+y^2+z^2} = \sqrt{5y^2+z^2}$. Given $|\vec{a}| = 5 \sqrt{2}$,so $5y^2+z^2 = (5 \sqrt{2})^2 = 50$. Since $\vec{a}$ makes an angle of $135^{\circ}$ with the $z$-axis,the component along the $z$-axis is $z = |\vec{a}| \cos 135^{\circ}$. $z = 5 \sqrt{2} 	imes (-\frac{1}{\sqrt{2}}) = -5$. Substituting $z = -5$ into the equation $5y^2+z^2 = 50$: $5y^2 + (-5)^2 = 50 \Rightarrow 5y^2 + 25 = 50 \Rightarrow 5y^2 = 25 \Rightarrow y^2 = 5 \Rightarrow y = \pm \sqrt{5}$. Since $x = 2y$,we have $x = \pm 2 \sqrt{5}$. Thus,$\vec{a} = \pm 2 \sqrt{5} \hat{i} \pm \sqrt{5} \hat{j} - 5 \hat{k}$. Comparing with the options,the correct vector is $2 \sqrt{5} \hat{i} + \sqrt{5} \hat{j} - 5 \hat{k}$.

Let $\vec{a}=x \hat{i}+y \hat{j}+z \hat{k}$ and $x=2 y$. If $|\vec{a}|=5 \sqrt{2}$ and $\vec{a}$ makes an angle of $135^{\circ}$ with the $z$-axis,then $\vec{a}=$

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