A vector of magnitude sqrt{51} which makes eq | vedclass

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(C) Let the required vector be $\bar{r} = x\bar{i} + y\bar{j} + z\bar{k}$.Given that $\bar{a}$,$\bar{b}$,and $\bar{c}$ are unit vectors. $\bar{a} = \f

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$-5 \bar{i}+\bar{j}+5 \bar{k}$

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(C) Let the required vector be $\bar{r} = x\bar{i} + y\bar{j} + z\bar{k}$.Given that $\bar{a}$,$\bar{b}$,and $\bar{c}$ are unit vectors. $\bar{a} = \frac{1}{3}(\bar{i}-2\bar{j}+2\bar{k})$,$|\bar{a}| = \frac{1}{3}\sqrt{1^2+(-2)^2+2^2} = 1$.$\bar{b} = \frac{1}{5}(-4\bar{i}-3\bar{k})$,$|\bar{b}| = \frac{1}{5}\sqrt{(-4)^2+0^2+(-3)^2} = 1$.$\bar{c} = \bar{j}$,$|\bar{c}| = 1$.Let $	heta$ be the angle between $\bar{r}$ and each of $\bar{a}, \bar{b}, \bar{c}$. Then $\bar{r} \cdot \bar{a} = \bar{r} \cdot \bar{b} = \bar{r} \cdot \bar{c} = |\bar{r}| \cos 	heta = \sqrt{51} \cos 	heta = k$.$\frac{1}{3}(x-2y+2z) = k \implies x-2y+2z = 3k$.$\frac{1}{5}(-4x-3z) = k \implies -4x-3z = 5k$.$y = k$.Substituting $y=k$ in the first equation: $x-2k+2z = 3k \implies x+2z = 5k$.From the second equation: $-4x-3z = 5k$.Solving for $x$ and $z$: $2x+4z = 10k$ and $-4x-3z = 5k$. Adding $2 	imes (x+2z=5k)$ and $(-4x-3z=5k)$ gives $z = 15k$ and $x = -25k$.Since $|\bar{r}|^2 = 51$,we have $(-25k)^2 + k^2 + (15k)^2 = 51 \implies (625+1+225)k^2 = 51 \implies 851k^2 = 51$. This suggests a re-evaluation of the vector components. Checking option $C$: $\bar{r} = -5\bar{i}+\bar{j}+5\bar{k}$,$|\bar{r}| = \sqrt{25+1+25} = \sqrt{51}$.$\bar{r} \cdot \bar{a} = \frac{1}{3}(-5+2+10) = 1$. $\bar{r} \cdot \bar{b} = \frac{1}{5}(20-15) = 1$. $\bar{r} \cdot \bar{c} = 1$. Since all dot products are equal,the vector is $-5\bar{i}+\bar{j}+5\bar{k}$.

$A$ vector of magnitude $\sqrt{51}$ which makes equal angles with the vectors $\bar{a}=\frac{1}{3}(\bar{i}-2 \bar{j}+2 \bar{k})$,$\bar{b}=\frac{1}{5}(-4 \bar{i}-3 \bar{k})$ and $\bar{c}=\bar{j}$,is

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