A unit vector in the plane of the vectors 2i | vedclass

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(B) Let the required unit vector be $\hat{a}$. Since $\hat{a}$ lies in the plane of $\vec{u} = 2i + j + k$ and $\vec{v} = i - j + k$,it can be express

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$\frac{3j - k}{\sqrt{10}}$

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(B) Let the required unit vector be $\hat{a}$. Since $\hat{a}$ lies in the plane of $\vec{u} = 2i + j + k$ and $\vec{v} = i - j + k$,it can be expressed as $\hat{a} = \alpha(2i + j + k) + \beta(i - j + k)$.Simplifying,we get $\hat{a} = (2\alpha + \beta)i + (\alpha - \beta)j + (\alpha + \beta)k$.Since $\hat{a}$ is a unit vector,its magnitude is $1$,so $(2\alpha + \beta)^2 + (\alpha - \beta)^2 + (\alpha + \beta)^2 = 1$.Expanding this,we get $6\alpha^2 + 4\alpha\beta + 3\beta^2 = 1$ ... $(i)$.Since $\hat{a}$ is orthogonal to $\vec{w} = 5i + 2j + 6k$,their dot product is zero: $5(2\alpha + \beta) + 2(\alpha - \beta) + 6(\alpha + \beta) = 0$.This simplifies to $18\alpha + 9\beta = 0$,which gives $\beta = -2\alpha$.Substituting $\beta = -2\alpha$ into equation $(i)$,we get $6\alpha^2 + 4\alpha(-2\alpha) + 3(-2\alpha)^2 = 1$.$6\alpha^2 - 8\alpha^2 + 12\alpha^2 = 1$,which implies $10\alpha^2 = 1$,so $\alpha = \pm \frac{1}{\sqrt{10}}$.Then $\beta = -2(\pm \frac{1}{\sqrt{10}}) = \mp \frac{2}{\sqrt{10}}$.Substituting these values into the expression for $\hat{a}$,we get $\hat{a} = \pm \frac{1}{\sqrt{10}} [ (2 - 2)i + (1 + 2)j + (1 - 2)k ] = \pm \frac{3j - k}{\sqrt{10}}$.

$A$ unit vector in the plane of the vectors $2i + j + k$ and $i - j + k$ and orthogonal to $5i + 2j + 6k$ is

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