A unit vector which is coplanar to vectors i | vedclass

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(B) Let the required vector be $\vec{v} = ai + bj + ck$. For $\vec{v}$ to be coplanar with $\vec{u_1} = i + j + 2k$ and $\vec{u_2} = i + 2j + k$,it mu

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$\pm \frac{j - k}{\sqrt{2}}$

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(B) Let the required vector be $\vec{v} = ai + bj + ck$. For $\vec{v}$ to be coplanar with $\vec{u_1} = i + j + 2k$ and $\vec{u_2} = i + 2j + k$,it must be a linear combination of $\vec{u_1}$ and $\vec{u_2}$. So,$\vec{v} = p(i + j + 2k) + r(i + 2j + k) = (p+r)i + (p+2r)j + (2p+r)k$. Comparing components,we get $a = p+r$,$b = p+2r$,and $c = 2p+r$. Since $\vec{v}$ is perpendicular to $\vec{w} = i + j + k$,their dot product is zero: $(ai + bj + ck) \cdot (i + j + k) = a + b + c = 0$. Substituting the expressions for $a, b, c$: $(p+r) + (p+2r) + (2p+r) = 4p + 4r = 0$,which implies $p = -r$. Substituting $p = -r$ into the expressions for $a, b, c$: $a = -r + r = 0$,$b = -r + 2r = r$,$c = 2(-r) + r = -r$. Thus,$\vec{v} = r(j - k)$. For $\vec{v}$ to be a unit vector,$|\vec{v}| = 1$: $\sqrt{0^2 + r^2 + (-r)^2} = 1 \Rightarrow \sqrt{2r^2} = 1 \Rightarrow |r|\sqrt{2} = 1 \Rightarrow r = \pm \frac{1}{\sqrt{2}}$. Therefore,the required unit vector is $\pm \frac{1}{\sqrt{2}}(j - k)$.

$A$ unit vector which is coplanar to vectors $i + j + 2k$ and $i + 2j + k$ and perpendicular to $i + j + k$ is

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