A unit vector in the plane of i + 2j + k and | vedclass

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Question

(D) Let the two vectors be $\vec{a} = i + 2j + k$ and $\vec{b} = i + j + 2k$. Any vector in the plane of $\vec{a}$ and $\vec{b}$ is given by $\vec{v}

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let the two vectors be $\vec{a} = i + 2j + k$ and $\vec{b} = i + j + 2k$. Any vector in the plane of $\vec{a}$ and $\vec{b}$ is given by $\vec{v} = \vec{a} + \lambda \vec{b}$ or $\vec{v} = \vec{a} 	imes \vec{b}$. However,a vector perpendicular to both $\vec{a}$ and $\vec{b}$ is $\vec{n} = \vec{a} 	imes \vec{b}$. We need a vector $\vec{r}$ in the plane of $\vec{a}$ and $\vec{b}$ such that $\vec{r} \perp (2i + j + k)$. Let $\vec{r} = \vec{a} 	imes (\vec{a} 	imes \vec{b}) = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$. Alternatively,consider the vector $\vec{v} = \vec{a} - \vec{b} = (i + 2j + k) - (i + j + 2k) = j - k$. Check if $\vec{v}$ is perpendicular to $2i + j + k$: $(j - k) \cdot (2i + j + k) = (0)(2) + (1)(1) + (-1)(1) = 1 - 1 = 0$. Since the dot product is $0$,the vector $j - k$ lies in the plane and is perpendicular to $2i + j + k$. The unit vector is $\frac{j - k}{|j - k|} = \frac{j - k}{\sqrt{0^2 + 1^2 + (-1)^2}} = \frac{j - k}{\sqrt{2}}$.

$A$ unit vector in the plane of $i + 2j + k$ and $i + j + 2k$ which is perpendicular to $2i + j + k$ is

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