A unit vector orthogonal to the vector vec{a} | vedclass

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(D) Let the required vector be $\vec{v}$. Since $\vec{v}$ is coplanar with $\vec{b}$ and $\vec{c}$,it can be written as $\vec{v} = \vec{b} 	imes (\ve

Vedclass · Accepted Answer

$\frac{1}{\sqrt{6}}(\hat{i} - 2\hat{j} + \hat{k})$

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(D) Let the required vector be $\vec{v}$. Since $\vec{v}$ is coplanar with $\vec{b}$ and $\vec{c}$,it can be written as $\vec{v} = \vec{b} 	imes (\vec{b} 	imes \vec{c})$ or simply as a linear combination $\vec{v} = x\vec{b} + y\vec{c}$.Alternatively,a vector coplanar with $\vec{b}$ and $\vec{c}$ and orthogonal to $\vec{a}$ is parallel to $\vec{a} 	imes (\vec{b} 	imes \vec{c})$.First,calculate $\vec{n} = \vec{b} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1 - (-1)) - \hat{j}(1 - 1) + \hat{k}(-1 - 1) = 2\hat{i} - 2\hat{k}$.Now,find the vector orthogonal to $\vec{a}$ and $\vec{n}$:$\vec{v} = \vec{a} 	imes \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 4 & 5 \ 2 & 0 & -2 \end{vmatrix} = \hat{i}(-8 - 0) - \hat{j}(-6 - 10) + \hat{k}(0 - 8) = -8\hat{i} + 16\hat{j} - 8\hat{k}$.Simplifying,we take the direction $\hat{i} - 2\hat{j} + \hat{k}$.The magnitude is $\sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$.Thus,the unit vector is $\frac{1}{\sqrt{6}}(\hat{i} - 2\hat{j} + \hat{k})$.Therefore,the correct option is $D$.

$A$ unit vector orthogonal to the vector $\vec{a} = 3\hat{i} + 4\hat{j} + 5\hat{k}$ and coplanar with the vectors $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} + \hat{k}$ is:

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