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Question

(A) Let $\vec{a} = 2i - j + k$ and $\vec{b} = 3i + 4j - k$. $A$ vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by the cross product $\v

Vedclass · Accepted Answer

$\frac{-3i + 5j + 11k}{\sqrt{155}}$

Vedclass · Answer

(A) Let $\vec{a} = 2i - j + k$ and $\vec{b} = 3i + 4j - k$. $A$ vector perpendicular to both $\vec{a}$ and $\vec{b}$ is given by the cross product $\vec{a} 	imes \vec{b}$. $\vec{a} 	imes \vec{b} = \begin{vmatrix} i & j & k \ 2 & -1 & 1 \ 3 & 4 & -1 \end{vmatrix} = i(1 - 4) - j(-2 - 3) + k(8 + 3) = -3i + 5j + 11k$. The magnitude of this vector is $|\vec{a} 	imes \vec{b}| = \sqrt{(-3)^2 + 5^2 + 11^2} = \sqrt{9 + 25 + 121} = \sqrt{155}$. Therefore,the unit vector perpendicular to both is $\frac{\vec{a} 	imes \vec{b}}{|\vec{a} 	imes \vec{b}|} = \frac{-3i + 5j + 11k}{\sqrt{155}}$.

$A$ unit vector perpendicular to each of the vectors $2i - j + k$ and $3i + 4j - k$ is equal to

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