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(A) Let the required vector be $\vec{v} = x\,i + y\,j + z\,k$. Since $\vec{v}$ is perpendicular to $\vec{a} = 3\,i + j - 4\,k$ and $\vec{b} = 6\,i + 5

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$2\,i - 2\,j + k$

Vedclass · Answer

(A) Let the required vector be $\vec{v} = x\,i + y\,j + z\,k$. Since $\vec{v}$ is perpendicular to $\vec{a} = 3\,i + j - 4\,k$ and $\vec{b} = 6\,i + 5\,j - 2\,k$,it must be parallel to $\vec{a} 	imes \vec{b}$. $\vec{a} 	imes \vec{b} = \begin{vmatrix} i & j & k \ 3 & 1 & -4 \ 6 & 5 & -2 \end{vmatrix} = i(-2 - (-20)) - j(-6 - (-24)) + k(15 - 6) = 18\,i - 18\,j + 9\,k$. Simplifying,we can take the direction vector $\vec{u} = 2\,i - 2\,j + k$. The magnitude of $\vec{u}$ is $|\vec{u}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$. Since the required vector has length $3$,the vector is $\pm(2\,i - 2\,j + k)$. Comparing with the given options,$2\,i - 2\,j + k$ is the correct choice.

$A$ vector of length $3$ perpendicular to each of the vectors $3\,i + j - 4\,k$ and $6\,i + 5\,j - 2\,k$ is

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