If the position vectors of three points A, B | vedclass

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(B) The position vectors are $\vec{A} = i + j + k$,$\vec{B} = 2i + 3j - 4k$,and $\vec{C} = 7i + 4j + 9k$. First,we find the vectors $\vec{AB}$ and $\v

Vedclass · Accepted Answer

$\frac{31i - 38j - 9k}{\sqrt{2486}}$

Vedclass · Answer

(B) The position vectors are $\vec{A} = i + j + k$,$\vec{B} = 2i + 3j - 4k$,and $\vec{C} = 7i + 4j + 9k$. First,we find the vectors $\vec{AB}$ and $\vec{AC}$: $\vec{AB} = \vec{B} - \vec{A} = (2-1)i + (3-1)j + (-4-1)k = i + 2j - 5k$. $\vec{AC} = \vec{C} - \vec{A} = (7-1)i + (4-1)j + (9-1)k = 6i + 3j + 8k$. The vector perpendicular to the plane is given by the cross product $\vec{n} = \vec{AB} 	imes \vec{AC}$: $\vec{n} = \begin{vmatrix} i & j & k \ 1 & 2 & -5 \ 6 & 3 & 8 \end{vmatrix} = i(16 - (-15)) - j(8 - (-30)) + k(3 - 12) = 31i - 38j - 9k$. The magnitude of this vector is $|\vec{n}| = \sqrt{31^2 + (-38)^2 + (-9)^2} = \sqrt{961 + 1444 + 81} = \sqrt{2486}$. The unit vector perpendicular to the plane is $\hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{31i - 38j - 9k}{\sqrt{2486}}$.

If the position vectors of three points $A, B$ and $C$ are respectively $i + j + k, 2i + 3j - 4k$ and $7i + 4j + 9k$,then the unit vector perpendicular to the plane containing the triangle $ABC$ is

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