Question

Determine the unit vector perpendicular to the plane of A = 3î - 4j -k and B = 4 î - 3j – 2k vectors
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Answer 1

Answer:

$\dfrac{5i + 2j-k}{\sqrt{30 } }$

Explanation:

Get a vector perpendicular to both the given vectors using cross product in the direction of plane.

$\overrightarrow{A} \times \overrightarrow{B} \\= (3i - 4j -k) \times (4i - 3j -2k)$

$\left[\begin{array}{ccc} i & j & k\\3&-4&-1\\4&-3&-2\end{array}\right]$

$\left[\begin{array}{ccc}-4&-1\\-3&-2\end{array}\right] i -\left[\begin{array}{ccc}3&-1\\4&-2\end{array}\right] j + \left[\begin{array}{ccc}3&-4\\4&-3\end{array}\right] k$

$(8-3)i - (-6 +4)j + (-9 + 8)k\\=5i + 2j-k$

To find unit vector in the direction divide that vector with it's magnitude.

$\dfrac{5i + 2j-k}{\sqrt{25 + 4 +1 } } \\\\= \dfrac{5i + 2j-k}{\sqrt{30 } }$