Question

Find unit vector perpendicular to the plane determined by the vectors a = 4i + 3j - k and b = 2i - 6j -3k.

Answer 1

Answer:

$\bigg(\dfrac{-3}{7} , \dfrac{2}{7}, \dfrac{-6}{7} \bigg) \text { or } \bigg(\dfrac{3}{7} , \dfrac{-2}{7}, \dfrac{6}{7} \bigg)$

Step-by-step explanation:

$\text{Find } \overrightarrow a \times \overrightarrow b:$

$\overrightarrow a \times \overrightarrow b = \left|\begin{array}{ccc}i&j&k\\4&3&-1\\2&-6&-3\end{array}\right|$

$\overrightarrow a \times \overrightarrow b = (3 \times (-3) - (-1 )\times (-6))i - (4 \times (-3) - (-1) \times (2))j + (4 \times (-6) - (3 \times 2)k$

$\overrightarrow a \times \overrightarrow b = (-15)i - (-10)j + (-30)k$

$\overrightarrow a \times \overrightarrow b = -15k + 10j - 30k$

$\overrightarrow a \times \overrightarrow b = \bigg(-15, 10, - 30 \bigg)$

$\text{Find } |\overrightarrow a \times \overrightarrow b| :$

$| a \times \overrightarrow b| = \sqrt{(-15)^2 + (10)^2 + (-30)^2}$

$| a \times \overrightarrow b| = \sqrt{1225}$

$| a \times \overrightarrow b| = 35$

$\text{Find unit vector} :$

$\dfrac{\overrightarrow a \times \overrightarrow b}{|\overrightarrow a \times \overrightarrow b|} = \bigg(\dfrac{-15}{35} , \dfrac{10}{35}, \dfrac{-30}{35} \bigg) \text { or } \bigg(\dfrac{15}{35} , \dfrac{-10}{35}, \dfrac{30}{35}\bigg)$

$\dfrac{\overrightarrow a \times \overrightarrow b}{|\overrightarrow a \times \overrightarrow b|}= \bigg(\dfrac{-3}{7} , \dfrac{2}{7}, \dfrac{-6}{7} \bigg) \text { or } \bigg(\dfrac{3}{7} , \dfrac{-2}{7}, \dfrac{6}{7} \bigg)$