Question

Find unit vector perpendicular to both i + j + k and 2i + j + 3k.

Answer 1

Answer:

$\bigg( \dfrac{2}{\sqrt{6} } , \dfrac{-1}{\sqrt{6} }, \dfrac{-1}{\sqrt{6} } \bigg) \text { or } \bigg( \dfrac{-2}{\sqrt{6} } , \dfrac{1}{\sqrt{6} }, \dfrac{1}{\sqrt{6} } \bigg)$

Step-by-step explanation:

$\text {Find } \overrightarrow A \times \overrightarrow B:$

$\overrightarrow A \times \overrightarrow B = \overrightarrow A \times \overrightarrow B$

$\overrightarrow A \times \overrightarrow B = \left|\begin{array}{ccc}i&j&k\\1&1&1\\2&1&3\end{array}\right|$

$\overrightarrow A \times \overrightarrow B = (1 \times 3 - 1 \times 1)i - (1 \times 3 - 2 \times 1)j + (1 \times 1 - 1 \times 2)k$

$\overrightarrow A \times \overrightarrow B = 2i - j - k$

$\overrightarrow A \times \overrightarrow B = \bigg(2, -1, -1 \bigg)$

$\text {Find } | \ \overrightarrow A \times \overrightarrow B| :$

$| \overrightarrow A \times \overrightarrow B| = \sqrt{(2)^2 + (-1)^2 + (-1)^2}$

$| \overrightarrow A \times \overrightarrow B| = \sqrt{6}$

$\text {Unit vector } = \dfrac{ \overrightarrow A \times \overrightarrow B}{| \overrightarrow A \times \overrightarrow B|}$

$\text {Unit vector } = \bigg( \dfrac{2}{\sqrt{6} } , \dfrac{-1}{\sqrt{6} }, \dfrac{-1}{\sqrt{6} } \bigg) \text { or } \bigg( \dfrac{-2}{\sqrt{6} } , \dfrac{1}{\sqrt{6} }, \dfrac{1}{\sqrt{6} } \bigg)$