Q.Find a unit vector perpendicular to each of the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$, where $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.

A vector normal to both given vectors is (a scalar multiple of) their cross product.

Let $\vec p = \hat i+\hat k=(1,0,1)$ and $\vec q=\hat i+\hat j=(1,1,0)$.

$\vec p \times \vec q = \begin{vmatrix}\hat i & \hat j & \hat k \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{vmatrix} = \hat i(0\cdot0-1\cdot1) - \hat j(1\cdot0-1\cdot1) + \hat k(1\cdot1-0\cdot1)$

$= -\hat i + \hat j + \hat k$

Any nonzero scalar multiple of this is also normal to both vectors, including its negative: …

To find a vector perpendicular to two given vectors, we use their cross product. Normalizing this resulting vector gives the unit vector. The unit vector perpendicular to $\hat{i} + \hat{k}$ and $\hat{i} - \hat{k}$ is $\boxed{\hat{j}}$.

A vector perpendicular to both given vectors is found via their cross product; normalizing it gives the unit vector.

Let $\vec a = \hat j - \hat k = (0,1,-1)$ and $\vec b = \hat i + \hat j = (1,1,0)$.

A vector perpendicular to both is $\vec a \times \vec b$:

$\vec a\times\vec b = \begin{vmatrix}\hat i & \hat j & \hat k\\ 0 & 1 & -1\\ 1 & 1 & 0\end{vmatrix} = \hat i(1\cdot0-(-1)\cdot1) - \hat j(0\cdot0-(-1)\cdot1) + \hat k(0\cdot1-1\cdot1)$

$= \hat i(1) - \hat j(1) + \hat k(-1) = \hat i - \hat j - \hat k$

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The unit vector perpendicular to both $\vec a$ and $\vec b$ is $\dfrac{\vec a\times\vec b}{|\vec a\times\vec b|}$.

$\vec a = \hat i - 2\hat j + 3\hat k$, $\vec b = \hat i + 2\hat j - \hat k$

$\vec a\times\vec b = \begin{vmatrix}\hat i & \hat j & \hat k\\ 1 & -2 & 3\\ 1 & 2 & -1\end{vmatrix}$

$= \hat i[(-2)(-1)-(3)(2)] - \hat j[(1)(-1)-(3)(1)] + \hat k[(1)(2)-(-2)(1)]$

$= \hat i(2-6) - \hat j(-1-3) + \hat k(2+2) = -4\hat i + 4\hat j + 4\hat k$

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