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Find the unit vector perpendicular to both

Mathematics

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To find the unit vector perpendicular to two vectors, we first calculate their cross product, which gives a vector perpendicular to both. Then, we divide this resultant vector by its magnitude to get the unit vector.

From the image, the two vectors appear to be:
a⃗=i⃗+2j⃗+k⃗\vec{a} = \vec{i} + 2\vec{j} + \vec{k}
b⃗=5i⃗−j⃗+2k⃗\vec{b} = 5\vec{i} - \vec{j} + 2\vec{k}

  1. Calculate the cross product a⃗×b⃗\vec{a} \times \vec{b}.

    a⃗×b⃗=∣i⃗j⃗k⃗1215−12∣\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 2 & 1 \\ 5 & -1 & 2 \end{vmatrix} =i⃗((2)(2)−(1)(−1))−j⃗((1)(2)−(1)(5))+k⃗((1)(−1)−(2)(5))= \vec{i}((2)(2) - (1)(-1)) - \vec{j}((1)(2) - (1)(5)) + \vec{k}((1)(-1) - (2)(5)) =i⃗(4−(−1))−j⃗(2−5)+k⃗(−1−10)= \vec{i}(4 - (-1)) - \vec{j}(2 - 5) + \vec{k}(-1 - 10) =i⃗(5)−j⃗(−3)+k⃗(−11)= \vec{i}(5) - \vec{j}(-3) + \vec{k}(-11) =5i⃗+3j⃗−11k⃗= 5\vec{i} + 3\vec{j} - 11\vec{k}
  2. Calculate the magnitude of the cross product ∣a⃗×b⃗∣|\vec{a} \times \vec{b}|.

    Let c⃗=5i⃗+3j⃗−11k⃗\vec{c} = 5\vec{i} + 3\vec{j} - 11\vec{k}.

    ∣c⃗∣=(5)2+(3)2+(−11)2|\vec{c}| = \sqrt{(5)^2 + (3)^2 + (-11)^2} =25+9+121= \sqrt{25 + 9 + 121} =155= \sqrt{155}
  3. Compute the unit vector perpendicular to both a⃗\vec{a} and b⃗\vec{b}.

    The unit vector n^\hat{n} is given by a⃗×b⃗∣a⃗×b⃗∣\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}.

    n^=5i⃗+3j⃗−11k⃗155\hat{n} = \frac{5\vec{i} + 3\vec{j} - 11\vec{k}}{\sqrt{155}} n^=5155i⃗+3155j⃗−11155k⃗\hat{n} = \frac{5}{\sqrt{155}}\vec{i} + \frac{3}{\sqrt{155}}\vec{j} - \frac{11}{\sqrt{155}}\vec{k}

    The unit vector perpendicular to both vectors is 5155i⃗+3155j⃗−11155k⃗\boxed{\frac{5}{\sqrt{155}}\vec{i} + \frac{3}{\sqrt{155}}\vec{j} - \frac{11}{\sqrt{155}}\vec{k}}.

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