Question

Найти координаты вектора А который ортогонален векторам

Answer 1

Ответ:

$\vec{a}=(\, x\, ;\, y\, ;\, z\, )\ \ ,\ \ \vec{b}=(-3;-9;-7)\ \ ,\ \ \vec{c}=(-9;-3;-9)\\\\\vec{a}\perp \vec{b}\ \ ,\ \ \vec{a}\perp \vec{c}\ \ \Rightarrow \ \ \vec{a}=\lambda \cdot [\, b\veca}\, \times \, \vec{c}\, ]\\\\\\{}[\ \vec{b}\times \vec{c}\ ]=\left|\begin{array}{ccc}i&j&k\\-3&-9&-7\\-9&-3&-9\end{array}\right|=60\, \vec{i}+36\, \vec{j}-72\, \vec{k}\\\\\\\vec{a}=(\ 60\lambda \, ;\, 36\lambda \, ;\, -72\lambda \, )$

$|\vec{a}|=\sqrt{(60\lambda )^2+(36\lambda )^2+(-72\lambda )^2}=\sqrt{10080\, \lambda ^2}=12\, \lambda \sqrt{70}\\\\|\vec{a}|=\sqrt{70}\ \ ,\ \ \ 12\, \lambda \sqrt{70}=\sqrt{70}\ \ \to \ \ \ \lambda =\dfrac{1}{12}\\\\\\\vec{a}=\Big(\ \dfrac{60}{12}\, ;\, \dfrac{36}{12}\ ;\ -\dfrac{72}{12}\, \Big)\ \ \ ,\ \ \ \boxed{\ \vec{a}=(\ 5\, ;\, 3\, -6\, )\ }$