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Find the point in the plane $$3x - y + z = 7$$ that is closest to the origin.

May 10, 2020

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Find the point in the plane $$3x - y + z = 7$$ that is closest to the origin.

We find the closest point to the origin in plane $$ax+by+cz = d$$
with the normal vector of the plane $$\vec{n} =(a,~b,~c)$$ by Formula: $$\dfrac{d}{\left(\vec{n}\right)^2} * \vec{n}$$.

$$\begin{array}{|rcll|} \hline \vec{n} &=& (a,~b,~c) \quad | \quad a=3,~b=-1,~c=1 \\ \mathbf{\vec{n}} &=& \mathbf{(3,~-1,~1)} \\\\ \left(\vec{n}\right)^2 &=& 3^2+(-1)^2+1^2 \\ \mathbf{\left(\vec{n}\right)^2} &=& \mathbf{11} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline && \dfrac{d}{\left(\vec{n}\right)^2} * \vec{n} \quad | \quad d=7,~ \mathbf{\left(\vec{n}\right)^2=11} \\\\ &=& \dfrac{7}{11} * (3,~-1,~1) \\\\ &=& \left(3*\dfrac{7}{11},~-1*\dfrac{7}{11},~1*\dfrac{7}{11}\right) \\\\ \dfrac{d}{\left(\vec{n}\right)^2} * \vec{n} &=& \left(\dfrac{21}{11},~\dfrac{-7}{11},~\dfrac{7}{11}\right) \\ \hline \end{array}$$

The vector that gives us the closest point is $$\left(\dfrac{21}{11},~\dfrac{-7}{11},~\dfrac{7}{11}\right)$$. May 11, 2020