LinAlg4.2.25

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Find the perpendicular distance from the origin to the plane passing through the points $(1,-2,5),(0,-5,-1),(-3,5,0)\,$

Let $a=(1,-2,5),b=(0,-5,-1),c=(-3,5,0)\,$ be the given points.

$\bar{b}\times\bar{c}=\begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \\ 0 & -5 & -1 \\ -3 & 5 & 0 \end{vmatrix}=\bar{i}(0+5)-\bar{j}(0-3)+\bar{k}(0-15)=5\bar{i}+3\bar{j}-15\bar{k}\,$

$\bar{c}\times\bar{a}=\begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \\ -3 & 5 & 0 \\ 1 & -2 & 5 \end{vmatrix}=\bar{i}(25-0)-\bar{j}(-15-0)+\bar{k}(6-5)=25\bar{i}+15\bar{j}+\bar{k}\,$

$\bar{a}\times\bar{b}=\begin{vmatrix} \bar{i} & \bar{j} & \bar{k} \\ 1 & -2 & 5 \\ 0 & -5 & -1 \end{vmatrix}=\bar{i}(2+25)-\bar{j}(-1-0)+\bar{k}(-5-0)=27\bar{i}+\bar{j}-5\bar{k}\,$

Therefore $(\bar{b}\times\bar{c})+(\bar{c}\times\bar{a})+(\bar{a}\times\bar{b})=57\bar{i}+19\bar{j}-19\bar{k}=19(3\bar{i}+\bar{j}-\bar{k})\,$

Now $(\bar{b}\times\bar{c})+(\bar{c}\times\bar{a})+(\bar{a}\times\bar{b})=19\sqrt{9+1+1}=19\sqrt{11}\,$

Also $[a b c]=\begin{vmatrix} 1 & -2 & 5 \\ 0 & -5 & -1 \\ -3 & 5 & 0 \end{vmatrix}=1(0+5)+2(0-3)+5(0-15)=5-6-75=-76\,$

Therefore,the perpendicular distance from the origin to the given plane is $\frac{|-76|}{|19\sqrt{11}|}=\frac{4}{\sqrt{11}}\,$

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LinAlg4.2.25

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