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 [abc]={\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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