Let M1 (1, 0, 0) and M2 (1, 2, 0).
We should find
d = ∣ P M 1 → × s → ∣ ∣ s → ∣ d=\frac{|\overrightarrow{PM_1}×\overrightarrow{s}|}{|\overrightarrow{s}|} d = ∣ s ∣ ∣ P M 1 × s ∣ , where s → = M 1 M 2 → = { 1 − 1 , 2 − 0 , 0 − 0 } = { 0 , 2 , 0 } \overrightarrow{s}=\overrightarrow{M_1M_2}=\{ 1-1, 2-0, 0-0\}=\{ 0, 2, 0\} s = M 1 M 2 = { 1 − 1 , 2 − 0 , 0 − 0 } = { 0 , 2 , 0 } and P M 1 → = { 1 − 1 , 0 − 3 , 0 − 2 } = { 0 , − 3 , − 2 } \overrightarrow{PM_1}=\{ 1-1, 0-3, 0-2\}=\{ 0, -3, -2\} P M 1 = { 1 − 1 , 0 − 3 , 0 − 2 } = { 0 , − 3 , − 2 } .
P M 1 → × s → = ∣ i → j → k → 0 − 3 − 2 0 2 0 ∣ = i → ( 0 + 4 ) − j → ( 0 − 0 ) + k → ( 0 − 0 ) = 4 i → = { 4 , 0 , 0 } \overrightarrow{PM_1}×\overrightarrow{s}=\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j}&\overrightarrow{k} \\
0 & -3&-2\\
0&2&0
\end{vmatrix}=\overrightarrow{i}(0+4)-\overrightarrow{j}(0-0)+\overrightarrow{k}(0-0)=4\overrightarrow{i}=\{ 4, 0, 0\} P M 1 × s = ∣ ∣ i 0 0 j − 3 2 k − 2 0 ∣ ∣ = i ( 0 + 4 ) − j ( 0 − 0 ) + k ( 0 − 0 ) = 4 i = { 4 , 0 , 0 } .
Then d = 4 2 + 0 2 + 0 2 0 2 + 2 2 + 0 2 = 4 2 = 2 d=\frac{\sqrt{4^2+0^2+0^2}}{\sqrt{0^2+2^2+0^2}}=\frac{4}{2}=2 d = 0 2 + 2 2 + 0 2 4 2 + 0 2 + 0 2 = 2 4 = 2 .
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