The distance from the point P(1,3,2) to the line through (1,0,0), and (1,2,0) is given by ?
Let M1(1, 0, 0) and M2(1, 2, 0).
We should find
d=∣PM1→×s→∣∣s→∣d=\frac{|\overrightarrow{PM_1}×\overrightarrow{s}|}{|\overrightarrow{s}|}d=∣s∣∣PM1×s∣ , where s→=M1M2→={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=M1M2={1−1,2−0,0−0}={0,2,0} and PM1→={1−1,0−3,0−2}={0,−3,−2}\overrightarrow{PM_1}=\{ 1-1, 0-3, 0-2\}=\{ 0, -3, -2\}PM1={1−1,0−3,0−2}={0,−3,−2} .
PM1→×s→=∣i→j→k→0−3−2020∣=i→(0+4)−j→(0−0)+k→(0−0)=4i→={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\}PM1×s=∣∣i00j−32k−20∣∣=i(0+4)−j(0−0)+k(0−0)=4i={4,0,0} .
Then d=42+02+0202+22+02=42=2d=\frac{\sqrt{4^2+0^2+0^2}}{\sqrt{0^2+2^2+0^2}}=\frac{4}{2}=2d=02+22+0242+02+02=24=2 .
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