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प्रश्न
The perpendicular distance of the point P (1, 2, 3) from the line \[\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}\] is
विकल्प
7
5
0
none of these
उत्तर
7
We have ,
\[\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}\]
Let point (1, 2, 3) be P and the point through which the line passes be Q (6, 7, 7). Also, the line is parallel to the vector
\[\overrightarrow{b} = 3 \hat{i} + 2 \hat{j} - 2 \hat{k}\]
Now,
\[\overrightarrow{PQ} = 5 \hat{i} + 5 \hat{j} + 4 \hat{k} \]
\[\therefore \overrightarrow{b} \times \overrightarrow{PQ} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & - 2 \\ 5 & 5 & 4\end{vmatrix}\]
\[ = 18 \hat{i} - 22 \hat{j} + 5 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b} \times \overrightarrow{PQ} \right| = \sqrt{{18}^2 + \left( - 22 \right)^2 + 5^2}\]
\[ = \sqrt{324 + 484 + 25}\]
\[ = \sqrt{833}\]
\[ \therefore d = \frac{\left| \overrightarrow{b} \times \overrightarrow{PQ} \right|}{\left| \overrightarrow{b} \right|}\]
\[ = \frac{\sqrt{833}}{\sqrt{17}}\]
\[ = \sqrt{49}\]
\[ = 7\]
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