Question

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 

 

Options

• 7

• 5

• 0

•  none of these 

Answer 1

 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,

\[\overrightarrow{PQ} = 5 \hat{i} + 5 \hat{j} + 4 \hat{k} \]

\[\therefore \overrightarrow{b} \times \overrightarrow{PQ} = \begin{vmatrix}\hat{i}  & \hat{j}  & \hat{k}  \\ 3 & 2 & - 2 \\ 5 & 5 & 4\end{vmatrix}\]

\[ = 18 \hat{i}  - 22 \hat{j}  + 5 \hat{k}  \]

\[ \Rightarrow \left| \overrightarrow{b} \times \overrightarrow{PQ} \right| = \sqrt{{18}^2 + \left( - 22 \right)^2 + 5^2}\]

\[ = \sqrt{324 + 484 + 25}\]

\[ = \sqrt{833}\]

\[ \therefore d = \frac{\left| \overrightarrow{b} \times \overrightarrow{PQ} \right|}{\left| \overrightarrow{b} \right|}\]

\[ = \frac{\sqrt{833}}{\sqrt{17}}\]

\[ = \sqrt{49}\]

\[ = 7\]